Chapter 6.3, Problem 64E

Rain Gutter A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. (See the figure on the next page.)

1. (a) Show that the cross-sectional area of the gutter is modeled by the function $A(\theta )=100\sin \theta +100\sin \theta \cos \theta$
2. (b) Graph the function A for 0 ≤ θ ≤ π/2.
3. (c) For what angle θ is the largest cross-sectional area achieved?

To determine

To show: The cross-sectional area of the rain gutter is modeled by the function $A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta$.

Explanation of Solution

Given:

A rain gutter constructed from a metal sheet of width $30\text{cm}$ by bending up one-third of sheet on each side by an angle $\theta$.

Proof:

The below figure shows the labeling of the rain gutter.

Figure (1)

The cross sectional area of the rain gutter is equal to the area of trapezium formed by Figure (1).

Formula to find the area of trapezium is,

$\text{Area}=\frac{1}{2}\times \left(a+b\right)\times h$. (I)

Where

• $a$ and $b$ are parallel sides.
• $h$ is the height of the trapezium.

Formula to find the opposite side $h$ in right angle triangle $\Delta \text{BAF}$.

$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$

Substitute $h$ for opposite and $10$ for hypotenuse.

$\begin{array}{c}\sin \theta =\frac{h}{\text{10}}\\ h=10\sin \theta \end{array}$

Therefore the height, $h$ of the Figure (1) is $10\sin \theta$.

Formula to find the side $\text{AF}$ in right angle triangle $\Delta \text{BAF}$.

$\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$

Substitute $\text{AF}$ for adjacent and $10$ for hypotenuse.

$\begin{array}{c}\cos \theta =\frac{\text{AF}}{\text{10}}\\ \text{AF}=10\cos \theta \end{array}$

Formula to find the side $\text{ED}$ in the right angle triangle $\Delta \text{CDE}$.

$\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$

Substitute $\text{ED}$ for adjacent and $10$ for hypotenuse.

$\begin{array}{c}\cos \theta =\frac{\text{ED}}{\text{10}}\\ \text{ED}=10\cos \theta \end{array}$

The side $\text{AD}$ from figure (1) is,

$\text{AD}=\text{AF}+\text{FE}+\text{ED}$

Substitute $10\cos \theta$ for $\text{AF}$, $10$ for $\text{FE}$ and $10\cos \theta$ for $\text{ED}$ in above equation to find $\text{AD}$.

$\begin{array}{c}\text{AD}=10\cos \theta +10+10\cos \theta \\ \text{AD}=20\cos \theta +10\end{array}$

The sides $\text{AD}$ and $\text{BC}$ are parallel in Figure (1) with height $h$.

Substitute $20\cos \theta +10$ for $a$ , $10$ for $b$, $10\sin \theta$ for $h$, and area by $A\left(\theta \right)$ in equation (I)

$\begin{array}{c}A\left(\theta \right)=\frac{1}{2}\times \left(20\cos \theta +10+10\right)\times 10\sin \theta \\ =\left(20\cos \theta +20\right)\times 5\sin \theta \\ =100\sin \theta +100\sin \theta \cos \theta \end{array}$

Hence, the cross sectional area of the rain gutter is modeled by the function $A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta$.

To determine

To sketch: The graph of given function $A\left(\theta \right)$ in part (a) for $0\le \theta \le \frac{\pi }{2}$.

Explanation of Solution

The function for cross sectional area is given by, $A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta$.

The below figure shows the graph of the function $A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta$ for $0\le \theta \le \frac{\pi }{2}$.

Figure (2)

The graph of the function in Figure (2) is parabolic by nature for $0\le \theta \le \frac{\pi }{2}$.

To determine

To find: The angle $\theta$, for which the function $A\left(\theta \right)$ in part (a) achieved largest cross sectional area.

Answer to Problem 64E

The angle $\theta$ for which the function $A\left(\theta \right)$ achieved the largest cross sectional area of the rain gutter is $60°$.

Explanation of Solution

Given:

The cross sectional area of the rain gutter is given by the function $A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta$.

Calculation:

The function $A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta$ has both trigonometric function $\sin \theta$ and $\cos \theta$.

Case (i):

The value of trigonometric function $\sin \theta$ is $0$ and $\cos \theta$ is $1$ at $\theta =0°$.

The function $A\left(\theta \right)$ becomes,

$\begin{array}{c}A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta \\ =0\end{array}$

Case (ii):

The value of trigonometric function $\sin \theta$ is $1$ and $\cos \theta$ is $0$ at $\theta =90°$.

The function $A\left(\theta \right)$ becomes,

$\begin{array}{c}A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta \\ =100+0\\ =100\end{array}$

Case (iii):

The value of trigonometric function $\sin \theta$ is $\frac{1}{2}$ and $\cos \theta$ is $\frac{\sqrt{3}}{2}$ at $\theta =30°$.

The function $A\left(\theta \right)$ becomes,

$\begin{array}{c}A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta \\ =100\times \frac{1}{2}+100\times \frac{1}{2}\times \frac{\sqrt{3}}{2}\\ =50+25\sqrt{3}\\ \approx 93.3\end{array}$

Case (iv):

The value of trigonometric function $\sin \theta$ is $\frac{1}{\sqrt{2}}$ and $\cos \theta$ is $\frac{1}{\sqrt{2}}$ at $\theta =45°$.

The function $A\left(\theta \right)$ becomes,

$\begin{array}{c}A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta \\ =\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}\cdot \sqrt{2}}\\ =50\sqrt{2}+50\\ \approx 121\end{array}$

Case (v):

The value of trigonometric function $\sin \theta$ is $\frac{\sqrt{3}}{2}$ and $\cos \theta$ is $\frac{1}{2}$ at $\theta =60°$.

The function $A\left(\theta \right)$ becomes,

$\begin{array}{c}A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta \\ =\frac{100\sqrt{3}}{2}+\frac{100\sqrt{3}}{2\times 2}\\ =50\sqrt{3}+25\sqrt{3}\\ \approx 130\end{array}$

Thus, the function $A\left(\theta \right)$ gives largest cross sectional area in case (v).

Therefore the angle $\theta$ for which the function $A\left(\theta \right)$ achieved the largest cross sectional, area is $60°$.