A-Level Edexcel Maths: Pure Differentiation: Find the equation of the tangent

Find the equation of the tangent to the curve $x = ext{cos}(2y + heta)$ at $ \left( 0, \frac{\pi}{4} \right) $ - Edexcel - A-Level Maths: Pure - Question 5 - 2009 - Paper 2

Question 5

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Find the equation of the tangent to the curve $x = ext{cos}(2y + heta)$ at $ \left( 0, \frac{\pi}{4} \right) $. Give your answer in the form $y = ax + b$, where... show full transcript

Worked Solution & Example Answer:Find the equation of the tangent to the curve $x = ext{cos}(2y + heta)$ at $ \left( 0, \frac{\pi}{4} \right) $ - Edexcel - A-Level Maths: Pure - Question 5 - 2009 - Paper 2

Step 1

Differentiate with respect to y

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Answer

To find the slope of the tangent line, we first differentiate the given relationship with respect to $y$ :

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

Step 2

Use implicit differentiation to find dy/dx

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Answer

Next, we apply the chain rule to find the derivative $abla = \frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{1}{2 \sin(2y + \theta)}$

Step 3

Evaluate dy/dx at y = π/4

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Answer

Substituting $y = \frac{\pi}{4}$ into our derivative:

$\frac{dy}{dx} \bigg|_{y = \frac{\pi}{4}} = \frac{1}{2 \sin(\frac{\pi}{2} + \theta)} = \frac{1}{2}$

Step 4

Find the equation of the tangent line

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Answer

Now, using the point-slope form of a line:

$y - \frac{\pi}{4} = \frac{1}{2} (x - 0)$

Rearranging gives:

$y = \frac{1}{2}x + \frac{\pi}{4}$

Thus, we find: $a = \frac{1}{2}, b = \frac{\pi}{4}$

Find the equation of the tangent to the curve $x = ext{cos}(2y + heta)$ at \( \left( 0, \frac{\pi}{4} \right) \) - Edexcel - A-Level Maths: Pure - Question 5 - 2009 - Paper 2

Worked Solution & Example Answer:Find the equation of the tangent to the curve $x = ext{cos}(2y + heta)$ at \( \left( 0, \frac{\pi}{4} \right) \) - Edexcel - A-Level Maths: Pure - Question 5 - 2009 - Paper 2

Differentiate with respect to y

Use implicit differentiation to find dy/dx

Evaluate dy/dx at y = π/4

Find the equation of the tangent line

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