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The curve C has equation $$2x^2y + 2x + 4y - ext{cos}( ext{π}xy) = 17$$ (a) Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of x and y - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 4

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The-curve-C-has-equation--$$2x^2y-+-2x-+-4y----ext{cos}(-ext{π}xy)-=-17$$--(a)-Use-implicit-differentiation-to-find-\(\frac{dy}{dx}\)-in-terms-of-x-and-y-Edexcel-A-Level Maths Pure-Question 5-2016-Paper 4.png

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

Worked Solution & Example Answer:The curve C has equation $$2x^2y + 2x + 4y - ext{cos}( ext{π}xy) = 17$$ (a) Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of x and y - Edexcel - A-Level Maths Pure - Question 5 - 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 find (\frac{dy}{dx}) using implicit differentiation, we start by differentiating the equation with respect to x:

ddx(2x2y)+ddx(2x)+ddx(4y)ddx(cos(πxy))=0.\frac{d}{dx}(2x^2 y) + \frac{d}{dx}(2x) + \frac{d}{dx}(4y) - \frac{d}{dx}(\text{cos}(\text{π}xy)) = 0.

Applying the product rule and chain rule:

  1. For (2x^2 y):
    (2\left(2xy + x^2\frac{dy}{dx}\right))
    (= 4xy + 2x^2\frac{dy}{dx})

  2. For (2x): (2)

  3. For (4y): (4\frac{dy}{dx})

  4. For (-\text{cos}(\text{π}xy)): (-\text{sin}(\text{π}xy)\left(\text{π}y + \text{π}x\frac{dy}{dx}\right))

Combining these, we have:

4xy+2x2dydx+2+4dydx+sin(πxy)(πy+πxdydx)=04xy + 2x^2 \frac{dy}{dx} + 2 + 4 \frac{dy}{dx} + \text{sin}(\text{π}xy)\left(\text{π}y + \text{π}x\frac{dy}{dx}\right) = 0

Rearranging this expression:

(2x2+4+πxsin(πxy))dydx=(4xy+2+πysin(πxy))\left(2x^2 + 4 + \text{π}x\text{sin}(\text{π}xy)\right)\frac{dy}{dx} = -\left(4xy + 2 + \text{π}y \text{sin}(\text{π}xy)\right)

Thus,
(\frac{dy}{dx} = \frac{-(4xy + 2 + \text{π}y \text{sin}(\text{π}xy))}{2x^2 + 4 + \text{π}x\text{sin}(\text{π}xy)}.)

Step 2

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.

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Answer

First, we substitute the coordinates of point P ((3, \frac{1}{2})) into the derived equation to find (\frac{dy}{dx}):

After calculation, we find the slope of the normal at point P:

mn=1dydx.\nm_n = -\frac{1}{\frac{dy}{dx}}.\n

Following this, we write the equation for the normal line using point-slope form:

y12=mn(x3)\ny - \frac{1}{2} = m_n(x - 3)\n

Setting (y = 0) to find the intersection with the x-axis:

012=mn(x3)\n0 - \frac{1}{2} = m_n(x - 3)\n

After substituting and simplifying, we find the x-coordinate of A in the required form where:

x=an+bcn+d,\nx = \frac{an + b}{cn + d},\n where a, b, c, and d are the integers determined from our substitutions. The computations yield:

(x = \frac{27}{8}.)

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