Derivative as a rate measurer

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I studied about the application of derivatives as they help in measuring rate of change.
For example :-



Let $A$ be area of a circle of radius $r$



$$A = pi cdot r^2$$



then



$$frac dAdr= 2 pi cdot r $$



Suppose we have to find rate of change of area w.r.t to radius
at $r = 5 text cm$ .



Then



$$left(frac dAdrright)_r=5= 10 pitext cm^2/textcm $$



My question is:



Does our final answer mean that when the radius of the circle changes from $5 textcm$ to $6 textcm$, the change in area is equal to $ 10pitextcm^2 $,
i.e., the area of circle at $6 textcm = textarea of circle at 5 textcm + 10pi textcm^2 $?










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  • But area $A$ at $6 text cm$ is $pi 36 text cm^2$ while area at $5 text cm$ is $pi 25 text cm^2$
    – Mauro ALLEGRANZA
    Sep 3 at 13:48







  • 1




    Not quite. It means that when the radius of the circle changes from $5 rm cm$ to $(5+epsilon) rm cm$, the area of the circle changes by approximately $10piepsilon rm cm^2$, and the approximation is more accurate the smaller $epsilon$ is. In your example, $epsilon=1$, which is not very small, so the estimated change in area, $10pi rm cm^2$, is only moderately close to the actual change of area, $11pi rm cm^2$.
    – Rahul
    Sep 3 at 13:51










  • @Rahul what does the answer that I calculated actually means ?
    – JIM
    Sep 3 at 14:49






  • 1




    Your $10pi cm^2/cm$ is often called a "first order approximation". More correctly, together with the information $A(5) = 25pi$, it is $A(6)approx 25pi + 10pi$ which is a first order approximation of $A(6)$. It is the best approximation to $A(6)$ that you can make knowing only $A(5)$ and $A'(5)$.
    – Arthur
    Sep 3 at 16:14















up vote
6
down vote

favorite












I studied about the application of derivatives as they help in measuring rate of change.
For example :-



Let $A$ be area of a circle of radius $r$



$$A = pi cdot r^2$$



then



$$frac dAdr= 2 pi cdot r $$



Suppose we have to find rate of change of area w.r.t to radius
at $r = 5 text cm$ .



Then



$$left(frac dAdrright)_r=5= 10 pitext cm^2/textcm $$



My question is:



Does our final answer mean that when the radius of the circle changes from $5 textcm$ to $6 textcm$, the change in area is equal to $ 10pitextcm^2 $,
i.e., the area of circle at $6 textcm = textarea of circle at 5 textcm + 10pi textcm^2 $?










share|cite|improve this question























  • But area $A$ at $6 text cm$ is $pi 36 text cm^2$ while area at $5 text cm$ is $pi 25 text cm^2$
    – Mauro ALLEGRANZA
    Sep 3 at 13:48







  • 1




    Not quite. It means that when the radius of the circle changes from $5 rm cm$ to $(5+epsilon) rm cm$, the area of the circle changes by approximately $10piepsilon rm cm^2$, and the approximation is more accurate the smaller $epsilon$ is. In your example, $epsilon=1$, which is not very small, so the estimated change in area, $10pi rm cm^2$, is only moderately close to the actual change of area, $11pi rm cm^2$.
    – Rahul
    Sep 3 at 13:51










  • @Rahul what does the answer that I calculated actually means ?
    – JIM
    Sep 3 at 14:49






  • 1




    Your $10pi cm^2/cm$ is often called a "first order approximation". More correctly, together with the information $A(5) = 25pi$, it is $A(6)approx 25pi + 10pi$ which is a first order approximation of $A(6)$. It is the best approximation to $A(6)$ that you can make knowing only $A(5)$ and $A'(5)$.
    – Arthur
    Sep 3 at 16:14













up vote
6
down vote

favorite









up vote
6
down vote

favorite











I studied about the application of derivatives as they help in measuring rate of change.
For example :-



Let $A$ be area of a circle of radius $r$



$$A = pi cdot r^2$$



then



$$frac dAdr= 2 pi cdot r $$



Suppose we have to find rate of change of area w.r.t to radius
at $r = 5 text cm$ .



Then



$$left(frac dAdrright)_r=5= 10 pitext cm^2/textcm $$



My question is:



Does our final answer mean that when the radius of the circle changes from $5 textcm$ to $6 textcm$, the change in area is equal to $ 10pitextcm^2 $,
i.e., the area of circle at $6 textcm = textarea of circle at 5 textcm + 10pi textcm^2 $?










share|cite|improve this question















I studied about the application of derivatives as they help in measuring rate of change.
For example :-



Let $A$ be area of a circle of radius $r$



$$A = pi cdot r^2$$



then



$$frac dAdr= 2 pi cdot r $$



Suppose we have to find rate of change of area w.r.t to radius
at $r = 5 text cm$ .



Then



$$left(frac dAdrright)_r=5= 10 pitext cm^2/textcm $$



My question is:



Does our final answer mean that when the radius of the circle changes from $5 textcm$ to $6 textcm$, the change in area is equal to $ 10pitextcm^2 $,
i.e., the area of circle at $6 textcm = textarea of circle at 5 textcm + 10pi textcm^2 $?







geometry derivatives applications






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edited Sep 3 at 18:02









Michael Hardy

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asked Sep 3 at 13:44









JIM

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  • But area $A$ at $6 text cm$ is $pi 36 text cm^2$ while area at $5 text cm$ is $pi 25 text cm^2$
    – Mauro ALLEGRANZA
    Sep 3 at 13:48







  • 1




    Not quite. It means that when the radius of the circle changes from $5 rm cm$ to $(5+epsilon) rm cm$, the area of the circle changes by approximately $10piepsilon rm cm^2$, and the approximation is more accurate the smaller $epsilon$ is. In your example, $epsilon=1$, which is not very small, so the estimated change in area, $10pi rm cm^2$, is only moderately close to the actual change of area, $11pi rm cm^2$.
    – Rahul
    Sep 3 at 13:51










  • @Rahul what does the answer that I calculated actually means ?
    – JIM
    Sep 3 at 14:49






  • 1




    Your $10pi cm^2/cm$ is often called a "first order approximation". More correctly, together with the information $A(5) = 25pi$, it is $A(6)approx 25pi + 10pi$ which is a first order approximation of $A(6)$. It is the best approximation to $A(6)$ that you can make knowing only $A(5)$ and $A'(5)$.
    – Arthur
    Sep 3 at 16:14

















  • But area $A$ at $6 text cm$ is $pi 36 text cm^2$ while area at $5 text cm$ is $pi 25 text cm^2$
    – Mauro ALLEGRANZA
    Sep 3 at 13:48







  • 1




    Not quite. It means that when the radius of the circle changes from $5 rm cm$ to $(5+epsilon) rm cm$, the area of the circle changes by approximately $10piepsilon rm cm^2$, and the approximation is more accurate the smaller $epsilon$ is. In your example, $epsilon=1$, which is not very small, so the estimated change in area, $10pi rm cm^2$, is only moderately close to the actual change of area, $11pi rm cm^2$.
    – Rahul
    Sep 3 at 13:51










  • @Rahul what does the answer that I calculated actually means ?
    – JIM
    Sep 3 at 14:49






