The curve C has equation $$2x^2y + 2x + 4y - ext{cos}(πy) = 17$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y. The point P with coordinates $ \left( 3, \frac{1}{2} \right) $ lies on C. The normal to C at P meets the x-axis at the point A. (b) Find the x coordinate of A, giving your answer in the form $ \frac{ax + b}{cx + d} $, where a, b, c and d are integers to be determined.

A-Level Edexcel Maths Pure Exponential & Logarithms: The curve C has equation $$2x^2

The curve C has equation $$2x^2y + 2x + 4y - ext{cos}(πy) = 17$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 4

Question 4

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The curve C has equation $$2x^2y + 2x + 4y - ext{cos}(πy) = 17$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y. The point P wi... show full transcript

Worked Solution & Example Answer:The curve C has equation $$2x^2y + 2x + 4y - ext{cos}(πy) = 17$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 4

Step 1

Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y.

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Answer

To differentiate the equation implicitly, we start with:

$2x^2y + 2x + 4y - \text{cos}(\pi y) = 17$

Differentiate both sides with respect to (x): [ \frac{d}{dx}(2x^2y) + \frac{d}{dx}(2x) + \frac{d}{dx}(4y) - \frac{d}{dx}(\text{cos}(\pi y)) = 0 ]
Applying the product and chain rules, we have: [ 2y \cdot \frac{d}{dx}(x^2) + 2x \cdot \frac{dy}{dx} + 4 \cdot \frac{dy}{dx} - (-\sin(\pi y) \cdot \pi \cdot \frac{dy}{dx}) = 0 ]
This simplifies to: [ 2x^2 \frac{dy}{dx} + 4 \frac{dy}{dx} + \pi \sin(\pi y) \frac{dy}{dx} = -2y - 2 ]
Factor out ( \frac{dy}{dx} ): [ \frac{dy}{dx} (2x^2 + 4 + \pi \sin(\pi y)) = -2y - 2 ]
Thus, we solve for ( \frac{dy}{dx} ): [ \frac{dy}{dx} = \frac{-2y - 2}{2x^2 + 4 + \pi \sin(\pi y)} ]

Step 2

Find the x coordinate of A, giving your answer in the form $ \frac{ax + b}{cx + d} $.

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Answer

We first substitute the point ( P \left( 3, \frac{1}{2} \right) ) into our expression for ( \frac{dy}{dx} ):

Substitute into the normal's slope: [ m_{n} = -\frac{dy}{dx} \text{ at } P = -\frac{-2 \cdot \frac{1}{2} - 2}{2 \cdot 3^2 + 4 + \pi \sin(\frac{\pi}{2})} = \frac{1 - 2}{18 + 4 + \pi} = \frac{-1}{22 + \pi} ]
The equation of the normal line can be written as: [ y - \frac{1}{2} = m_{n}(x - 3) ]
Set ( y = 0 ) to find the intersection with the x-axis: [ 0 - \frac{1}{2} = \frac{-1}{22 + \pi}(x - 3) ]
Rearranging gives: [ x - 3 = -\frac{1}{22 + \pi}(-\frac{1}{2}) ]
Solving for x, we get: [ x = 3 + \frac{1.5}{22 + \pi} ]
Formatting this into the required form: [ \frac{\text{numerator}}{\text{denominator}} = \frac{3(22 + \pi) + 1.5}{22 + \pi} = \frac{66 + 3\pi + 1.5}{22 + \pi} ]

Therefore, we can express the final answer as desired.

The curve C has equation $$2x^2y + 2x + 4y - ext{cos}(πy) = 17$$ (a) Use implicit differentiation to find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 4

Worked Solution & Example Answer:The curve C has equation $$2x^2y + 2x + 4y - ext{cos}(πy) = 17$$ (a) Use implicit differentiation to find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 4

Use implicit differentiation to find \( \frac{dy}{dx} \) in terms of x and y.

Find the x coordinate of A, giving your answer in the form \( \frac{ax + b}{cx + d} \).

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