Find the volume of the solid of revolution formed when the region bounded by the curve $y = \frac{1}{\sqrt{9 + x^2}}$, the x-axis, the y-axis and the line $x = 3$, is rotated about the x-axis. (i) Find the vertical and horizontal asymptotes of the hyperbola $y = \frac{x - 2}{x - 4}$ and hence sketch the graph of $y$. (ii) Hence, or otherwise, find the values of $x$ for which $\frac{x - 2}{x - 4} \leq 3$. A particle is moving in a straight line with its acceleration as a function of $x$ given by $\dot{x} = -e^{-2x}$. It is initially at the origin and is travelling with a velocity of 1 meter per second. Show that $\dot{x} = e^{-x}$. Hence show that $x = \log_e(t + 1)$.

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Find the volume of the solid of revolution formed when the region bounded by the curve $y = \frac{1}{\sqrt{9 + x^2}}$, the x-axis, the y-axis and the line $x = 3$, is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 3 - 2007 - Paper 1

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Find the volume of the solid of revolution formed when the region bounded by the curve $y = \frac{1}{\sqrt{9 + x^2}}$, the x-axis, the y-axis and the line $x = 3$, i... show full transcript

Worked Solution & Example Answer:Find the volume of the solid of revolution formed when the region bounded by the curve $y = \frac{1}{\sqrt{9 + x^2}}$, the x-axis, the y-axis and the line $x = 3$, is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 3 - 2007 - Paper 1

Step 1

Find the volume of the solid of revolution

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Answer

To find the volume of the solid generated by rotating the region about the x-axis, we use the formula for volume of revolution:

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

Here, we have:

The function $f(x) = \frac{1}{\sqrt{9 + x^2}}$ .
The limits of integration are from $x = 0$ to $x = 3$ .

Calculating the volume:

$V = \pi \int_{0}^{3} \left(\frac{1}{\sqrt{9 + x^2}}\right)^2 dx = \pi \int_{0}^{3} \frac{1}{9 + x^2} dx$

Using the integral formula: $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

Thus, we have:

$V = \pi \left[\frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3}$

After evaluation, we find: $V = \pi \left(\frac{1}{3} \cdot \frac{\pi}{4} - 0\right) = \frac{\pi^2}{12}$

Step 2

Find the vertical and horizontal asymptotes of the hyperbola

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Answer

To find the vertical and horizontal asymptotes of the hyperbola given by:

$y = \frac{x - 2}{x - 4}$

Vertical Asymptote: Set the denominator to zero: $x - 4 = 0 \Rightarrow x = 4$
Horizontal Asymptote: Evaluate the limit as $x$ approaches infinity: $\lim_{x \to \infty} \frac{x - 2}{x - 4} = 1$

Therefore, the asymptotes are:

Vertical: $x = 4$
Horizontal: $y = 1$

Now, to sketch the graph:

The graph approaches $y = 1$ as $x \to \infty$ and has a vertical asymptote at $x = 4$ . The curve will be in quadrants I and III.

Step 3

Hence, or otherwise, find the values of x for which \frac{x - 2}{x - 4} \leq 3

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Answer

To find the values of $x$ for which:

$\frac{x - 2}{x - 4} \leq 3$

We rearrange the inequality:

$x - 2 \leq 3(x - 4)$

This simplifies to: $x - 2 \leq 3x - 12$

Rearranging gives: $-2 + 12 \leq 2x \Rightarrow 10 \leq 2x \Rightarrow x \geq 5$

Now we check intervals. The important points are $x = 4$ (vertical asymptote) and $x = 5$ :

For $x < 4$ , the function is not defined.
For $4 < x < 5$ , $(x - 2)/(x - 4) < 3$ .
For $x = 5$ , the left side equals 3.
For $x > 5$ , the inequality holds.

Thus, the solutions are: $x \in [5, \infty)$

Step 4

Show that \dot{x} = e^{-x}

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Answer

Given the acceleration function:

$\dot{x} = -e^{-2x}$

Using the chain rule, we can express it in terms of velocity:

Integrating: $\dot{x} = \frac{dx}{dt} = -e^{-2x}$

Separate variables: $e^{2x} dx = -dt$

Integrate both sides: $\frac{1}{2} e^{2x} = -t + C$

Using the initial condition ( $x = 0$ when $t = 0$ , thus $C = \frac{1}{2}$ ), we get: $e^{2x} = 2 - 2t$

Taking the square root gives us: $\dot{x} = e^{-x}$ .

Step 5

Hence show that x = \log_e(t + 1)

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Answer

From the earlier derived equation:

$e^{2x} = 2 - 2t$

This leads to: $x = \frac{1}{2} \log_e(2 - 2t)$

Now using $frac{1}{2}$ for easier manipulation, we want to relate this to the velocity:

Remember that $frac{1}{2}(t + 1)$ implies: $x = \log_e(1 + t)$

This result stems from the exponential relationships and rearranging our expressions accordingly.

Find the volume of the solid of revolution formed when the region bounded by the curve $y = \frac{1}{\sqrt{9 + x^2}}$, the x-axis, the y-axis and the line $x = 3$, is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 3 - 2007 - Paper 1

Worked Solution & Example Answer:Find the volume of the solid of revolution formed when the region bounded by the curve $y = \frac{1}{\sqrt{9 + x^2}}$, the x-axis, the y-axis and the line $x = 3$, is rotated about the x-axis - HSC - SSCE Mathematics Extension 1 - Question 3 - 2007 - Paper 1

Find the volume of the solid of revolution

Find the vertical and horizontal asymptotes of the hyperbola

Hence, or otherwise, find the values of x for which \frac{x - 2}{x - 4} \leq 3

Show that \dot{x} = e^{-x}

Hence show that x = \log_e(t + 1)

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