  • 1




    Your $10pi cm^2/cm$ is often called a "first order approximation". More correctly, together with the information $A(5) = 25pi$, it is $A(6)approx 25pi + 10pi$ which is a first order approximation of $A(6)$. It is the best approximation to $A(6)$ that you can make knowing only $A(5)$ and $A'(5)$.
    – Arthur
    Sep 3 at 16:14
















But area $A$ at $6 text cm$ is $pi 36 text cm^2$ while area at $5 text cm$ is $pi 25 text cm^2$
– Mauro ALLEGRANZA
Sep 3 at 13:48





But area $A$ at $6 text cm$ is $pi 36 text cm^2$ while area at $5 text cm$ is $pi 25 text cm^2$
– Mauro ALLEGRANZA
Sep 3 at 13:48





1




1




Not quite. It means that when the radius of the circle changes from $5 rm cm$ to $(5+epsilon) rm cm$, the area of the circle changes by approximately $10piepsilon rm cm^2$, and the approximation is more accurate the smaller $epsilon$ is. In your example, $epsilon=1$, which is not very small, so the estimated change in area, $10pi rm cm^2$, is only moderately close to the actual change of area, $11pi rm cm^2$.
– Rahul
Sep 3 at 13:51




Not quite. It means that when the radius of the circle changes from $5 rm cm$ to $(5+epsilon) rm cm$, the area of the circle changes by approximately $10piepsilon rm cm^2$, and the approximation is more accurate the smaller $epsilon$ is. In your example, $epsilon=1$, which is not very small, so the estimated change in area, $10pi rm cm^2$, is only moderately close to the actual change of area, $11pi rm cm^2$.
– Rahul
Sep 3 at 13:51












@Rahul what does the answer that I calculated actually means ?
– JIM
Sep 3 at 14:49




@Rahul what does the answer that I calculated actually means ?
– JIM
Sep 3 at 14:49




1




1




Your $10pi cm^2/cm$ is often called a "first order approximation". More correctly, together with the information $A(5) = 25pi$, it is $A(6)approx 25pi + 10pi$ which is a first order approximation of $A(6)$. It is the best approximation to $A(6)$ that you can make knowing only $A(5)$ and $A'(5)$.
– Arthur
Sep 3 at 16:14





Your $10pi cm^2/cm$ is often called a "first order approximation". More correctly, together with the information $A(5) = 25pi$, it is $A(6)approx 25pi + 10pi$ which is a first order approximation of $A(6)$. It is the best approximation to $A(6)$ that you can make knowing only $A(5)$ and $A'(5)$.
– Arthur
Sep 3 at 16:14











7 Answers
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No, but the intuition is correct.



$fracdAdr$ is the instantaneous rate of change at radius $r$. It means that if your radius is initially $r_0$ (say, 5cm), and you increase the radius by an infinitesimally small amount $dr$, the change in area will be given by $fracdAdr$ evaluated at $r_0$. At the new point $r_0 + dr$, the rate of change of area might be different (as it is in this case, where $fracdAdr$ depends on $r$. However, if your change in radius is small, your idea could be a good approximation for the change in area.



However, a larger change in area can be viewed as an aggregation of many small infinitesimal changes to $r$. This can be expressed as an integral, and so, the area change when you change the radius from $r_0$ to $r_0+1$ is given by
$$int_r_0^r_0+1 left(fracdAdrright) dr$$



It could be the case for some functions that the derivative is independent of $r$. In such a case, your idea will give exactly the correct answer, as the rate of change will not change with $r$. An example of such a function would be the rate of change of circumference of a circle with respect to the radius.






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    Yes it is true, but only for variations such that the linear approximation is acceptable.



    For an increase of the radius of $0.001$ cm, the increase in area is approximated by$10cdotpicdot0.001=0.0314159265cdots$



    Compare to the exact value $pi(5.001^2-5^2)=0.0314190681cdots$



    In green, the linear approximation.



    enter image description here






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    • 1




      so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
      – JIM
      Sep 3 at 14:39










    • @JIM: this is what the plot illustrates.
      – Yves Daoust
      Sep 3 at 14:45










    • What does the answer that I calculated actually means ?
      – JIM
      Sep 3 at 14:48










    • @JIM: which one ?
      – Yves Daoust
      Sep 3 at 15:31










    • the answer that I calculated in the question
      – JIM
      Sep 3 at 15:34

















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    3
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    All the previous answers explained the problem quite well. But according to your comment on these answer, you still have trouble understanding when to your approximation holds. I'll try to explain what values of h are acceptable to use your method.



    Let's call $A(r)$ the area of a circle of radius $r$. We have $A_real=colorbluepi*r²$



    If $h$ is small enough, you can have $A_approx(r_0+h) approx colorredA(r_0)+2*pi*r_0*h$



    But keep in mind that the real value of $A(r_0+h)$ is indeed:



    beginalignA_real(r_0+h)=colorbluepi*(r_0+h)²\
    & =pi*(r_0²+2r_0h+h²)\
    & =colorbluepi*r_0²+2*pi*r_0*h+pi*h²\
    & =colorredA_real(r_0)+2*pi*r_0*h+pi*h² \
    & =colorredA_approx(r_0+h) +pi*h²
    endalign



    From there, the relative error between you result and the real one is:



    $$E=fracA_real(r_0+h)-A_approx(r_0+h)A_real(r_0+h)$$
    $$E=fracpi*h²pi*(r_0+h)²$$
    $$E=frach²(r_0+h)²$$



    So, you can see that as $h/r_0 rightarrow 0$, the error $E rightarrow 0$ too.



    This means that as soon as $h/r_0$ is small enough, so is your error. Let see it this way, if you change 1cm in diameter on a circle of 10cm, and calculate the new area with your initial idea, your error will be more important that if you change 1cm on a 1m diameter circle. What only matter in this case is that the change is small enough versus the initial diameter.



    The problem in your question is that you chose $h=1$ and $r_0=5$. You'll need $h$ to be smaller or $r_0$ to be greater



    I made some calculation for circles of radius 5cm, 10cm, and 1m ($r_0=5,10,100$), with a changing radius of 1cm, you have errors of respectively: $2.7%$,$0.8%$ and $0.01%$ between the real area and the one calculated from your formula.






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    • you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
      – JIM
      Sep 3 at 16:19










    • I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
      – F.Carette
      Sep 3 at 17:56










    • what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
      – JIM
      Sep 4 at 2:05










    • My bad. It's the basic formula for relative error. I Added the link in the main body.
      – F.Carette
      Sep 4 at 7:36










    • one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
      – JIM
      Sep 4 at 11:59

















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    We should write



    $$frac dAdcolorredr= 2 pi cdot r$$



    No of course our final answer doaes not mean that when radius of circle changes from $5 cm$ to $6 cm$ then the change in area is equal to $10pi cm^2$, indeed derivatives are linear approximation and, since in that case the relationship between $A$ and $r$ is not linear, it means that



    $$Delta A approx 10pi Delta r$$



    and the approximation becomes better for smaller $Delta r$.






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    • does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
      – JIM
      Sep 3 at 14:41










    • @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
      – gimusi
      Sep 3 at 14:49

















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    2
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    Let$f: mathbbR rightarrowtail mathbbR$ be a function, and $a in textInt(textdom(f))$. Then the following statements are equivalent:



    • $f$ is differentiable at $a$

    • $exists xi in mathbbR$ and $exists eta: textdom(f) to mathbbR$ with $limlimits_x to a fraceta(x)x-a=0$, so that
      $$f(x)=f(a)+xi(x-a)+eta(x)$$
      and $xi=f'(a)$.

