mjhjmtu

A particle moves along the top of the
parabola $y^2 = 2x$ from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point $(2, 2)$.

So I isolate $y$, giving me $y=sqrt2x$. I then find the derivative of $y$, which is $1/sqrt2x$. And $sqrt2x=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.

In that case is useful to use parametric equation that is

$$y^2=2x implies vec s(t)=(2c^2t^2,2ct) implies vec v(t)=s'(t)=(4c^2t,2c)$$

and since the point $(2,2)$ is reached at $t=1/c$, that is $vec s(1/c)=(2,2)$, the velocity at that point is $vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have

$$|v(1/c)|=sqrt16c^2+4c^2=5 implies c=fracsqrt 52$$

therefore the velocity at $(2,2)$ is

$$v(t)=left(2sqrt 5,sqrt 5right)$$

As an alternative

$$y^2=2x implies 2ydy=2dx implies fracdydx=frac1y$$

therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find

$$5=ksqrt16+4implies k=fracsqrt 52$$

$(2y) dy/dx=2$; $dy/dx =1/y$.

Slope at $(2,2)$: $dy/dx =1/2= tan alpha$.

$cos alpha =2/√5$, $sin alpha =1/√5.$
(Pythagoras).

$v_x= v cos alpha$; $v_y = v sin alpha$, where $v = |vec v|$ .

$vec v = (v_x,v_y) $.

Consider the curve in a parametric form $vecz=(x(t),y(t))$, in that case we know that

$$
vecv=fracmathrmdveczmathrmdt=(x'(t),y'(t))
$$

Additionally, we know that the speed is $|vecv|=5$. Therefore, we need to find the direction of $vecv$. Given the equation of the curve

$$
y^2=2x
$$
we can write the parametric from in terms of $y$ alone:
$$
vecz=left(fracy(t)^22,y(t) right)
$$
Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then:
$$
vecz=left(fracc^2t^22,ct right)
$$
Then the velocity is
$$
vecv=(c^2t,c)
$$
At the point $vecz=(2,2)$, $t=2/c$. Therefore,

$$
vecv=(2c,c) Rightarrow |vecv|=sqrt4c^2+c^2=5
$$
Then, $c=sqrt5$. Lets check our solution,
$$
beginalign
vecz(t)=&left(frac5t^22,sqrt5tright)\
vecv(t)=&left(5t,sqrt5right)
endalign
$$
at $t=frac2sqrt5$:
$$
beginalign
veczleft(frac2sqrt5right)=&left(frac52cdotfrac45,sqrt5frac2sqrt5right)=(2,2)\
vecvleft(frac2sqrt5right)=&left(5frac2sqrt5,sqrt5right)=(2sqrt5,sqrt5)
endalign
$$

in cartesian co-ordinate sysytem :

$(speed)^2=left(dfracdxdtright)^2+left(dfracdydtright)^2=25$ also,

$left(dfracdydtright)=dfrac1yleft(dfracdxdtright) $ put it in above equation

to get ,

$left(dfracdxdtright)=dfrac5ysqrt1+y^2$ and $left(dfracdydtright)=dfrac5sqrt1+y^2$

thus, at point (2,2) speed will be

$sqrt (2sqrt5)^2+(sqrt 5)^2=5 $ unit per second

in above calculations gravity is not taken into account