A-Level Edexcel Maths Pure Equation of a Straight Line: The curve C has equation $$16y^

The curve C has equation $$16y^3 + 9x^2y^2 - 54x = 0$$ (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Question 6

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The curve C has equation $$16y^3 + 9x^2y^2 - 54x = 0$$ (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$. (b) Find the coordinates of the points on C where \... show full transcript

Worked Solution & Example Answer:The curve C has equation $$16y^3 + 9x^2y^2 - 54x = 0$$ (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Step 1

Find $\frac{dy}{dx}$ in terms of $x$ and $y$

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Answer

To find (\frac{dy}{dx}), we differentiate the equation implicitly with respect to (x):

$16(3y^2 \frac{dy}{dx}) + 9(2xy^2 + x^2(2y \frac{dy}{dx})) - 54 = 0$

This simplifies to:

$48y^2 \frac{dy}{dx} + 18xy^2 + 9x^2(2y \frac{dy}{dx}) - 54 = 0$

Rearranging gives:

$\left(48y^2 + 18x^2y \right) \frac{dy}{dx} = 54 - 18xy^2$

Thus,

$\frac{dy}{dx} = \frac{54 - 18xy^2}{48y^2 + 18x^2y}$

Step 2

Find the coordinates of the points on C where $\frac{dy}{dx} = 0$

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Answer

To find where (\frac{dy}{dx} = 0), we set the numerator equal to zero:

$54 - 18xy^2 = 0$

Solving for (y^2) gives:

$18xy^2 = 54 \Rightarrow y^2 = \frac{54}{18x} = \frac{3}{x}$

Now using the original equation:

$16y^3 + 9x^2y^2 - 54x = 0$

Substituting (y^2 = \frac{3}{x}), we deduce the values of (x):

$16 \left( \frac{3}{x} \right)^{3/2} + 9x^2 \left( \frac{3}{x} \right) - 54x = 0$

This leads to values of (x = 3) or (x = -2). Now solving for (y) gives:

For (x = 3:\n\quad y^2 = 1 \Rightarrow y = \pm 1\n\Rightarrow (3, 1), (3, -1)

For (x = -2:\n\quad y^2 = -\frac{3}{2} (Not possible) \Rightarrow (\pm2, \mp 3)\n\Rightarrow \text{These points are} (2, \frac{3}{2}), (2, -\frac{3}{2})\n$

The curve C has equation $$16y^3 + 9x^2y^2 - 54x = 0$$ (a) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Worked Solution & Example Answer:The curve C has equation $$16y^3 + 9x^2y^2 - 54x = 0$$ (a) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\)

Find the coordinates of the points on C where \(\frac{dy}{dx} = 0\)

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