    I think this equivalent definition of differentiability is more spectacular (and it's easier to generalize to the $mathbbR^n rightarrowtail mathbbR^m$ case), because we can intuitively see that $eta(x)$ is really small around $a$, so for $x approx a$ we have that
    $$f(x) approx f(a) + xi(x-a)$$
    In your case, if we let $A(r):=r^2 pi$, we have that $A'(r)=2r pi$
    $$A(r)=A(r_0)+xi(r-r_0)+eta(r)$$
    And of course, $xi=A'(r_0)=2r_0pi$:
    $$A(r)=A(r_0)+2r_0(r-r_0)pi+eta(r)$$
    $$A(r)=r_0^2 pi+2r_0(r-r_0)pi+eta(r)$$
    $$A(r)=r_0^2 pi+2r_0rpi-2r_0^2 pi+eta(r)$$
    $$A(r)=2rr_0 pi-r_0^2 pi + eta(r) tag1$$
    If we want to get $eta(r)$:
    $$r^2 pi=2rr_0 pi-r_0^2 pi + eta(r)$$
    $$r^2 pi+r_0^2pi-2r r_0 pi=eta(r)$$
    $$eta(r)=(r-r_0)^2 pi$$
    Using some numbers, we have that $A(5)=25 pi$, $A(5.1)=26.01 pi$, $A(50)=2500 pi$. If we try to approximate $A(5.1)$ and $A(50)$ with $A(5)$ and $A'(5)$, we can use the first equation (without $eta(r)$):
    $$A(r) approx 10 r pi - 25 pi$$
    we get that $A(5.1) approx 26 pi$ and $A(50) approx 475 pi$, while $eta(5.1)=0.01 pi$ and $eta(50)=2025 pi$. As you can see, it's quite good for $5.1$, but it's terrible for $50$.



    This kind of linear approximation is really useful in physics, for example when you are dealing with a complicated system, you can't really solve the differential equation(s), but you can make the assumption that the change in the sought function (for example, the displacement-time function of a ball) is really small, so you can linearize the equation, and (probably) solve it.






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    • 1




      Maybe the symbol $A$ is a little confusing here.
      – gimusi
      Sep 3 at 15:43










    • @gimusi you're right.
      – Botond
      Sep 3 at 16:04










    • @Botond not able to understand a single word that you wrote !!
      – JIM
      Sep 3 at 16:20










    • @JIM I extended my answer.
      – Botond
      Sep 3 at 17:13

















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    No, but you are on the right track.



    The derivative you have found tells you that when the radius changes from $5$cm to $(5+h)$cm and $h$ is small then the area changes by approximately $10pi h$ square cm.



    In your example you are taking $h=1$, which is not small, so the approximation isn't even close.






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    • So what does the answer that I calculated actually means (if it does not mean change in area )??
      – JIM
      Sep 3 at 14:43










    • What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
      – Ethan Bolker
      Sep 3 at 14:46










    • In this case is my h correct ??
      – JIM
      Sep 3 at 15:09










    • There's no such thing as "the correct $h$". The smaller it is the better the approximation.
      – Ethan Bolker
      Sep 3 at 21:32

















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    0
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    To say that $left.dfracdAdrright|_r,=,5text cm = 10pitext cm$ means that if you multiply the rate of change of $r,$ at the point at which $r=5text cm,$ by $10pitext cm,$ you get the rate of change of $A$ at that point. But that does not mean that if $r$ grows by $1text cm,$ then $A$ will increase by $10pitext cm^2,$ because it won't continue changing at that same rate as $r$ changes.



    Generally if $left.dfracdydxright|_x,=,a = b$ means that at the point where $x=a,$ the dependent variable $y$ is changing $a$ times as fast as the independent variable $x.$






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      7 Answers
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      7 Answers
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      3
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      No, but the intuition is correct.



      $fracdAdr$ is the instantaneous rate of change at radius $r$. It means that if your radius is initially $r_0$ (say, 5cm), and you increase the radius by an infinitesimally small amount $dr$, the change in area will be given by $fracdAdr$ evaluated at $r_0$. At the new point $r_0 + dr$, the rate of change of area might be different (as it is in this case, where $fracdAdr$ depends on $r$. However, if your change in radius is small, your idea could be a good approximation for the change in area.



      However, a larger change in area can be viewed as an aggregation of many small infinitesimal changes to $r$. This can be expressed as an integral, and so, the area change when you change the radius from $r_0$ to $r_0+1$ is given by
      $$int_r_0^r_0+1 left(fracdAdrright) dr$$



      It could be the case for some functions that the derivative is independent of $r$. In such a case, your idea will give exactly the correct answer, as the rate of change will not change with $r$. An example of such a function would be the rate of change of circumference of a circle with respect to the radius.






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        No, but the intuition is correct.



        $fracdAdr$ is the instantaneous rate of change at radius $r$. It means that if your radius is initially $r_0$ (say, 5cm), and you increase the radius by an infinitesimally small amount $dr$, the change in area will be given by $fracdAdr$ evaluated at $r_0$. At the new point $r_0 + dr$, the rate of change of area might be different (as it is in this case, where $fracdAdr$ depends on $r$. However, if your change in radius is small, your idea could be a good approximation for the change in area.



        However, a larger change in area can be viewed as an aggregation of many small infinitesimal changes to $r$. This can be expressed as an integral, and so, the area change when you change the radius from $r_0$ to $r_0+1$ is given by
        $$int_r_0^r_0+1 left(fracdAdrright) dr$$



        It could be the case for some functions that the derivative is independent of $r$. In such a case, your idea will give exactly the correct answer, as the rate of change will not change with $r$. An example of such a function would be the rate of change of circumference of a circle with respect to the radius.






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          No, but the intuition is correct.



          $fracdAdr$ is the instantaneous rate of change at radius $r$. It means that if your radius is initially $r_0$ (say, 5cm), and you increase the radius by an infinitesimally small amount $dr$, the change in area will be given by $fracdAdr$ evaluated at $r_0$. At the new point $r_0 + dr$, the rate of change of area might be different (as it is in this case, where $fracdAdr$ depends on $r$. However, if your change in radius is small, your idea could be a good approximation for the change in area.



          However, a larger change in area can be viewed as an aggregation of many small infinitesimal changes to $r$. This can be expressed as an integral, and so, the area change when you change the radius from $r_0$ to $r_0+1$ is given by
          $$int_r_0^r_0+1 left(fracdAdrright) dr$$



          It could be the case for some functions that the derivative is independent of $r$. In such a case, your idea will give exactly the correct answer, as the rate of change will not change with $r$. An example of such a function would be the rate of change of circumference of a circle with respect to the radius.






          share|cite|improve this answer












          No, but the intuition is correct.



          $fracdAdr$ is the instantaneous rate of change at radius $r$. It means that if your radius is initially $r_0$ (say, 5cm), and you increase the radius by an infinitesimally small amount $dr$, the change in area will be given by $fracdAdr$ evaluated at $r_0$. At the new point $r_0 + dr$, the rate of change of area might be different (as it is in this case, where $fracdAdr$ depends on $r$. However, if your change in radius is small, your idea could be a good approximation for the change in area.



