Question

If $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2at}{1 + t^2}$, then $\frac{dy}{dx} =$

• A. $\frac{a(t^2 + 1)}{2t}$
• B. $\frac{a(t^2 - 1)}{t}$
• C. $\frac{a(1 - t^2)}{2t}$
• D. $\frac{a(t^2 - 1)}{2t}$

Hint:

Use trigonometric substitution to bypass the quotient rule entirely! Let $t = \tan\theta$.
Then $x = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos(2\theta)$ and $y = \frac{2a\tan\theta}{1 + \tan^2\theta} = a\sin(2\theta)$.
Differentiating with respect to $\theta$: $\frac{dx}{d\theta} = -2\sin(2\theta)$ and $\frac{dy}{d\theta} = 2a\cos(2\theta)$.
Thus, $\frac{dy}{dx} = \frac{2a\cos(2\theta)}{-2\sin(2\theta)} = -a\cot(2\theta) = -a \left( \frac{1 - \tan^2\theta}{2\tan\theta} \right) = \frac{a(\tan^2\theta - 1)}{2\tan\theta} = \frac{a(t^2 - 1)}{2t}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Use trigonometric substitution to avoid the quotient rule when differentiating parametric equations involving $t$.

Step 2: Key Formula or Approach:
Substitute $t = \tan\theta$, transforming the given rational expressions into simpler trigonometric functions using double-angle identities: $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$. Then differentiate with respect to $\theta$ and apply the chain rule $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

Step 3: Detailed Explanation:
Setting $t = \tan\theta$ converts $x = \frac{1-t^2}{1+t^2} = \cos(2\theta)$ and $y = \frac{2at}{1+t^2} = a\sin(2\theta)$. Differentiating each with respect to $\theta$: $\frac{dx}{d\theta} = -2\sin(2\theta)$ and $\frac{dy}{d\theta} = 2a\cos(2\theta)$. Therefore, $\frac{dy}{dx} = \frac{2a\cos(2\theta)}{-2\sin(2\theta)} = -a\cot(2\theta) = -a\left(\frac{1-\tan^2\theta}{2\tan\theta}\right) = \frac{a(t^2-1)}{2t}$.

Step 4: Final Answer:
The derivative simplifies neatly to $\frac{a(t^2-1)}{2t}$ without ever invoking the quotient rule.