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

The volume of the solid can be calculated using the formula for the volume of solids of revolution. The area of the equilateral triangle cross-section can be expressed in terms of the height, which is determined by the parabola's equation.

The length of the side of the equilateral triangle at height $y$ is given by the distance between the parabola and the $y$ -axis:

$x = 1 - y^2.$ Thus, the side length $s = 1 - y^2$ .
The area $A$ of the equilateral triangle is given by:

$A = \frac{s^2 \sqrt{3}}{4} = \frac{(1 - y^2)^2 \sqrt{3}}{4}.$
To find the volume, we integrate the area with respect to $y$ from the lower limit $y = -1$ to the upper limit $y = 1$ :

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

$V = \frac{\sqrt{3}}{4} \int_{-1}^{1} (1 - 2y^2 + y^4) \, dy = \frac{\sqrt{3}}{4} \left[ y - \frac{2}{3}y^3 + \frac{1}{5}y^5 \right]_{-1}^{1} = \frac{\sqrt{3}}{4} \left( 2 - \frac{2}{3}(2) + \frac{2}{5}(2) \right) = \frac{4\sqrt{3}}{15}.$

Step 2

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

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Answer

To find ( \frac{dy}{dx} ) using implicit differentiation, we start with the equation:

$x^2 + xy + y^2 = 3.$

Differentiating both sides with respect to $x$ gives:

$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0.$

Rearranging leads to:

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

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 derived for ( \frac{dy}{dx} ), to find where this equals zero, set the numerator equal to zero:

$-2x + y = 0 \Rightarrow y = 2x.$

Substitute $y = 2x$ back into the original curve equation:

$x^2 + x(2x) + (2x)^2 = 3 \Rightarrow x^2 + 2x^2 + 4x^2 = 3 \Rightarrow 7x^2 = 3 \Rightarrow x^2 = \frac{3}{7} \Rightarrow x = \pm \sqrt{\frac{3}{7}}.$

Therefore, the points are:

$x = \sqrt{\frac{3}{7}} \Rightarrow y = 2\sqrt{\frac{3}{7}} \text{ and } x = -\sqrt{\frac{3}{7}} \Rightarrow y = -2\sqrt{\frac{3}{7}}.$

The coordinates are ( \left( \sqrt{\frac{3}{7}}, 2\sqrt{\frac{3}{7}} \right) \text{ and } \left( -\sqrt{\frac{3}{7}}, -2\sqrt{\frac{3}{7}} \right). \

Step 4

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

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Answer

To evaluate the integral:

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

we can use polynomial long division or rewrite the integrand:

Rewrite it as:

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

The first integral evaluates to:

$x.$

For the second integral, complete the square in the denominator:

$x^2 + 2x + 5 = (x + 1)^2 + 4.$

Thus,

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

Collecting the results:

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

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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