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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 $y = ext{sin}(2x)$, 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

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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-$y-=--ext{sin}(2x)$,-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.png

5. (a) Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve $y = ext{sin}(2x)$, the x-axis and the line $x = \f... 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 $y = ext{sin}(2x)$, 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 $y = ext{sin}(2x)$, 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 the solid of revolution, we can use the formula for the volume of revolution around the x-axis:

V=πabf(x)2dxV = \pi \int_{a}^{b} f(x)^2 \, dx

In this case, we have:

  • f(x)=sin(2x)f(x) = \text{sin}(2x)
  • a=0a = 0
  • b=π8b = \frac{\pi}{8}.

Thus, the volume becomes:

V=π0π8sin2(2x)dxV = \pi \int_{0}^{\frac{\pi}{8}} \text{sin}^2(2x) \, dx

Using the identity sin2(x)=1cos(2x)2\text{sin}^2(x) = \frac{1 - \text{cos}(2x)}{2}, we can rewrite sin2(2x)\text{sin}^2(2x) as:

sin2(2x)=1cos(4x)2\text{sin}^2(2x) = \frac{1 - \text{cos}(4x)}{2}

Then, substituting this back into the volume equation, we have:

V=π0π81cos(4x)2dxV = \pi \int_{0}^{\frac{\pi}{8}} \frac{1 - \text{cos}(4x)}{2} \, dx

Now, calculate the integral:

=π20π8(1cos(4x))dx= \frac{\pi}{2} \int_{0}^{\frac{\pi}{8}} (1 - \text{cos}(4x)) \, dx

This evaluates to:

=π2[xsin(4x)4]0π8= \frac{\pi}{2} \left[ x - \frac{\text{sin}(4x)}{4} \right]_{0}^{\frac{\pi}{8}}

Substituting in the limits gives:

=π2(π80)=π216= \frac{\pi}{2} \left( \frac{\pi}{8} - 0 \right) = \frac{\pi^2}{16}

So, the volume of the solid of revolution is:

V=π216.V = \frac{\pi^2}{16}.

Step 2

Explain why DPAE is a cyclic quadrilateral.

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Answer

A cyclic quadrilateral is one where all vertices lie on the circumference of a circle. In quadrilateral DPAE, the opposite angles add up to 180 degrees. Specifically, we have:

  • DAP+DEP=180\angle DAP + \angle DEP = 180^\circ
  • DPE+AEP=180\angle DPE + \angle AEP = 180^\circ As such, DPAE satisfies the property of cyclic quadrilaterals.

Step 3

Prove that $\angle APE = \angle ABC.$

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Answer

In triangle ABE, angle ABE subtends the chord AE at points A and E. By the property of angles subtended by the same chord, we have: ABE=APE\angle ABE = \angle APE In triangle ABC, angle ABC also subtends the same chord BC at points B and C. Therefore, we can conclude that: APE=ABC.\angle APE = \angle ABC.

Step 4

Deduce that PQ is perpendicular to BC.

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Answer

Since APE=ABC\angle APE = \angle ABC and we know that both angles are equal, triangles APE and ABC are similar by AA criterion. Thus, if angle APE is equal to angle ABC and given that line segments ARE both perpendicular (due to the properties of cyclic quadrilaterals), we can deduce that PQ must be perpendicular to BC.

Step 5

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

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Answer

To convert 3sin(3t)\sqrt{3}\text{sin}(3t) into the form R\text{sin}(3t - \alpha), we identify R and \alpha such that: Rsin(3tα)=R(sin(3t)cos(α)cos(3t)sin(α))R\text{sin}(3t - \alpha) = R(\text{sin}(3t)\text{cos}(\alpha) - \text{cos}(3t)\text{sin}(\alpha)) Comparing coefficients:

  • Rcos(α)=3R\text{cos}(\alpha) = \sqrt{3}
  • Rsin(α)=1R\text{sin}(\alpha) = 1 To find R, use: R=(3)2+(1)2=2R = \sqrt{(\sqrt{3})^2 + (1)^2} = 2 Now to find \alpha: tan(α)=13α=π6\tan(\alpha) = \frac{1}{\sqrt{3}} \\ \Rightarrow \alpha = \frac{\pi}{6} \\ Thus, the expression becomes: 2sin(3tπ6).2\text{sin}(3t - \frac{\pi}{6}).

Step 6

The particle assumes simple harmonic motion. Find the amplitude and the circumference of the motion.

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Answer

For a particle exhibiting simple harmonic motion described by: x(t)=5+3sin(3t)cos(3t),x(t) = 5 + \sqrt{3}\text{sin}(3t) - \text{cos}(3t), The amplitude is given by the maximum value of the terms that multiply the sinusoidal functions. Hence, the amplitude can be calculated as: A=(3)2+(1)2=2.A = \sqrt{(\sqrt{3})^2 + (1)^2} = 2. The circumference of the motion is given by the formula: C=2πA=2π(2)=4π.C = 2\pi A = 2\pi (2) = 4\pi.

Step 7

When does the particle first reach its maximum speed?

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Answer

The speed of the particle is given by the derivative of position: v(t)=dxdtv(t) = \frac{dx}{dt} For maximum speed, set the derivative of speed equal to zero while observing max speed times: v(t)=33cos(3t)+3sin(3t)=0.v(t) = 3\sqrt{3}\text{cos}(3t) + 3\text{sin}(3t) = 0. Thus, when 3t3t is such that: 3t=kπ+π2,kZ3t = k\pi + \frac{\pi}{2}, k \in \mathbb{Z} From this, solving for tt yields: t=kπ+π23.t = \frac{k\pi + \frac{\pi}{2}}{3}. Thus the first occurrence happens at: t=π6t = \frac{\pi}{6}. This corresponds to the conditions set in parts (i) and (ii).

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