          However, a larger change in area can be viewed as an aggregation of many small infinitesimal changes to $r$. This can be expressed as an integral, and so, the area change when you change the radius from $r_0$ to $r_0+1$ is given by
          $$int_r_0^r_0+1 left(fracdAdrright) dr$$



          It could be the case for some functions that the derivative is independent of $r$. In such a case, your idea will give exactly the correct answer, as the rate of change will not change with $r$. An example of such a function would be the rate of change of circumference of a circle with respect to the radius.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 3 at 13:52









          GoodDeeds

          10.2k21335




          10.2k21335




















              up vote
              3
              down vote













              Yes it is true, but only for variations such that the linear approximation is acceptable.



              For an increase of the radius of $0.001$ cm, the increase in area is approximated by$10cdotpicdot0.001=0.0314159265cdots$



              Compare to the exact value $pi(5.001^2-5^2)=0.0314190681cdots$



              In green, the linear approximation.



              enter image description here






              share|cite|improve this answer
















              • 1




                so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
                – JIM
                Sep 3 at 14:39










              • @JIM: this is what the plot illustrates.
                – Yves Daoust
                Sep 3 at 14:45










              • What does the answer that I calculated actually means ?
                – JIM
                Sep 3 at 14:48










              • @JIM: which one ?
                – Yves Daoust
                Sep 3 at 15:31










              • the answer that I calculated in the question
                – JIM
                Sep 3 at 15:34














              up vote
              3
              down vote













              Yes it is true, but only for variations such that the linear approximation is acceptable.



              For an increase of the radius of $0.001$ cm, the increase in area is approximated by$10cdotpicdot0.001=0.0314159265cdots$



              Compare to the exact value $pi(5.001^2-5^2)=0.0314190681cdots$



              In green, the linear approximation.



              enter image description here






              share|cite|improve this answer
















              • 1




                so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
                – JIM
                Sep 3 at 14:39










              • @JIM: this is what the plot illustrates.
                – Yves Daoust
                Sep 3 at 14:45










              • What does the answer that I calculated actually means ?
                – JIM
                Sep 3 at 14:48










              • @JIM: which one ?
                – Yves Daoust
                Sep 3 at 15:31










              • the answer that I calculated in the question
                – JIM
                Sep 3 at 15:34












              up vote
              3
              down vote










              up vote
              3
              down vote









              Yes it is true, but only for variations such that the linear approximation is acceptable.



              For an increase of the radius of $0.001$ cm, the increase in area is approximated by$10cdotpicdot0.001=0.0314159265cdots$



              Compare to the exact value $pi(5.001^2-5^2)=0.0314190681cdots$



              In green, the linear approximation.



              enter image description here






              share|cite|improve this answer












              Yes it is true, but only for variations such that the linear approximation is acceptable.



              For an increase of the radius of $0.001$ cm, the increase in area is approximated by$10cdotpicdot0.001=0.0314159265cdots$



              Compare to the exact value $pi(5.001^2-5^2)=0.0314190681cdots$



              In green, the linear approximation.



              enter image description here







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 3 at 13:54









              Yves Daoust

              116k667211




              116k667211







              • 1




                so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
                – JIM
                Sep 3 at 14:39










              • @JIM: this is what the plot illustrates.
                – Yves Daoust
                Sep 3 at 14:45










              • What does the answer that I calculated actually means ?
                – JIM
                Sep 3 at 14:48










              • @JIM: which one ?
                – Yves Daoust
                Sep 3 at 15:31










              • the answer that I calculated in the question
                – JIM
                Sep 3 at 15:34












              • 1




                so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
                – JIM
                Sep 3 at 14:39










              • @JIM: this is what the plot illustrates.
                – Yves Daoust
                Sep 3 at 14:45










              • What does the answer that I calculated actually means ?
                – JIM
                Sep 3 at 14:48










              • @JIM: which one ?
                – Yves Daoust
                Sep 3 at 15:31










              • the answer that I calculated in the question
                – JIM
                Sep 3 at 15:34







              1




              1




              so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
              – JIM
              Sep 3 at 14:39




              so does that mean if we take the change in radius tending to zero , we would start getting more accurate value ?
              – JIM
              Sep 3 at 14:39












              @JIM: this is what the plot illustrates.
              – Yves Daoust
              Sep 3 at 14:45




              @JIM: this is what the plot illustrates.
              – Yves Daoust
              Sep 3 at 14:45












              What does the answer that I calculated actually means ?
              – JIM
              Sep 3 at 14:48




              What does the answer that I calculated actually means ?
              – JIM
              Sep 3 at 14:48












              @JIM: which one ?
              – Yves Daoust
              Sep 3 at 15:31




              @JIM: which one ?
              – Yves Daoust
              Sep 3 at 15:31












              the answer that I calculated in the question
              – JIM
              Sep 3 at 15:34




              the answer that I calculated in the question
              – JIM
              Sep 3 at 15:34










              up vote
              3
              down vote













              All the previous answers explained the problem quite well. But according to your comment on these answer, you still have trouble understanding when to your approximation holds. I'll try to explain what values of h are acceptable to use your method.



              Let's call $A(r)$ the area of a circle of radius $r$. We have $A_real=colorbluepi*r²$



              If $h$ is small enough, you can have $A_approx(r_0+h) approx colorredA(r_0)+2*pi*r_0*h$



              But keep in mind that the real value of $A(r_0+h)$ is indeed:



              beginalignA_real(r_0+h)=colorbluepi*(r_0+h)²\
              & =pi*(r_0²+2r_0h+h²)\
              & =colorbluepi*r_0²+2*pi*r_0*h+pi*h²\
              & =colorredA_real(r_0)+2*pi*r_0*h+pi*h² \
              & =colorredA_approx(r_0+h) +pi*h²
              endalign



              From there, the relative error between you result and the real one is:



              $$E=fracA_real(r_0+h)-A_approx(r_0+h)A_real(r_0+h)$$
              $$E=fracpi*h²pi*(r_0+h)²$$
              $$E=frach²(r_0+h)²$$



              So, you can see that as $h/r_0 rightarrow 0$, the error $E rightarrow 0$ too.



              This means that as soon as $h/r_0$ is small enough, so is your error. Let see it this way, if you change 1cm in diameter on a circle of 10cm, and calculate the new area with your initial idea, your error will be more important that if you change 1cm on a 1m diameter circle. What only matter in this case is that the change is small enough versus the initial diameter.



              The problem in your question is that you chose $h=1$ and $r_0=5$. You'll need $h$ to be smaller or $r_0$ to be greater



              I made some calculation for circles of radius 5cm, 10cm, and 1m ($r_0=5,10,100$), with a changing radius of 1cm, you have errors of respectively: $2.7%$,$0.8%$ and $0.01%$ between the real area and the one calculated from your formula.






              share|cite|improve this answer






















              • you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
                – JIM
                Sep 3 at 16:19










              • I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
                – F.Carette
                Sep 3 at 17:56










              • what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
                – JIM
                Sep 4 at 2:05










              • My bad. It's the basic formula for relative error. I Added the link in the main body.
                – F.Carette
                Sep 4 at 7:36










              • one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
                – JIM
                Sep 4 at 11:59














              up vote
              3
              down vote













              All the previous answers explained the problem quite well. But according to your comment on these answer, you still have trouble understanding when to your approximation holds. I'll try to explain what values of h are acceptable to use your method.



