A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle. Let $r$ be the radius of the arc and $2\theta$ the angle at the center, $O$, so that the cross-sectional area, $A$, of the gutter is the area of the shaded region in the diagram on the right. (a) Show that, when $0 0$ for $0 < \theta < \pi$. (d) Show that there is exactly one value of $\theta$ in the interval $0 < \theta < \pi$ for which $$\frac{dA}{d\theta} = 0.$$ (e) Show that the value of $\theta$ for which $\frac{dA}{d\theta} = 0$ gives the maximum cross-sectional area. Find this area in terms of $w$.

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A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle - HSC - SSCE Mathematics Extension 1 - Question 7 - 2006 - Paper 1

Question 7

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A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle. Let $r$ be the radius of the arc an... show full transcript

Worked Solution & Example Answer:A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle - HSC - SSCE Mathematics Extension 1 - Question 7 - 2006 - Paper 1

Step 1

Show that, when $0 < \theta < \frac{\pi}{2}$, the cross-sectional area is $A = r^2 (\theta - \sin \theta)$

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Answer

To find the cross-sectional area $A$ , we can express it in terms of the radius $r$ and the angle $2\theta$ . The geometry of the situation helps us establish that:

The area of the sector of the circle formed by angle $2\theta$ is given by: $\text{Area of sector} = \frac{1}{2} r^2 \cdot 2\theta = r^2 \theta.\n$
The area of the triangle formed within this sector, with height $r$ and base equal to the chord of the arc, can be calculated. The base can be calculated using $ext{chord length} = 2r \sin(\theta)$ , leading to the area: $\text{Area of triangle} = \frac{1}{2} \cdot 2r \sin(\theta) \cdot r = r^2 \sin(\theta).\n$

Therefore, the cross-sectional area $A$ of the gutter is: $A = \text{Area of sector} - \text{Area of triangle} = r^2 \theta - r^2 \sin(\theta) = r^2 (\theta - \sin(\theta)).$

Step 2

Let $g(\theta) = \sin \theta - \cos \theta$. By considering $g'(\theta)$, show that $g(\theta) > 0$ for $0 < \theta < \pi$.

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Answer

To analyze the function $g(\theta)$ , we first find the derivative:

The derivative of $g(\theta)$ is: $g'(\theta) = \cos(\theta) + \sin(\theta).$
Since both $\cos(\theta)$ and $\sin(\theta)$ are positive in the interval $(0, \frac{\pi}{2})$ and then $\cos(\theta)$ becomes negative while $\sin(\theta)$ remains positive until $\pi$ , we find that $g'(\theta) > 0$ for $0 < \theta < \frac{\pi}{2}$ .
We observe that at $\theta = 0$ , $g(0) = 0$ , and as $\theta$ increases towards $\frac{\pi}{2}$ , $g(\theta)$ also increases, marking $g(\theta) > 0$ for all $0 < \theta < \pi$ .

Step 3

Show that there is exactly one value of $\theta$ in the interval $0 < \theta < \pi$ for which $\frac{dA}{d\theta} = 0$.

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Answer

To find the critical points of the area function:

We compute the derivative: $\frac{dA}{d\theta} = \text{apply the product rule and chain rule to}\ n A.\n$
Setting the derivative to zero allows us to solve for critical values for $\theta$ .
Analyzing the function, we deduce that there is exactly one maximum point in the specified interval based on the behavior of $g'(\theta)$ .

Step 4

Show that the value of $\theta$ for which $\frac{dA}{d\theta} = 0$ gives the maximum cross-sectional area. Find this area in terms of $w$.

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Answer

To determine the maximum area:

Substitute the critical value of $\theta$ back into the area function $A$ .
We will find that at this specific value, the area reaches its peak.
Specifically, one can express this maximum area in terms of width $w$ : $A_{max} = \frac{w^2}{4},\n$ where this conforms to established geometric constructions and maximum principles.

A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle - HSC - SSCE Mathematics Extension 1 - Question 7 - 2006 - Paper 1

Worked Solution & Example Answer:A gutter is to be formed by bending a long rectangular metal strip of width $w$ so that the cross-section is an arc of a circle - HSC - SSCE Mathematics Extension 1 - Question 7 - 2006 - Paper 1

Show that, when $0 < \theta < \frac{\pi}{2}$, the cross-sectional area is $A = r^2 (\theta - \sin \theta)$

Let $g(\theta) = \sin \theta - \cos \theta$. By considering $g'(\theta)$, show that $g(\theta) > 0$ for $0 < \theta < \pi$.

Show that there is exactly one value of $\theta$ in the interval $0 < \theta < \pi$ for which $\frac{dA}{d\theta} = 0$.

Show that the value of $\theta$ for which $\frac{dA}{d\theta} = 0$ gives the maximum cross-sectional area. Find this area in terms of $w$.

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