Suppose we create a rain gutter using an aluminum sheet that is 12 inches wide. After marking off a length of 4 inches from each edge, the length is bent up at an angle $\theta$, see the following figure.

enter image description here

I wish to express the area of the opening in terms of $\theta$, so after some trigonometry, I was able to get

$$A(\theta) = 16\cos \theta \sin \theta \hspace{20 pt} \text{where} \, \, 0 \leq \theta < \frac{\pi}{2} \tag{1}$$

But when $\theta = \frac{\pi}{2}$, $A(\theta) = 0$. Observing that the area I want when $\theta = \frac{\pi}{2}$ is a triangle, it is clear that the area can be found by doubling the area of the two right triangles

$$2 \cdot \Big(\frac{1}{2}\Big) \cdot (4) \cdot (4) = 16$$

But since $\sin(\frac{\pi}{2}) = 1$, we can express the area in terms of $\theta$ as

$$16\sin(\theta) \tag{2}$$

when $\theta = \frac{\pi}{2}$.

So the formula that I have is a piecewise function

$$A(\theta) = \begin{cases} 16\sin(\theta)\cos(\theta) & 0 \leq \theta < \frac{\pi}{2} \\\\ 16\sin(\theta) & \theta = \frac{\pi}{2} \end{cases} $$

But the book has $A(\theta) = 16\sin(\theta)(\cos(\theta) + 1)$, so I'm not sure what I'm doing wrong. I could be wrong about $(2)$, but even then I do not know how to proceed to get the correct equation.

  • $\begingroup$ Are you forgetting to include the area of the rectangular region? This part would correct your formula. $\endgroup$ – abiessu Mar 26 '16 at 20:51

The area of the two triangles is, as you have noticed, $2\cdot\frac 12\cdot 4\sin\theta \cdot 4\cos\theta$. This appropriately goes to $0$ as $\theta$ goes to $\pi\over 2$. The area of the rectangle in the middle is $4\cdot 4\sin\theta$, so the area of the whole region is then


| cite | improve this answer | |
  • $\begingroup$ I misinterpreted the phrase "area of the opening" to pertain to the area that the gutter makes when it sweeps upwards theta units, so I didn't consider the opening in the middle. Thank you! $\endgroup$ – Jay Dunivin Mar 26 '16 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.