# What is the total cost to heat the greenhouse during this 24-hour period?

The temperature outside the greenhouse during a $$24$$-hour period is given by $$F(t) = 22 + 20\cos(\frac{\pi t}{12}), 0 \le t \le 24$$, 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!

• You can make your integral easier by just considering when $\cos{\frac{\pi t}{12}}$ is not greater than $\frac{7}{10}$. – John Douma Dec 24 '18 at 4:55
• You have a typo : $3.082$ should be $3.0382$ – Claude Leibovici Dec 24 '18 at 10:44

The temperature outside is $$F(t)=22+20\cos(\frac{\pi t}{12})$$ 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 $$0.06\Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral. $$\mathrm{Cost}=\int_{t_1}^{t_2}0.06(36-22-20\cos(\frac{\pi t}{12}))dt$$ Here $$t_1$$ is when the temperature dips below 36 degrees, and $$t_2$$ is when it raises to 36 again.

• I like your explanation! – Larry Dec 24 '18 at 4:51

This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed. So you have found $$t\in[3.082,20.962]$$ Then the setup for part e. will be the following: \begin{align} C_{total~cost} &= 0.06\int_{3.082}^{20.962}\left(36-\left[22+20\cos\left(\frac{\pi t}{12}\right)\right]\right)dt\\ &=0.06(360.03979)=21.602\approx21.6 \end{align}

Just for the fun of it.

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

$$C_{total~cost} = \frac 6 {100} \int_{12 \sqrt{\frac{3}{47}}}^{24-12 \sqrt{\frac{3}{47}}}\left(36-\left[22+20\cos\left(\frac{\pi t}{12}\right)\right]\right)dt$$ 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 $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $$1400$$ years ago) $$\sin \left(\sqrt{\frac{3}{47}} \pi \right)\simeq \frac{16 \left(235 \sqrt{141}-177\right)}{58753}$$ $$C_{total~cost}=\frac{504 \left(47-\sqrt{141}\right)}{1175}+\frac{2304 \left(235 \sqrt{141}-177\right)}{293765 \pi }\approx 21.5912$$ 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.