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 8 - 2007 - Paper 7

Evaluate $x$ and $y$ at $t = \frac{\pi}{4}$:

• $x = \tan^2 \frac{\pi}{4} = 1$
• $y = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.

Find $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$:

• Substitute into the expression:
  $\frac{dy}{dx} = \frac{\cos^3 \frac{\pi}{4}}{2 \sin \frac{\pi}{4}} = \frac{(\frac{\sqrt{2}}{2})^3}{2 \cdot \frac{\sqrt{2}}{2}} = \frac{1}{2}.$

Use the point-slope form of the equation:

• The slope is $\frac{1}{2}$ and point is $(1, \frac{\sqrt{2}}{2})$:
• $y - \frac{\sqrt{2}}{2} = \frac{1}{2}(x - 1)$
• This rearranges to: $y = \frac{1}{2}x + \left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right).$

From the parametric equations, substitute $x = \tan^2 t$:

• We have $y = \sin t$.
• To eliminate $t$, we express $\sin t$ in terms of $x$:
• Since $\tan t = \sqrt{x}$, we can use the identity:
  $1 + \tan^2 t = \sec^2 t \implies 1 + x = \sec^2 t.$

Now, using the relation $\sin^2 t + \cos^2 t = 1$:

• We find $\sin^2 t = 1 - \cos^2 t$.
• Since $\cos^2 t = \frac{1}{\sec^2 t}$, we substitute:
  $\sin^2 t = 1 - \frac{1}{1+x} = \frac{x}{1+x}.$

Therefore, we express $y^2$ in terms of $x$:

• The Cartesian equation is:
  $y^2 = \frac{x}{1+x}.$