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A curve C has parametric equations $x = ext{sin}^2 t, y = 2 an t, 0 < t < \frac{\pi}{2}$ (a) Find \( \frac{dy}{dx} \) in terms of t - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 6
Question 6
A curve C has parametric equations $x = ext{sin}^2 t, y = 2 an t, 0 < t < \frac{\pi}{2}$ (a) Find \( \frac{dy}{dx} \) in terms of t. The tangent to C at the... show full transcript
Worked Solution & Example Answer:A curve C has parametric equations $x = ext{sin}^2 t, y = 2 an t, 0 < t < \frac{\pi}{2}$ (a) Find \( \frac{dy}{dx} \) in terms of t - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 6
Step 1
Find \( \frac{dy}{dx} \) in terms of t.
Answer
To find ( \frac{dy}{dx} ) in terms of t, we first differentiate x and y with respect to t:
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Differentiate x: [ \frac{dx}{dt} = 2 \sin t \cos t = \sin(2t) ]
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Differentiate y: [ \frac{dy}{dt} = 2 \sec^2 t ]
Using the chain rule, we can express ( \frac{dy}{dx} ) as:
[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \sec^2 t}{\sin(2t)} ]
Step 2
Find the x-coordinate of P.
Answer
To find the x-coordinate of P, we need to evaluate at ( t = \frac{\pi}{3} ).
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First, substitute ( t = \frac{\pi}{3} ) into the derivative to find y: [ \frac{dy}{dx} \Big|_{t=\frac{\pi}{3}} = \frac{2 \sec^2(\frac{\pi}{3})}{\sin(\frac{2\pi}{3})} = \frac{2 \cdot 4/3}{\sqrt{3}/2} = \frac{16/3}{\sqrt{3}/2} = \frac{32}{3\sqrt{3}} ]
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Next, find the y-coordinate at this value of t: [ y = 2 \tan(\frac{\pi}{3}) = 2 \cdot \sqrt{3} ]
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Set y = 0 for the x-axis: [ 0 = 2 \tan t - \frac{16}{9} \Rightarrow 2 \tan(\frac{\pi}{3}) -\frac{16}{9} = 0 ]
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Finally, substitute t back to find the x-coordinate: [ x = \text{sin}^2(\frac{\pi}{3}) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} ]