              Let's call $A(r)$ the area of a circle of radius $r$. We have $A_real=colorbluepi*r²$



              If $h$ is small enough, you can have $A_approx(r_0+h) approx colorredA(r_0)+2*pi*r_0*h$



              But keep in mind that the real value of $A(r_0+h)$ is indeed:



              beginalignA_real(r_0+h)=colorbluepi*(r_0+h)²\
              & =pi*(r_0²+2r_0h+h²)\
              & =colorbluepi*r_0²+2*pi*r_0*h+pi*h²\
              & =colorredA_real(r_0)+2*pi*r_0*h+pi*h² \
              & =colorredA_approx(r_0+h) +pi*h²
              endalign



              From there, the relative error between you result and the real one is:



              $$E=fracA_real(r_0+h)-A_approx(r_0+h)A_real(r_0+h)$$
              $$E=fracpi*h²pi*(r_0+h)²$$
              $$E=frach²(r_0+h)²$$



              So, you can see that as $h/r_0 rightarrow 0$, the error $E rightarrow 0$ too.



              This means that as soon as $h/r_0$ is small enough, so is your error. Let see it this way, if you change 1cm in diameter on a circle of 10cm, and calculate the new area with your initial idea, your error will be more important that if you change 1cm on a 1m diameter circle. What only matter in this case is that the change is small enough versus the initial diameter.



              The problem in your question is that you chose $h=1$ and $r_0=5$. You'll need $h$ to be smaller or $r_0$ to be greater



              I made some calculation for circles of radius 5cm, 10cm, and 1m ($r_0=5,10,100$), with a changing radius of 1cm, you have errors of respectively: $2.7%$,$0.8%$ and $0.01%$ between the real area and the one calculated from your formula.






              share|cite|improve this answer






















              • you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
                – JIM
                Sep 3 at 16:19










              • I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
                – F.Carette
                Sep 3 at 17:56










              • what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
                – JIM
                Sep 4 at 2:05










              • My bad. It's the basic formula for relative error. I Added the link in the main body.
                – F.Carette
                Sep 4 at 7:36










              • one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
                – JIM
                Sep 4 at 11:59












              up vote
              3
              down vote










              up vote
              3
              down vote









              All the previous answers explained the problem quite well. But according to your comment on these answer, you still have trouble understanding when to your approximation holds. I'll try to explain what values of h are acceptable to use your method.



              Let's call $A(r)$ the area of a circle of radius $r$. We have $A_real=colorbluepi*r²$



              If $h$ is small enough, you can have $A_approx(r_0+h) approx colorredA(r_0)+2*pi*r_0*h$



              But keep in mind that the real value of $A(r_0+h)$ is indeed:



              beginalignA_real(r_0+h)=colorbluepi*(r_0+h)²\
              & =pi*(r_0²+2r_0h+h²)\
              & =colorbluepi*r_0²+2*pi*r_0*h+pi*h²\
              & =colorredA_real(r_0)+2*pi*r_0*h+pi*h² \
              & =colorredA_approx(r_0+h) +pi*h²
              endalign



              From there, the relative error between you result and the real one is:



              $$E=fracA_real(r_0+h)-A_approx(r_0+h)A_real(r_0+h)$$
              $$E=fracpi*h²pi*(r_0+h)²$$
              $$E=frach²(r_0+h)²$$



              So, you can see that as $h/r_0 rightarrow 0$, the error $E rightarrow 0$ too.



              This means that as soon as $h/r_0$ is small enough, so is your error. Let see it this way, if you change 1cm in diameter on a circle of 10cm, and calculate the new area with your initial idea, your error will be more important that if you change 1cm on a 1m diameter circle. What only matter in this case is that the change is small enough versus the initial diameter.



              The problem in your question is that you chose $h=1$ and $r_0=5$. You'll need $h$ to be smaller or $r_0$ to be greater



              I made some calculation for circles of radius 5cm, 10cm, and 1m ($r_0=5,10,100$), with a changing radius of 1cm, you have errors of respectively: $2.7%$,$0.8%$ and $0.01%$ between the real area and the one calculated from your formula.






              share|cite|improve this answer














              All the previous answers explained the problem quite well. But according to your comment on these answer, you still have trouble understanding when to your approximation holds. I'll try to explain what values of h are acceptable to use your method.



              Let's call $A(r)$ the area of a circle of radius $r$. We have $A_real=colorbluepi*r²$



              If $h$ is small enough, you can have $A_approx(r_0+h) approx colorredA(r_0)+2*pi*r_0*h$



              But keep in mind that the real value of $A(r_0+h)$ is indeed:



              beginalignA_real(r_0+h)=colorbluepi*(r_0+h)²\
              & =pi*(r_0²+2r_0h+h²)\
              & =colorbluepi*r_0²+2*pi*r_0*h+pi*h²\
              & =colorredA_real(r_0)+2*pi*r_0*h+pi*h² \
              & =colorredA_approx(r_0+h) +pi*h²
              endalign



              From there, the relative error between you result and the real one is:



              $$E=fracA_real(r_0+h)-A_approx(r_0+h)A_real(r_0+h)$$
              $$E=fracpi*h²pi*(r_0+h)²$$
              $$E=frach²(r_0+h)²$$



              So, you can see that as $h/r_0 rightarrow 0$, the error $E rightarrow 0$ too.



              This means that as soon as $h/r_0$ is small enough, so is your error. Let see it this way, if you change 1cm in diameter on a circle of 10cm, and calculate the new area with your initial idea, your error will be more important that if you change 1cm on a 1m diameter circle. What only matter in this case is that the change is small enough versus the initial diameter.



              The problem in your question is that you chose $h=1$ and $r_0=5$. You'll need $h$ to be smaller or $r_0$ to be greater



              I made some calculation for circles of radius 5cm, 10cm, and 1m ($r_0=5,10,100$), with a changing radius of 1cm, you have errors of respectively: $2.7%$,$0.8%$ and $0.01%$ between the real area and the one calculated from your formula.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Sep 4 at 6:25

























              answered Sep 3 at 16:06









              F.Carette

              6088




              6088











              • you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
                – JIM
                Sep 3 at 16:19










              • I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
                – F.Carette
                Sep 3 at 17:56










              • what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
                – JIM
                Sep 4 at 2:05










              • My bad. It's the basic formula for relative error. I Added the link in the main body.
                – F.Carette
                Sep 4 at 7:36










              • one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
                – JIM
                Sep 4 at 11:59
















              • you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
                – JIM
                Sep 3 at 16:19










              • I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
                – F.Carette
                Sep 3 at 17:56










              • what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
                – JIM
                Sep 4 at 2:05










              • My bad. It's the basic formula for relative error. I Added the link in the main body.
                – F.Carette
                Sep 4 at 7:36










              • one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
                – JIM
                Sep 4 at 11:59















              you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
              – JIM
              Sep 3 at 16:19




              you really answered the question well !! And I am able to understand most part of question except from "From There . the relative error .............. is small enough , so is your error ."
              – JIM
              Sep 3 at 16:19












              I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
              – F.Carette
              Sep 3 at 17:56




              I edited so it may appear clearer. Anyway, I just wrote this as a side answer because you kept asking if you $h$ was small enough. Please note that the whole $h/r_0$ thing is really specific to THAT case. The general reason why derivative can't be used that way is covered by other answers
              – F.Carette
              Sep 3 at 17:56












              what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
              – JIM
              Sep 4 at 2:05




              what i meant to ask was that how did you get the formula for E=Areal(r0+h)−Aapprox(r0+h)Areal(r0+h)
              – JIM
              Sep 4 at 2:05












