A-Level Edexcel Maths Pure Integration: The curve C has equation $$16 y

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

Question 7

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

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

Differentiate both sides: $\frac{d}{dx}(16 y^3) + \frac{d}{dx}(9 x^2 y) - \frac{d}{dx}(54 x) = 0$
Apply the product and chain rules:
- For (16 y^3): (48 y^2 \frac{dy}{dx})
- For (9 x^2 y): Using the product rule: (9(2x y + x^2 \frac{dy}{dx}))
- For (54 x): Differentiating gives (54)
The differentiated equation becomes: $48 y^2 \frac{dy}{dx} + 9(2 x y + x^2 \frac{dy}{dx}) - 54 = 0$
Rearranging gives: $(48 y^2 + 9 x^2) \frac{dy}{dx} = 54 - 18 x y$
Finally, isolate ( \frac{dy}{dx} ): $\frac{dy}{dx} = \frac{54 - 18 x y}{48 y^2 + 9 x^2}$

Step 2

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

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Answer

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

Solve: $54 - 18 x y = 0$ This gives: $18 x y = 54 \Longrightarrow xy = 3$
Substitute (y = \frac{3}{x}) into the original equation: $16 \left(\frac{3}{x}\right)^3 + 9 x^2 \left(\frac{3}{x}\right) - 54 x = 0$
This simplifies to: $\frac{432}{x^3} + 27 - 54 x = 0$ Multiplying through by (x^3) to eliminate the fraction: $432 + 27 x^3 - 54 x^4 = 0$ Rearranging gives: $54 x^4 - 27 x^3 - 432 = 0\Longrightarrow 2 x^4 - x^3 - 16 = 0$
Testing possible rational roots leads us to find:
- For (x = 2): (y = \frac{3}{2}) yielding point ((2, \frac{3}{2}))
- For (x = -2): (y = -\frac{3}{2}) yielding point ((-2, -\frac{3}{2}))
Therefore, the coordinates of the points on C where ( \frac{dy}{dx} = 0 ) are: $\left(2, \frac{3}{2}\right) \quad \text{and} \quad \left(-2, -\frac{3}{2}\right).$

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

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