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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!

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

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  • $\begingroup$ I like your explanation! $\endgroup$ – Larry Dec 24 '18 at 4:51
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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: $$\begin{align} C_{total~cost} &= 0.06\int_{3.082}^{20.962}\left(36-\left[22+20\cos\left(\frac{\pi t}{12}\right)\right]\right)dt\\ &=0.06(360.03979)=21.602\approx$21.6 \end{align}$$

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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.

Merry Xmas

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