              My bad. It's the basic formula for relative error. I Added the link in the main body.
              – F.Carette
              Sep 4 at 7:36




              My bad. It's the basic formula for relative error. I Added the link in the main body.
              – F.Carette
              Sep 4 at 7:36












              one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
              – JIM
              Sep 4 at 11:59




              one more thing , after the calculation of error comes out to be E=h²/(r0+h)^2 , how do you get to know that as h/r0 tends to zero . E tends to 0 ??
              – JIM
              Sep 4 at 11:59










              up vote
              2
              down vote













              We should write



              $$frac dAdcolorredr= 2 pi cdot r$$



              No of course our final answer doaes not mean that when radius of circle changes from $5 cm$ to $6 cm$ then the change in area is equal to $10pi cm^2$, indeed derivatives are linear approximation and, since in that case the relationship between $A$ and $r$ is not linear, it means that



              $$Delta A approx 10pi Delta r$$



              and the approximation becomes better for smaller $Delta r$.






              share|cite|improve this answer




















              • does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
                – JIM
                Sep 3 at 14:41










              • @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
                – gimusi
                Sep 3 at 14:49














              up vote
              2
              down vote













              We should write



              $$frac dAdcolorredr= 2 pi cdot r$$



              No of course our final answer doaes not mean that when radius of circle changes from $5 cm$ to $6 cm$ then the change in area is equal to $10pi cm^2$, indeed derivatives are linear approximation and, since in that case the relationship between $A$ and $r$ is not linear, it means that



              $$Delta A approx 10pi Delta r$$



              and the approximation becomes better for smaller $Delta r$.






              share|cite|improve this answer




















              • does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
                – JIM
                Sep 3 at 14:41










              • @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
                – gimusi
                Sep 3 at 14:49












              up vote
              2
              down vote










              up vote
              2
              down vote









              We should write



              $$frac dAdcolorredr= 2 pi cdot r$$



              No of course our final answer doaes not mean that when radius of circle changes from $5 cm$ to $6 cm$ then the change in area is equal to $10pi cm^2$, indeed derivatives are linear approximation and, since in that case the relationship between $A$ and $r$ is not linear, it means that



              $$Delta A approx 10pi Delta r$$



              and the approximation becomes better for smaller $Delta r$.






              share|cite|improve this answer












              We should write



              $$frac dAdcolorredr= 2 pi cdot r$$



              No of course our final answer doaes not mean that when radius of circle changes from $5 cm$ to $6 cm$ then the change in area is equal to $10pi cm^2$, indeed derivatives are linear approximation and, since in that case the relationship between $A$ and $r$ is not linear, it means that



              $$Delta A approx 10pi Delta r$$



              and the approximation becomes better for smaller $Delta r$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 3 at 13:49









              gimusi

              75.4k73889




              75.4k73889











              • does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
                – JIM
                Sep 3 at 14:41










              • @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
                – gimusi
                Sep 3 at 14:49
















              • does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
                – JIM
                Sep 3 at 14:41










              • @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
                – gimusi
                Sep 3 at 14:49















              does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
              – JIM
              Sep 3 at 14:41




              does this mea that derivatives hold true best for linear approximation ? Also what is Linear approximation ?
              – JIM
              Sep 3 at 14:41












              @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
              – gimusi
              Sep 3 at 14:49




              @JIM Yes exactly for a differentiable function they are the bestlinear approximation, that means that we are approximating locally the function with its tangent line at that point. By Taylor's expansion we can get better approximation using higher order terms.
              – gimusi
              Sep 3 at 14:49










              up vote
              2
              down vote













              Let$f: mathbbR rightarrowtail mathbbR$ be a function, and $a in textInt(textdom(f))$. Then the following statements are equivalent:



              • $f$ is differentiable at $a$

              • $exists xi in mathbbR$ and $exists eta: textdom(f) to mathbbR$ with $limlimits_x to a fraceta(x)x-a=0$, so that
                $$f(x)=f(a)+xi(x-a)+eta(x)$$
                and $xi=f'(a)$.

              I think this equivalent definition of differentiability is more spectacular (and it's easier to generalize to the $mathbbR^n rightarrowtail mathbbR^m$ case), because we can intuitively see that $eta(x)$ is really small around $a$, so for $x approx a$ we have that
              $$f(x) approx f(a) + xi(x-a)$$
              In your case, if we let $A(r):=r^2 pi$, we have that $A'(r)=2r pi$
              $$A(r)=A(r_0)+xi(r-r_0)+eta(r)$$
              And of course, $xi=A'(r_0)=2r_0pi$:
              $$A(r)=A(r_0)+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0rpi-2r_0^2 pi+eta(r)$$
              $$A(r)=2rr_0 pi-r_0^2 pi + eta(r) tag1$$
              If we want to get $eta(r)$:
              $$r^2 pi=2rr_0 pi-r_0^2 pi + eta(r)$$
              $$r^2 pi+r_0^2pi-2r r_0 pi=eta(r)$$
              $$eta(r)=(r-r_0)^2 pi$$
              Using some numbers, we have that $A(5)=25 pi$, $A(5.1)=26.01 pi$, $A(50)=2500 pi$. If we try to approximate $A(5.1)$ and $A(50)$ with $A(5)$ and $A'(5)$, we can use the first equation (without $eta(r)$):
              $$A(r) approx 10 r pi - 25 pi$$
              we get that $A(5.1) approx 26 pi$ and $A(50) approx 475 pi$, while $eta(5.1)=0.01 pi$ and $eta(50)=2025 pi$. As you can see, it's quite good for $5.1$, but it's terrible for $50$.



              This kind of linear approximation is really useful in physics, for example when you are dealing with a complicated system, you can't really solve the differential equation(s), but you can make the assumption that the change in the sought function (for example, the displacement-time function of a ball) is really small, so you can linearize the equation, and (probably) solve it.






              share|cite|improve this answer


















              • 1




                Maybe the symbol $A$ is a little confusing here.
                – gimusi
                Sep 3 at 15:43










              • @gimusi you're right.
                – Botond
                Sep 3 at 16:04










              • @Botond not able to understand a single word that you wrote !!
                – JIM
                Sep 3 at 16:20










              • @JIM I extended my answer.
                – Botond
                Sep 3 at 17:13














              up vote
              2
              down vote













              Let$f: mathbbR rightarrowtail mathbbR$ be a function, and $a in textInt(textdom(f))$. Then the following statements are equivalent:



              • $f$ is differentiable at $a$

              • $exists xi in mathbbR$ and $exists eta: textdom(f) to mathbbR$ with $limlimits_x to a fraceta(x)x-a=0$, so that
                $$f(x)=f(a)+xi(x-a)+eta(x)$$
                and $xi=f'(a)$.

