The base of a solid is the region enclosed by the parabola $x = 1 - y^2$ and the $y$-axis. Each cross-section perpendicular to the $y$-axis is an equilateral triangle, as shown in the diagram. Find the volume of the solid. A curve has equation $x^2 + xy + y^2 = 3$. (i) Use implicit differentiation to show that $ \frac{dy}{dx} = \frac{2x+y}{x+2y} $. (ii) Hence, or otherwise, find the coordinates of the points on the curve where $ \frac{dy}{dx} = 0 $. (c) Find $ \int\frac{x^2 + 2x}{x^2 + 2x + 5}dx $.

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The base of a solid is the region enclosed by the parabola $x = 1 - y^2$ and the $y$-axis - HSC - SSCE Mathematics Extension 2 - Question 12 - 2018 - Paper 1

Question 12

The base of a solid is the region enclosed by the parabola $x = 1 - y^2$ and the $y$-axis. Each cross-section perpendicular to the $y$-axis is an equilateral triangl... show full transcript

Worked Solution & Example Answer:The base of a solid is the region enclosed by the parabola $x = 1 - y^2$ and the $y$-axis - HSC - SSCE Mathematics Extension 2 - Question 12 - 2018 - Paper 1

Step 1

Find the volume of the solid

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Answer

To find the volume of the solid, we first need to determine the area of a typical cross-section. The cross-section is an equilateral triangle with each side corresponding to a distance between the curve and the $y$ -axis.

The height of each triangle can be expressed as $h = 1 - y^2$ .

The area of an equilateral triangle is given by:

$A = \frac{\sqrt{3}}{4} s^2$

where $s$ is the side length. For our case, the side length $s$ is equal to the height $h$ , thus:

$A = \frac{\sqrt{3}}{4} (1 - y^2)^2.$

Now, we need to determine the bounds for $y$ . Since the parabola intersects the $y$ -axis at $y = -1$ and $y = 1$ , the volume can be found by integrating the area from $y = -1$ to $y = 1$ :

$V = \int_{-1}^{1} A \, dy = \int_{-1}^{1} \frac{\sqrt{3}}{4} (1 - y^2)^2 \, dy.$

Calculating the integral:

Expand $(1 - y^2)^2 = 1 - 2y^2 + y^4$ .
The integral becomes: $V = \frac{\sqrt{3}}{4} \int_{-1}^{1} (1 - 2y^2 + y^4) \, dy.$
Calculate each part:
- $\int_{-1}^{1} 1 \, dy = 2$ ,
- $\int_{-1}^{1} 2y^2 \, dy = \frac{4}{3}$ ,
- $\int_{-1}^{1} y^4 \, dy = \frac{2}{5}$ .
Substitute back: $V = \frac{\sqrt{3}}{4} \left( 2 - \frac{4}{3} + \frac{2}{5} \right) = \frac{\sqrt{3}}{4} \left( \frac{30 - 20 + 8}{15} \right) = \frac{\sqrt{3}}{4} \left( \frac{18}{15} \right) = \frac{3\sqrt{3}}{10}.$

Step 2

Use implicit differentiation to show that $ \frac{dy}{dx} = \frac{2x+y}{x+2y} $

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Answer

Starting with the equation of the curve: $x^2 + xy + y^2 = 3.$

Differentiating both sides with respect to $x$ , we have:

differentiating gives: $\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(3),$ which results in: $2x + \left( x \frac{dy}{dx} + y \right) + 2y \frac{dy}{dx} = 0.$

Rearranging, we isolate ( \frac{dy}{dx} ) as follows: $\frac{dy}{dx} (x + 2y) = -2x - y.$

Thus, $$\frac{dy}{dx} = \frac{-2x - y}{x + 2y} = \frac{2x + y}{x + 2y}.$

Step 3

Hence, or otherwise, find the coordinates of the points on the curve where $ \frac{dy}{dx} = 0 $

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Answer

From the result of part (i), we find ( \frac{dy}{dx} = 0 ) when the numerator equals zero: $2x + y = 0 \Rightarrow y = -2x.$

Now, we substitute $y = -2x$ back into the original equation of the curve: $x^2 + x(-2x) + (-2x)^2 = 3.$

This simplifies to: $x^2 - 2x^2 + 4x^2 = 3 \Rightarrow 3x^2 = 3 \Rightarrow x^2 = 1 \Rightarrow x = \\pm 1.$

Substituting these values back, we get the corresponding $y$ values:

For $x = 1$ , $y = -2(1) = -2$ .
For $x = -1$ , $y = -2(-1) = 2.$

Thus, the points on the curve where ( \frac{dy}{dx} = 0 ) are $(1, -2)$ and $(-1, 2)$ .

Step 4

Find $ \int \frac{x^2 + 2x}{x^2 + 2x + 5}dx $

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Answer

We rewrite the integral as: $\int \frac{x^2 + 2x}{x^2 + 2x + 5}dx = \int \frac{x^2 + 2x + 5 - 5}{x^2 + 2x + 5}dx = \int \left( 1 - \frac{5}{x^2 + 2x + 5} \right) dx.$

This can be separated into two integrals: $\int dx - 5 \int \frac{1}{x^2 + 2x + 5}dx.$

For the first part, we have: $\int dx = x.$

For the second part, note that: $x^2 + 2x + 5 = (x + 1)^2 + 4.$

Thus, $5 \int \frac{1}{(x + 1)^2 + 4}dx = \frac{5}{2} \tan^{-1}\left( \frac{x + 1}{2} \right) + C.$

Combining both parts yields the final result: $\int \frac{x^2 + 2x}{x^2 + 2x + 5}dx = x - \frac{5}{2} \tan^{-1}\left( \frac{x + 1}{2} \right) + C.$

The base of a solid is the region enclosed by the parabola $x = 1 - y^2$ and the $y$-axis - HSC - SSCE Mathematics Extension 2 - Question 12 - 2018 - Paper 1

Worked Solution & Example Answer:The base of a solid is the region enclosed by the parabola $x = 1 - y^2$ and the $y$-axis - HSC - SSCE Mathematics Extension 2 - Question 12 - 2018 - Paper 1

Find the volume of the solid

Use implicit differentiation to show that \( \frac{dy}{dx} = \frac{2x+y}{x+2y} \)

Hence, or otherwise, find the coordinates of the points on the curve where \( \frac{dy}{dx} = 0 \)

Find \( \int \frac{x^2 + 2x}{x^2 + 2x + 5}dx \)

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