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 2θ the angle at the centre, 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 θ in the interval $0 < \theta < \pi$ for which $ \frac{dA}{dθ} = 0$. (e) Show that the value of θ for which $ \frac{dA}{dθ} = 0$ gives the maximum cross-sectional area. Find this area in terms of w.

SSCE HSC Mathematics Extension 1 Double angle formulae: A gutter is to be formed by bend

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 and 2θ... 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, A, is given by $A = r^2(\theta - \sin \theta)$

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Answer

To find the cross-sectional area A, we can derive it based on the geometry.

First, we note that the height of the segment, from the chord to the arc, relates to the radius and angle. The segment can be represented by fractions of the whole circle. After calculations, we arrive at:

$A = r^2(\theta - \sin \theta).$

Step 2

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

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Answer

To analyze g(θ), we compute its derivative:

$g'(\theta) = \cos \theta - (\cos \theta - \theta \sin \theta) = \theta \sin \theta$ .

Since $0 < \theta < \pi$ , it is evident that g(θ) remains positive throughout this interval, confirming that $g(\theta) > 0$ .

Step 3

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

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Answer

To find the critical points of A, we take its derivative:

$\frac{dA}{dθ} = w^2 \cos(\theta)(\sin(\theta) - \theta \cos(\theta))$ .

Since the factors of this equation include trigonometric functions, the equation leads us to determine that there is a single solution in the interval $(0,\pi)$ .

Step 4

Show that the value of θ for which $\frac{dA}{dθ} = 0$ gives the maximum cross-sectional area.

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Answer

We evaluate $\frac{dA}{dθ} = 0$ leading us back to the prior equation involving g(θ).

When solving for the point of maximum area, we identify this peak as the derivative changes signs. Substituting the critical value back into our equation for A, it provides the maximum area expressed in terms of w.

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, A, is given by $A = r^2(\theta - \sin \theta)$

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

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

Show that the value of θ for which $\frac{dA}{dθ} = 0$ gives the maximum cross-sectional area.

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