              I think this equivalent definition of differentiability is more spectacular (and it's easier to generalize to the $mathbbR^n rightarrowtail mathbbR^m$ case), because we can intuitively see that $eta(x)$ is really small around $a$, so for $x approx a$ we have that
              $$f(x) approx f(a) + xi(x-a)$$
              In your case, if we let $A(r):=r^2 pi$, we have that $A'(r)=2r pi$
              $$A(r)=A(r_0)+xi(r-r_0)+eta(r)$$
              And of course, $xi=A'(r_0)=2r_0pi$:
              $$A(r)=A(r_0)+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0rpi-2r_0^2 pi+eta(r)$$
              $$A(r)=2rr_0 pi-r_0^2 pi + eta(r) tag1$$
              If we want to get $eta(r)$:
              $$r^2 pi=2rr_0 pi-r_0^2 pi + eta(r)$$
              $$r^2 pi+r_0^2pi-2r r_0 pi=eta(r)$$
              $$eta(r)=(r-r_0)^2 pi$$
              Using some numbers, we have that $A(5)=25 pi$, $A(5.1)=26.01 pi$, $A(50)=2500 pi$. If we try to approximate $A(5.1)$ and $A(50)$ with $A(5)$ and $A'(5)$, we can use the first equation (without $eta(r)$):
              $$A(r) approx 10 r pi - 25 pi$$
              we get that $A(5.1) approx 26 pi$ and $A(50) approx 475 pi$, while $eta(5.1)=0.01 pi$ and $eta(50)=2025 pi$. As you can see, it's quite good for $5.1$, but it's terrible for $50$.



              This kind of linear approximation is really useful in physics, for example when you are dealing with a complicated system, you can't really solve the differential equation(s), but you can make the assumption that the change in the sought function (for example, the displacement-time function of a ball) is really small, so you can linearize the equation, and (probably) solve it.






              share|cite|improve this answer


















              • 1




                Maybe the symbol $A$ is a little confusing here.
                – gimusi
                Sep 3 at 15:43










              • @gimusi you're right.
                – Botond
                Sep 3 at 16:04










              • @Botond not able to understand a single word that you wrote !!
                – JIM
                Sep 3 at 16:20










              • @JIM I extended my answer.
                – Botond
                Sep 3 at 17:13












              up vote
              2
              down vote










              up vote
              2
              down vote









              Let$f: mathbbR rightarrowtail mathbbR$ be a function, and $a in textInt(textdom(f))$. Then the following statements are equivalent:



              • $f$ is differentiable at $a$

              • $exists xi in mathbbR$ and $exists eta: textdom(f) to mathbbR$ with $limlimits_x to a fraceta(x)x-a=0$, so that
                $$f(x)=f(a)+xi(x-a)+eta(x)$$
                and $xi=f'(a)$.

              I think this equivalent definition of differentiability is more spectacular (and it's easier to generalize to the $mathbbR^n rightarrowtail mathbbR^m$ case), because we can intuitively see that $eta(x)$ is really small around $a$, so for $x approx a$ we have that
              $$f(x) approx f(a) + xi(x-a)$$
              In your case, if we let $A(r):=r^2 pi$, we have that $A'(r)=2r pi$
              $$A(r)=A(r_0)+xi(r-r_0)+eta(r)$$
              And of course, $xi=A'(r_0)=2r_0pi$:
              $$A(r)=A(r_0)+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0rpi-2r_0^2 pi+eta(r)$$
              $$A(r)=2rr_0 pi-r_0^2 pi + eta(r) tag1$$
              If we want to get $eta(r)$:
              $$r^2 pi=2rr_0 pi-r_0^2 pi + eta(r)$$
              $$r^2 pi+r_0^2pi-2r r_0 pi=eta(r)$$
              $$eta(r)=(r-r_0)^2 pi$$
              Using some numbers, we have that $A(5)=25 pi$, $A(5.1)=26.01 pi$, $A(50)=2500 pi$. If we try to approximate $A(5.1)$ and $A(50)$ with $A(5)$ and $A'(5)$, we can use the first equation (without $eta(r)$):
              $$A(r) approx 10 r pi - 25 pi$$
              we get that $A(5.1) approx 26 pi$ and $A(50) approx 475 pi$, while $eta(5.1)=0.01 pi$ and $eta(50)=2025 pi$. As you can see, it's quite good for $5.1$, but it's terrible for $50$.



              This kind of linear approximation is really useful in physics, for example when you are dealing with a complicated system, you can't really solve the differential equation(s), but you can make the assumption that the change in the sought function (for example, the displacement-time function of a ball) is really small, so you can linearize the equation, and (probably) solve it.






              share|cite|improve this answer














              Let$f: mathbbR rightarrowtail mathbbR$ be a function, and $a in textInt(textdom(f))$. Then the following statements are equivalent:



              • $f$ is differentiable at $a$

              • $exists xi in mathbbR$ and $exists eta: textdom(f) to mathbbR$ with $limlimits_x to a fraceta(x)x-a=0$, so that
                $$f(x)=f(a)+xi(x-a)+eta(x)$$
                and $xi=f'(a)$.

              I think this equivalent definition of differentiability is more spectacular (and it's easier to generalize to the $mathbbR^n rightarrowtail mathbbR^m$ case), because we can intuitively see that $eta(x)$ is really small around $a$, so for $x approx a$ we have that
              $$f(x) approx f(a) + xi(x-a)$$
              In your case, if we let $A(r):=r^2 pi$, we have that $A'(r)=2r pi$
              $$A(r)=A(r_0)+xi(r-r_0)+eta(r)$$
              And of course, $xi=A'(r_0)=2r_0pi$:
              $$A(r)=A(r_0)+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0(r-r_0)pi+eta(r)$$
              $$A(r)=r_0^2 pi+2r_0rpi-2r_0^2 pi+eta(r)$$
              $$A(r)=2rr_0 pi-r_0^2 pi + eta(r) tag1$$
              If we want to get $eta(r)$:
              $$r^2 pi=2rr_0 pi-r_0^2 pi + eta(r)$$
              $$r^2 pi+r_0^2pi-2r r_0 pi=eta(r)$$
              $$eta(r)=(r-r_0)^2 pi$$
              Using some numbers, we have that $A(5)=25 pi$, $A(5.1)=26.01 pi$, $A(50)=2500 pi$. If we try to approximate $A(5.1)$ and $A(50)$ with $A(5)$ and $A'(5)$, we can use the first equation (without $eta(r)$):
              $$A(r) approx 10 r pi - 25 pi$$
              we get that $A(5.1) approx 26 pi$ and $A(50) approx 475 pi$, while $eta(5.1)=0.01 pi$ and $eta(50)=2025 pi$. As you can see, it's quite good for $5.1$, but it's terrible for $50$.



              This kind of linear approximation is really useful in physics, for example when you are dealing with a complicated system, you can't really solve the differential equation(s), but you can make the assumption that the change in the sought function (for example, the displacement-time function of a ball) is really small, so you can linearize the equation, and (probably) solve it.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Sep 4 at 7:12

























              answered Sep 3 at 13:54









              Botond

              4,2332632




              4,2332632







              • 1




                Maybe the symbol $A$ is a little confusing here.
                – gimusi
                Sep 3 at 15:43










              • @gimusi you're right.
                – Botond
                Sep 3 at 16:04










              • @Botond not able to understand a single word that you wrote !!
                – JIM
                Sep 3 at 16:20










              • @JIM I extended my answer.
                – Botond
                Sep 3 at 17:13












              • 1




                Maybe the symbol $A$ is a little confusing here.
                – gimusi
                Sep 3 at 15:43










              • @gimusi you're right.
                – Botond
                Sep 3 at 16:04










              • @Botond not able to understand a single word that you wrote !!
                – JIM
                Sep 3 at 16:20










              • @JIM I extended my answer.
                – Botond
                Sep 3 at 17:13







              1




              1




              Maybe the symbol $A$ is a little confusing here.
              – gimusi
              Sep 3 at 15:43




              Maybe the symbol $A$ is a little confusing here.
              – gimusi
              Sep 3 at 15:43












              @gimusi you're right.
              – Botond
              Sep 3 at 16:04




              @gimusi you're right.
              – Botond
              Sep 3 at 16:04












              @Botond not able to understand a single word that you wrote !!
              – JIM
              Sep 3 at 16:20




              @Botond not able to understand a single word that you wrote !!
              – JIM
              Sep 3 at 16:20












              @JIM I extended my answer.
              – Botond
              Sep 3 at 17:13




              @JIM I extended my answer.
              – Botond
              Sep 3 at 17:13










              up vote
              0
              down vote













              No, but you are on the right track.



