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 start by determining the area of the equivalent equilateral triangle cross-section. Given that the base of the triangle at a height $y$ is the distance from the $y$ -axis to the parabola, we have:

$x = 1 - y^2$

Thus, the base of the triangle can be expressed as: $b = 1 - y^2$

Using the properties of equilateral triangles, the height $h$ can be calculated as: $h = \frac{\sqrt{3}}{2} b = \frac{\sqrt{3}}{2} (1 - y^2)$

The area $A$ of the cross-section is given by: $A = \frac{1}{2} b h = \frac{1}{2} (1 - y^2) \left( \frac{\sqrt{3}}{2} (1 - y^2) \right) = \frac{\sqrt{3}}{4} (1 - y^2)^2$

Now, we compute the volume by integrating the area with respect to $y$ from $-1$ to $1$ :

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

This gives us:

Expand $(1 - y^2)^2$ : $(1 - y^2)^2 = 1 - 2y^2 + y^4$
Therefore: $V = \frac{\sqrt{3}}{4} \int_{-1}^{1} (1 - 2y^2 + y^4) \, dy$
Evaluate the integral: $V = \frac{\sqrt{3}}{4} \left[ y - \frac{2y^3}{3} + \frac{y^5}{5} \right]_{-1}^{1} = \frac{\sqrt{3}}{4} \left( 2 - 0 \right) = \frac{\sqrt{3}}{2}$

Step 2

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

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Answer

Differentiating both sides of the equation $x^2 + xy + y^2 = 3$ with respect to $x$ gives:

Differentiate $x^2$ : $\frac{d}{dx}(x^2) = 2x$
Differentiate $xy$ using the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$
Differentiate $y^2$ : $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$

Putting this together, we have: $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

Rearranging terms, we find: $x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$

Factoring out (\frac{dy}{dx}): $\frac{dy}{dx}(x + 2y) = -2x - y$

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

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 expression we derived for ( \frac{dy}{dx} ), setting it equal to zero gives: $-2x - y = 0$

This implies: $y = -2x$

Now, substituting this into the original curve equation $x^2 + xy + y^2 = 3$ :

Substitute $y = -2x$ : $x^2 + x(-2x) + (-2x)^2 = 3$ Simplifying yields: $x^2 - 2x^2 + 4x^2 = 3\implies 3x^2 = 3$ Thus, ( x^2 = 1 ) leading to: $x = 1 \text{ or } x = -1$
Calculate $y$ for both $x$ values:
- If $x = 1$ , then ( y = -2(1) = -2 ).
- If $x = -1$ , then ( y = -2(-1) = 2 ).

Thus, the coordinates 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

To evaluate the integral, first simplify the integrand:

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

This results in: $\int 1 \, dx - 5 \int \frac{1}{x^2 + 2x + 5} \, dx$

The first integral is straightforward: $x$

Now for the second integral, complete the square in the denominator: $x^2 + 2x + 5 = (x + 1)^2 + 4$

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

Combining results, we have: $\int \frac{x^2 + 2x}{x^2 + 2x + 5} \, dx = x - \frac{5}{2} \tan^{-1}\left( \frac{x + 1}{2} \right) + C$

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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