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A curve has parametric equations $x = 2 \, ext{cot} \, t, \quad y = 2 \, ext{sin}^2 t, \quad 0 < t < \frac{\pi}{2}$ (a) Find an expression for \( \frac{dy}{dx} \) in terms of the parameter \( t \) - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 6
Question 8
A curve has parametric equations $x = 2 \, ext{cot} \, t, \quad y = 2 \, ext{sin}^2 t, \quad 0 < t < \frac{\pi}{2}$ (a) Find an expression for \( \frac{dy}{dx} \... show full transcript
Worked Solution & Example Answer:A curve has parametric equations $x = 2 \, ext{cot} \, t, \quad y = 2 \, ext{sin}^2 t, \quad 0 < t < \frac{\pi}{2}$ (a) Find an expression for \( \frac{dy}{dx} \) in terms of the parameter \( t \) - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 6
Step 1
Find an expression for \( \frac{dy}{dx} \) in terms of the parameter \( t \)
Answer
To find ( \frac{dy}{dx} ), we need to use the chain rule for parametric equations:
[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} ]
First, we compute ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ):
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For ( y = 2 \sin^2 t ): [ \frac{dy}{dt} = 2 \cdot 2 \sin t \cdot \cos t = 4 \sin t \cos t ]
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For ( x = 2 \cot t ): [ \frac{dx}{dt} = 2 \cdot (-\csc^2 t) = -2 \csc^2 t ]
Now, substituting these into the expression:
[ \frac{dy}{dx} = \frac{4 \sin t \cos t}{-2 \csc^2 t} = -2 \sin t \cos t \cdot \sin^2 t = -2 \frac{\sin^3 t}{\cos t} ]
Step 2
Find an equation of the tangent to the curve at the point where \( t = \frac{\pi}{4} \)
Answer
At ( t = \frac{\pi}{4} ):
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Calculate ( x ) and ( y ): [ x = 2 \cot \left( \frac{\pi}{4} \right) = 2 \cdot 1 = 2 ] [ y = 2 \sin^2 \left( \frac{\pi}{4} \right) = 2 \cdot \left( \frac{\sqrt{2}}{2} \right)^2 = 1 ]
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Now, find ( \frac{dy}{dx} ) at ( t = \frac{\pi}{4} ): [ \frac{dy}{dx} = -2 \frac{\sin^3 \left( \frac{\pi}{4} \right)}{\cos \left( \frac{\pi}{4} \right)} = -2 \frac{\left( \frac{\sqrt{2}}{2} \right)^3}{\frac{\sqrt{2}}{2}} = -1 ]
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The equation of the tangent line is given by: [ y - y_1 = m(x - x_1) ] Where ( (x_1, y_1) = (2, 1) ) and ( m = -1 ): [ y - 1 = -1(x - 2) ] Rearranging this gives: [ y = -x + 3 ]
Step 3
Find a cartesian equation of the curve in the form \( y = f(x) \). State the domain on which the curve is defined
Answer
From the parametric equations, we have:
[ x = 2 \cot t \Rightarrow \cot t = \frac{x}{2} \Rightarrow t = \cot^{-1} \left( \frac{x}{2} \right) ]
Substituting ( t ) in the equation for ( y ):
[ y = 2 \sin^2 t = 2 \cdot \sin^2 \left( \cot^{-1} \left( \frac{x}{2} \right) \right) ]
Using the identity: [ \sin^2 t = \frac{1}{1 + \cot^2 t} \Rightarrow \sin^2 \left( \cot^{-1} \left( \frac{x}{2} \right) \right) = \frac{1}{1 + \left( \frac{x}{2} \right)^2} = \frac{4}{4 + x^2} ]
Thus, the cartesian equation becomes:
[ y = 2 \cdot \frac{4}{4 + x^2} = \frac{8}{4 + x^2} ]
Domain: Since ( t ) varies from 0 to ( \frac{\pi}{2} ), and ( \cot t ) ranges from ( +\infty ) to 0, the domain of ( x ) is from 0 to ( +\infty ).