Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find dy/dx of each of the functions expressed in parametric form x = 1+ logt / t^2

Find dy/dx of each of the functions expressed in parametric form

\mathrm{x}=\frac{1+\log \mathrm{t}}{\mathrm{t}^{2}}, \mathrm{y}=\frac{3+2 \log \mathrm{t}}{\mathrm{t}}

Answers (1)

We have given, \mathrm{x}=\frac{1+\log \mathrm{t}}{\mathrm{t}^{2}}, \mathrm{y}=\frac{3+2 \log \mathrm{t}}{\mathrm{t}}

Now, differentiate w.r.t t

\\ \frac{d x}{d t}=\frac{t^{2} \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t}\left(t^{2}\right)}{\left(t^{2}\right)^{2}} \\ \frac{d x}{d t}=\frac{t^{2} \cdot \frac{1}{t}-(1+\log t) \cdot 2 t}{t^{4}} \\ \frac{d x}{d t}=\frac{t-(1+\log t) \cdot 2 t}{t^{4}} \\ \frac{d x}{d t}=\frac{-1-2 \log t}{t^{3}}

Also,

\\ \frac{d y}{d t}=\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t}(t)}{t^{2}} \\ \frac{d y}{d t}=\frac{t \cdot \frac{2}{t}-(3+2 \log t)}{t^{2}} \\ \frac{d y}{d t}=\frac{-1-2 \log t}{t^{2}} \\ \frac{d y}{d t}=\frac{-1-2 \log t}{t^{2}} \quad \ldots(i)

\\$Now$, \\\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}$ \\$\frac{d y}{d x}=\frac{-1-2 \log t}{t^{2}} \times \frac{t^{3}}{-1-2 \log t}$ \\Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=t

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question