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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Differentiation

NDA 2015 · Mathematics · Differentiation

Consider the parametric equation x=(a(1-t^2))/(1+t^2), y=(2 a...

Consider the parametric equation x=a(1−t2)1+t2,y=2at1+t2x=\frac{a\left(1-t^2\right)}{1+t^2}, y=\frac{2 a t}{1+t^2} What is d2ydx2\frac{d^2 y}{d x^2} equal to?

Options

  1. Aa2y2\frac{a^2}{y^2}
  2. Ba2x2\frac{a^2}{x^2}
  3. C−a2x2-\frac{a^2}{x^2}
  4. D−a2y3-\frac{a^2}{y^3}

Correct answer

D. −a2y3-\frac{a^2}{y^3}

Step-by-step solution

d2ydx2=yddx(−x)−(−x)dydxy2\frac{d^2 y}{d x^2}=\frac{y \frac{d}{d x}(-x)-(-x) \frac{d y}{d x}}{y^2}=−y+xdydxy2=−y+x(−xy)y2=−y2−x2y3=−(x2+y2)y3=−a2y3\begin{aligned} & =\frac{-y+x \frac{d y}{d x}}{y^2}=\frac{-y+x\left(\frac{-x}{y}\right)}{y^2} \\ & =\frac{-y^2-x^2}{y^3}=\frac{-\left(x^2+y^2\right)}{y^3}=-\frac{a^2}{y^3}\end{aligned}∴\therefore \quad Option (D) is correct.
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