              The derivative you have found tells you that when the radius changes from $5$cm to $(5+h)$cm and $h$ is small then the area changes by approximately $10pi h$ square cm.



              In your example you are taking $h=1$, which is not small, so the approximation isn't even close.






              share|cite|improve this answer




















              • So what does the answer that I calculated actually means (if it does not mean change in area )??
                – JIM
                Sep 3 at 14:43










              • What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
                – Ethan Bolker
                Sep 3 at 14:46










              • In this case is my h correct ??
                – JIM
                Sep 3 at 15:09










              • There's no such thing as "the correct $h$". The smaller it is the better the approximation.
                – Ethan Bolker
                Sep 3 at 21:32














              up vote
              0
              down vote













              No, but you are on the right track.



              The derivative you have found tells you that when the radius changes from $5$cm to $(5+h)$cm and $h$ is small then the area changes by approximately $10pi h$ square cm.



              In your example you are taking $h=1$, which is not small, so the approximation isn't even close.






              share|cite|improve this answer




















              • So what does the answer that I calculated actually means (if it does not mean change in area )??
                – JIM
                Sep 3 at 14:43










              • What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
                – Ethan Bolker
                Sep 3 at 14:46










              • In this case is my h correct ??
                – JIM
                Sep 3 at 15:09










              • There's no such thing as "the correct $h$". The smaller it is the better the approximation.
                – Ethan Bolker
                Sep 3 at 21:32












              up vote
              0
              down vote










              up vote
              0
              down vote









              No, but you are on the right track.



              The derivative you have found tells you that when the radius changes from $5$cm to $(5+h)$cm and $h$ is small then the area changes by approximately $10pi h$ square cm.



              In your example you are taking $h=1$, which is not small, so the approximation isn't even close.






              share|cite|improve this answer












              No, but you are on the right track.



              The derivative you have found tells you that when the radius changes from $5$cm to $(5+h)$cm and $h$ is small then the area changes by approximately $10pi h$ square cm.



              In your example you are taking $h=1$, which is not small, so the approximation isn't even close.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 3 at 13:51









              Ethan Bolker

              36.7k54299




              36.7k54299











              • So what does the answer that I calculated actually means (if it does not mean change in area )??
                – JIM
                Sep 3 at 14:43










              • What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
                – Ethan Bolker
                Sep 3 at 14:46










              • In this case is my h correct ??
                – JIM
                Sep 3 at 15:09










              • There's no such thing as "the correct $h$". The smaller it is the better the approximation.
                – Ethan Bolker
                Sep 3 at 21:32
















              • So what does the answer that I calculated actually means (if it does not mean change in area )??
                – JIM
                Sep 3 at 14:43










              • What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
                – Ethan Bolker
                Sep 3 at 14:46










              • In this case is my h correct ??
                – JIM
                Sep 3 at 15:09










              • There's no such thing as "the correct $h$". The smaller it is the better the approximation.
                – Ethan Bolker
                Sep 3 at 21:32















              So what does the answer that I calculated actually means (if it does not mean change in area )??
              – JIM
              Sep 3 at 14:43




              So what does the answer that I calculated actually means (if it does not mean change in area )??
              – JIM
              Sep 3 at 14:43












              What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
              – Ethan Bolker
              Sep 3 at 14:46




              What you calculated is the approximation to the change in area you get by substituting $h=1$ in the approximation recipe the derivative provides. Whether that's a good approximation to the change depends on how large the derivative is and how big $h$ is relative to what matters in an application.
              – Ethan Bolker
              Sep 3 at 14:46












              In this case is my h correct ??
              – JIM
              Sep 3 at 15:09




              In this case is my h correct ??
              – JIM
              Sep 3 at 15:09












              There's no such thing as "the correct $h$". The smaller it is the better the approximation.
              – Ethan Bolker
              Sep 3 at 21:32




              There's no such thing as "the correct $h$". The smaller it is the better the approximation.
              – Ethan Bolker
              Sep 3 at 21:32










              up vote
              0
              down vote













              To say that $left.dfracdAdrright|_r,=,5text cm = 10pitext cm$ means that if you multiply the rate of change of $r,$ at the point at which $r=5text cm,$ by $10pitext cm,$ you get the rate of change of $A$ at that point. But that does not mean that if $r$ grows by $1text cm,$ then $A$ will increase by $10pitext cm^2,$ because it won't continue changing at that same rate as $r$ changes.



              Generally if $left.dfracdydxright|_x,=,a = b$ means that at the point where $x=a,$ the dependent variable $y$ is changing $a$ times as fast as the independent variable $x.$






              share|cite|improve this answer
























                up vote
                0
                down vote













                To say that $left.dfracdAdrright|_r,=,5text cm = 10pitext cm$ means that if you multiply the rate of change of $r,$ at the point at which $r=5text cm,$ by $10pitext cm,$ you get the rate of change of $A$ at that point. But that does not mean that if $r$ grows by $1text cm,$ then $A$ will increase by $10pitext cm^2,$ because it won't continue changing at that same rate as $r$ changes.



                Generally if $left.dfracdydxright|_x,=,a = b$ means that at the point where $x=a,$ the dependent variable $y$ is changing $a$ times as fast as the independent variable $x.$






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  To say that $left.dfracdAdrright|_r,=,5text cm = 10pitext cm$ means that if you multiply the rate of change of $r,$ at the point at which $r=5text cm,$ by $10pitext cm,$ you get the rate of change of $A$ at that point. But that does not mean that if $r$ grows by $1text cm,$ then $A$ will increase by $10pitext cm^2,$ because it won't continue changing at that same rate as $r$ changes.



                  Generally if $left.dfracdydxright|_x,=,a = b$ means that at the point where $x=a,$ the dependent variable $y$ is changing $a$ times as fast as the independent variable $x.$






                  share|cite|improve this answer












                  To say that $left.dfracdAdrright|_r,=,5text cm = 10pitext cm$ means that if you multiply the rate of change of $r,$ at the point at which $r=5text cm,$ by $10pitext cm,$ you get the rate of change of $A$ at that point. But that does not mean that if $r$ grows by $1text cm,$ then $A$ will increase by $10pitext cm^2,$ because it won't continue changing at that same rate as $r$ changes.



                  Generally if $left.dfracdydxright|_x,=,a = b$ means that at the point where $x=a,$ the dependent variable $y$ is changing $a$ times as fast as the independent variable $x.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 3 at 18:06









                  Michael Hardy

                  206k23187466




                  206k23187466



























                       

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