The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20cos(frac{pi t}{12}), 0 le t le 24$, 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 $$F(t)=22+20cos(frac{pi t}{12})$$ 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.
$$mathrm{Cost}=int_{t_1}^{t_2}0.06(36-22-20cos(frac{pi t}{12}))dt$$
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:
$$begin{align}
C_{total~cost} &= 0.06int_{3.082}^{20.962}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt\
&=0.06(360.03979)=21.602approx$21.6
end{align}$$

Just for the fun of it.

You used a graphing calculator for solving
$$22 + 20cos(frac{pi t}{12})=36 implies cos(frac{pi t}{12})=frac 7{10}$$ You could have used the approximation
$$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=frac{pi t}{12}$, we end then with
$$frac{144-4 t^2 }{144+t^2}=frac 7{10}implies 47t^2=432implies t=12 sqrt{frac{3}{47}}approx 3.03175$$

$$C_{total~cost} = frac 6 {100} int_{12 sqrt{frac{3}{47}}}^{24-12 sqrt{frac{3}{47}}}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt$$ making the nice
$$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{144 }{5 pi }sin left(sqrt{frac{3}{47}} pi right)$$ We can continue using the magnificent approximation
$$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
$$sin left(sqrt{frac{3}{47}} pi right)simeq frac{16 left(235 sqrt{141}-177right)}{58753}$$
$$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{2304 left(235 sqrt{141}-177right)}{293765 pi }approx 21.5912$$ while a completely rigorous calculation would give
$$C_{total~cost}=frac{72 left(sqrt{51}+7 pi -7 cos ^{-1}left(frac{7}{10}right)right)}{25
pi }approx 21.6026$$

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

Merry Xmas