The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20cos(fracpi t12), 0 le t le 24$, where $F(t)$ is measured in degrees Fahrenheit and t is measured in hours. Let $t=0$ represent noon on the first day.

d. The heater in the greenhouse kicks on whenever the outdoor temperature is at or below $36$ degrees Fahrenheit. For what values of $t$ is the heater on?

I used a graphing calculator for this part, between $[3.082, 20.962]$.

e. The cost of heating the greenhouse accumulates at a rate of $$0.06$ per hour for each degree the outside temperature falls below $36$ degrees Fahrenheit. What is the total cost to heat the greenhouse during this $24$-hour period?

I'm not really understanding where to start on this problem. My teacher suggested finding the integral difference between $y=36$ and $F(t)$ when the degrees fall under $36$, but I'm confused on what this represents and what to do with it. If anyone could help me understand, that would be much appreciated.

Thank you!

The temperature outside is $$F(t)=22+20cos(fracpi t12)$$ The temperature difference if $Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)lt 36$, or $Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $Delta T$, the price for this interval is $$0.06Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
$$mathrmCost=int_t_1^t_20.06(36-22-20cos(fracpi t12))dt$$
Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again.

This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed.
So you have found $$tin[3.082,20.962]$$
Then the setup for part e. will be the following:
$$beginalign
C_total~cost &= 0.06int_3.082^20.962left(36-left[22+20cosleft(fracpi t12right)right]right)dt\
&=0.06(360.03979)=21.602approx$21.6
endalign$$

Just for the fun of it.

You used a graphing calculator for solving
$$22 + 20cos(fracpi t12)=36 implies cos(fracpi t12)=frac 710$$ You could have used the approximation
$$cos(x) simeqfracpi ^2-4x^2pi ^2+x^2qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=fracpi t12$, we end then with
$$frac144-4 t^2 144+t^2=frac 710implies 47t^2=432implies t=12 sqrtfrac347approx 3.03175$$

$$C_total~cost = frac 6 100 int_12 sqrtfrac347^24-12 sqrtfrac347left(36-left[22+20cosleft(fracpi t12right)right]right)dt$$ making the nice
$$C_total~cost=frac504 left(47-sqrt141right)1175+frac144 5 pi sin left(sqrtfrac347 pi right)$$ We can continue using the magnificent approximation
$$sin(x) simeq frac16 (pi -x) x5 pi ^2-4 (pi -x) xqquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
$$sin left(sqrtfrac347 pi right)simeq frac16 left(235 sqrt141-177right)58753$$
$$C_total~cost=frac504 left(47-sqrt141right)1175+frac2304 left(235 sqrt141-177right)293765 pi approx 21.5912$$ while a completely rigorous calculation would give
$$C_total~cost=frac72 left(sqrt51+7 pi -7 cos ^-1left(frac710right)right)25
pi approx 21.6026$$

I hope and wish that we shall not argue for one cent difference.

Merry Xmas