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A curve has parametric equations $x = \tan^2 t, \quad y = \sin t, \quad 0 < t < \frac{\pi}{2}.$ (a) Find an expression for $\frac{dy}{dx}$ in terms of $t$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 7
Question 7
A curve has parametric equations $x = \tan^2 t, \quad y = \sin t, \quad 0 < t < \frac{\pi}{2}.$ (a) Find an expression for $\frac{dy}{dx}$ in terms of $t$. You nee... show full transcript
Worked Solution & Example Answer:A curve has parametric equations $x = \tan^2 t, \quad y = \sin t, \quad 0 < t < \frac{\pi}{2}.$ (a) Find an expression for $\frac{dy}{dx}$ in terms of $t$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 7
Step 1
Find an expression for \( \frac{dy}{dx} \) in terms of \( t \)
Answer
To find ( \frac{dy}{dx} ), we use the chain rule:
[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} ]
First, we calculate ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ):
[ x = \tan^2 t \Rightarrow \frac{dx}{dt} = 2\tan t \sec^2 t ]
[ y = \sin t \Rightarrow \frac{dy}{dt} = \cos t ]
Now, we find ( \frac{dt}{dx} ):
[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2\tan t \sec^2 t} ]
Consequently, substituting gives:
[ \frac{dy}{dx} = \frac{\cos t}{2 \tan t \sec^2 t} = \frac{\cos^3 t}{2 \sin t} ]
Step 2
Find an equation of the tangent to the curve at the point where \( t = \frac{\pi}{4} \)
Answer
First, we substitute ( t = \frac{\pi}{4} ) into the equations for ( x ) and ( y ):
[ x = \tan^2\left(\frac{\pi}{4}\right) = 1, \quad y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} ]
Next, we compute ( \frac{dy}{dx} ) at this point:
[ \frac{dy}{dx} = \frac{\cos^3\left(\frac{\pi}{4}\right)}{2 \sin\left(\frac{\pi}{4}\right)} = \frac{\left(\frac{\sqrt{2}}{2}\right)^3}{2 \left(\frac{\sqrt{2}}{2}\right)} = \frac{\frac{2\sqrt{2}}{8}}{\sqrt{2}} = \frac{1}{4} ]
Now, using the point-slope form to find the tangent equation:
The equation is given by ( y - y_1 = m (x - x_1) )
Substituting the values ( (1, \frac{\sqrt{2}}{2}) ) and ( m = \frac{1}{4} ):
[ y - \frac{\sqrt{2}}{2} = \frac{1}{4} (x - 1) ]
Rearranging gives ( y = \frac{1}{4}x + b \quad (b = \frac{\sqrt{2}}{2} - \frac{1}{4}) ]
Step 3
Find a Cartesian equation of the curve in the form \( y^2 = f(x) \)
Answer
Starting from the parametric equations:
Using the identity ( 1 + \tan^2 t = \sec^2 t ), we can relate ( x ) and ( y ):
From ( x = \tan^2 t ), we have ( \tan t = \sqrt{x} ).
Using ( y = \sin t ) and the identity:
[ y^2 = \sin^2 t = 1 - \cos^2 t ]
To express ( \cos t ) in terms of ( x ):
[ \sec^2 t = 1 + \tan^2 t = 1 + x \Rightarrow \cos^2 t = \frac{1}{1+x} ]
Therefore, we can derive:
[ y^2 = 1 - \frac{1}{1+x} = \frac{x}{1+x} ]
This leads to the final form:
[ y^2 = \frac{x}{1+x} ]