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 and the line $x = 3$, is rotated about the x-axis. (b) (i) Find the vertical and horizontal asymptotes of the hyperbola $y = \frac{x-2}{x-4}$ and hence sketch the graph of $y = \frac{x-2}{x-4}$. (ii) Hence, or otherwise, find the values of $x$ for which $\frac{x-2}{x-4} \leq 3$. (c) 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. (i) Show that $\dot{x} = e^{-2x}$. (ii) 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 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 and the line $x = 3$, is rotated abo... 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 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 formed when the region bounded by the curve $y = \frac{1}{\sqrt{9+x^2}}$, the x-axis and the line $x = 3$, is rotated about the x-axis.

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Answer

To find the volume of the solid of revolution, we can use the disk method. The volume $V$ can be calculated using the formula:

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

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

Thus,

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

Now, applying the integral:

$$V = \pi \left[ \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3} = \pi \left(\frac{1}{3} \tan^{-1}(1) - 0 \right) = \pi \left(\frac{\pi}{12}\right) = \frac{\pi^2}{12}.$

Step 2

Find the vertical and horizontal asymptotes of the hyperbola $y = \frac{x-2}{x-4}$ and hence sketch the graph of $y = \frac{x-2}{x-4}$.

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Answer

To find the vertical asymptote, we set the denominator to zero:

$x - 4 = 0 \Rightarrow x = 4$

For the horizontal asymptote, we can analyze the end behavior as $x \to \infty$ :

$y = \frac{x-2}{x-4} \approx \frac{x}{x} = 1 \quad \text{(as } x \to \infty)$

Thus, the horizontal asymptote is $y = 1$ .

To sketch the graph, we note:

The vertical asymptote at $x=4$ .
The horizontal asymptote at $y=1$ .

The graph approaches these asymptotes and will intersect the x-axis when $y=0$ , solving:

$$0 = \frac{x-2}{x-4} \Rightarrow x=2.$

Step 3

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

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Answer

We start by rewriting the inequality:

$\frac{x-2}{x-4} - 3 \leq 0$

This simplifies to:

$\frac{x-2 - 3(x-4)}{x-4} \leq 0$

which is:

$\frac{-2x + 10}{x-4} \leq 0.$

Next, we find the critical points by setting the numerator and denominator to zero:

Numerator: $-2x + 10 = 0 \Rightarrow x = 5$ .
Denominator: $x - 4 = 0 \Rightarrow x = 4$ .

We can test intervals around these points to determine the sign of the expression. The solution set is:

$x \leq 4 \text{ or } 5 \leq x$

Step 4

Show that $\dot{x} = e^{-2x}$.

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Answer

Given that $\dot{x} = -e^{-2x}$ , to show that $\dot{x} = e^{-2x}$ with the change of variables, we can integrate:

$\int \dot{x} \, dx = \int -e^{-2x} \, dx = \frac{1}{2} e^{-2x} + C$

At $x=0$ , the velocity is 1 (initial condition), thus setting the equation:

$1 = \frac{1}{2} e^{0} + C \Rightarrow C = \frac{1}{2}$

Therefore, replacing back gives us:

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

Step 5

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

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Answer

Starting with the equation:

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

Setting $t = \int \dot{x} \, dt$ leads to:

$\dot{x} = -e^{-2x} o dx = -e^{-2x} dt$

Integrating both sides allows us to express:

$\int dx = - \int e^{-2x} dt = -\frac{1}{2} e^{-2x}$

Thus,

Integrating leads to:

$x = -\frac{1}{2} e^{-2x} + C$

Solving for $x$ with respect to $t$ leads to: $x = \log_e(t + 1)$ . This result follows from rearranging the equations and applying logarithmic identities.

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 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 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 formed when the region bounded by the curve $y = \frac{1}{\sqrt{9+x^2}}$, the x-axis and the line $x = 3$, is rotated about the x-axis.

Find the vertical and horizontal asymptotes of the hyperbola $y = \frac{x-2}{x-4}$ and hence sketch the graph of $y = \frac{x-2}{x-4}$.

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

Show that $\dot{x} = e^{-2x}$.

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

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