5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis. (b) Two chords of a circle, AB and CD, intersect at E. The perpendiculars to AB at A and CD at D intersect at P. The line PE meets BC at Q, as shown in the diagram. (i) Explain why DPAE is a cyclic quadrilateral. (ii) Prove that \angle APE = \angle ABC. (iii) Deduce that PQ is perpendicular to BC. (c) A particle moves in a straight line and its position at time t is given by $x = 5 + \sqrt{3}\sin 3t - \cos 3t$. (i) Express $\sqrt{3}\sin 3t - \cos 3t$ in the form R sin(3t - α), where α is in radians. (ii) Determine the particle's simple harmonic motion. Find the amplitude and the period of the motion. (iii) When the particle first reaches its maximum speed although some were unsure about using the condition $v = 0$, calculate when this occurs in terms of t.

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5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 5 - 2005 - Paper 1

Question 5

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5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\p... show full transcript

Worked Solution & Example Answer:5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 5 - 2005 - Paper 1

Step 1

Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis.

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Answer

To find the volume of revolution, we use the formula:

$V = \pi \int_{{a}}^{{b}} [f(x)]^2 \, dx$

In this case, $f(x) = \sin 2x$ and we integrate from $0$ to $\frac{\pi}{8}$ :

$V = \pi \int_0^{\frac{\pi}{8}} [\sin 2x]^2 \, dx$

Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ , we rewrite the integrand:

$V = \pi \int_0^{\frac{\pi}{8}} \frac{1 - \cos(4x)}{2} \, dx$

Now we can split the integral:

$V = \frac{\pi}{2} \int_0^{\frac{\pi}{8}} 1 \, dx - \frac{\pi}{2} \int_0^{\frac{\pi}{8}} \cos(4x) \, dx$

Evaluating the first integral:

$= \frac{\pi}{2} \left[ x \right]_0^{\frac{\pi}{8}} = \frac{\pi^2}{16}$

For the second integral:

$= -\frac{\pi}{8} [\sin(4x)]_0^{\frac{\pi}{8}} = -\frac{\pi}{8} (\sin(\frac{\pi}{2}) - \sin(0)) = -\frac{\pi}{8} imes 1 = -\frac{\pi}{8}$

Thus:

$V = \frac{\pi}{16} - \left(-\frac{\pi}{8}\right) = \frac{\pi}{16} + \frac{2\pi}{16} = \frac{3\pi}{16}$

Therefore, the exact value of the volume is:

$V = \frac{3\pi}{16}$

Step 2

Explain why DPAE is a cyclic quadrilateral.

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Answer

A quadrilateral is cyclic if the opposite angles are supplementary. In this case, since angle DPA and angle DEA subtend the same arc DE, they are supplementary. Thus, DPAE is a cyclic quadrilateral.

Step 3

Prove that \angle APE = \angle ABC.

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Answer

In cyclic quadrilaterals, opposite angles are equal. We have: $\angle APE + \angle AED = 180^\circ$ Since angle AED subtends the same arc AB as angle ABC, we have: $\angle AED = \angle ABC.$ Thus, $\angle APE = \angle ABC.$ This fulfills the proof.

Step 4

Deduce that PQ is perpendicular to BC.

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Answer

Since we have established that\angle APE = \angle ABC $, we can extend this reasoning. Given that$ \angle APE + \angle APQ = 90^\circ$, and that angles APE and ABC create angles that sum to 90 degrees, it follows that PQ must be perpendicular to BC.

Step 5

Express $\sqrt{3}\sin 3t - \cos 3t$ in the form R sin(3t - α), where α is in radians.

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Answer

To express this in the form $R \sin(3t - \alpha)$ , we can use the identity:

$R \sin(\theta) = R \sin(3t) \cos(\alpha) - R \cos(3t) \sin(\alpha)$

Setting up the equations:

$\sqrt{3} = R \cos(\alpha)$
$-1 = -R \sin(\alpha)$

From the second equation: $R \sin(\alpha) = 1$

Squaring both equations gives: $3 = R^2 \cos^2(\alpha)\quad (1)$ $1 = R^2 \sin^2(\alpha)\quad (2)$

Adding (1) and (2):

$4 = R^2\ \\ R = 2$

To find alpha:

From $\sqrt{3} = 2 \cos(\alpha) \rightarrow \cos(\alpha) = \frac{\sqrt{3}}{2} \rightarrow \alpha = \frac{\pi}{6}$

So, we get the transformation: $\sqrt{3}\sin 3t - \cos 3t = 2\sin(3t - \frac{\pi}{6})$ .

Step 6

Determine the particle's simple harmonic motion. Find the amplitude and the period of the motion.

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Answer

From the expression $x = 5 + 2 \sin(3t - \frac{\pi}{6})$ , we can see that the coefficient of sine represents the amplitude, so the amplitude is:

$A = 2.$

The angular frequency $\omega = 3$ , thus the period $T$ is given by:

$T = \frac{2\pi}{\omega} = \frac{2\pi}{3}.$

Step 7

When the particle first reaches its maximum speed although some were unsure about using the condition v = 0.

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Answer

To find when the particle reaches maximum speed, we first find the velocity:

$v = \frac{dx}{dt} = 3\cos(3t - \frac{\pi}{6}).$

Setting $v = 0$ ,

$3 \cos(3t - \frac{\pi}{6}) = 0 \Rightarrow \cos(3t - \frac{\pi}{6}) = 0.$

This occurs at:

where k\in \mathbb{Z}.$$ Solving for $t$ gives: $$t = \frac{\pi/2 + \pi/6 + k\pi}{3} = \frac{\pi/3 + k\pi}{3}.$$ To find when the first maximum speed occurs: $$t = \frac{\pi}{9} \text{ for } k=0.$$

5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 5 - 2005 - Paper 1

Worked Solution & Example Answer:5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 5 - 2005 - Paper 1

Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = \sin 2x$, the x-axis and the line $x = \frac{\pi}{8}$ is rotated about the x-axis.

Explain why DPAE is a cyclic quadrilateral.

Prove that \angle APE = \angle ABC.

Deduce that PQ is perpendicular to BC.

Express $\sqrt{3}\sin 3t - \cos 3t$ in the form R sin(3t - α), where α is in radians.

Determine the particle's simple harmonic motion. Find the amplitude and the period of the motion.

When the particle first reaches its maximum speed although some were unsure about using the condition v = 0.

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