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A curve has parametric equations $x = 2 \, ext{cot} \, t, \, y = 2 \, ext{sin}^2 t, \, 0 < t < \frac{\pi}{2}$ (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 $x = 2 \, ext{cot} \, t, \, y = 2 \, ext{sin}^2 t, \, 0 < t < \frac{\pi}{2}$ (a) Find an expression for \( \frac{dy}{dx} \) in t... show full transcript
Worked Solution & Example Answer:A curve has parametric equations $x = 2 \, ext{cot} \, t, \, y = 2 \, ext{sin}^2 t, \, 0 < t < \frac{\pi}{2}$ (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 use the chain rule:
[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} ]
Calculating ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ):
[ \frac{dy}{dt} = 4 \sin t \cos t \quad \text{(using the chain rule on } y = 2\sin^2 t \text{)} ] [ \frac{dx}{dt} = -2 \csc^2 t \quad \text{(using the chain rule on } x = 2\cot t \text{)} ]
Now substitute into the expression for ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = \frac{4 \sin t \cos t}{-2\csc^2 t} = -2 \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
First, we need to find the coordinates of the point at ( t = \frac{\pi}{4} ):
[ x = 2 \cot \left( \frac{\pi}{4} \right) = 2 \quad \text{and} \quad y = 2 \sin^2 \left( \frac{\pi}{4} \right) = 2 \cdot \left( \frac{\sqrt{2}}{2} \right)^2 = 1 ]
The point is ( (2, 1) ).
Now we find ( \frac{dy}{dx} ) at ( t = \frac{\pi}{4} ):
[ \frac{dy}{dx} = -2 \sin^3 \left( \frac{\pi}{4} \right) \cos \left( \frac{\pi}{4} \right) = -2 \cdot \left( \frac{\sqrt{2}}{2} \right)^3 \cdot \left( \frac{\sqrt{2}}{2} \right) = -1 ]
Using the point-slope form of the line:
[ y - y_1 = m (x - x_1) ]
Substituting in our values:
[ y - 1 = -1 (x - 2) ]
This simplifies to:
[ 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
We start with the parametric equations:
[ x = 2 \cot t \quad \text{and} \quad y = 2 \sin^2 t ]
From the first equation, we can express ( \cot t ) in terms of ( x ):
[ \cot t = \frac{x}{2} \quad \Rightarrow \quad 1 + \cot^2 t = 1 + \left( \frac{x}{2} \right)^2 ]
Using the Pythagorean identity for sine and cosine:
[ \sin^2 t + \cot^2 t \cdot \sin^2 t = 1 \quad \Rightarrow \quad \sin^2 t = \frac{1}{1 + \left( \frac{x}{2} \right)^2} ]
Substituting this into ( y = 2 \sin^2 t ):
[ y = 2 \cdot \frac{1}{1 + \left( \frac{x}{2} \right)^2} = \frac{4}{x^2 + 4} ]
Thus, the Cartesian equation is:
[ y = \frac{4}{x^2 + 4} ]
The domain of the curve is ( x \in (-\infty, \infty) ).