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Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 17

Answers (1)

Answer:

$\frac{d y}{d x}=\frac{x}{y}$

Hint:

Use $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given:

$\begin{aligned} &x=a\left(t+\frac{1}{t}\right) \\ &y=a\left(t-\frac{1}{t}\right) \end{aligned}$

Solution:

$x=a\left(t+\frac{1}{t}\right) \\$

$\begin{aligned} & &\frac{d x}{d t}=a \frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}$

$\begin{aligned} &=a\left[\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right] \\ &=a\left[1+\frac{d(t)^{-1}}{d t}\right] \end{aligned}$
$=a\left[1+\left\{(-1) t^{-1-1}\right\}\right] \\$ $\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$

$\begin{aligned} & &\frac{d x}{d t}=a\left(1-t^{2}\right) \end{aligned}$ (1)

$y=a\left(t-\frac{1}{t}\right) \\$

$\frac{d y}{d t}=a \frac{d\left(t-\frac{1}{t}\right)}{d t} \\$

$\begin{aligned} & &=a\left[\frac{d t}{d t}-\frac{d\left(\frac{1}{t}\right)}{d t}\right] \end{aligned}$

$=a\left[1-\frac{d t^{-1}}{d t}\right] \\$

$=a\left[1-(-1) t^{-1-1}\right] \\$

$\begin{aligned} & &\frac{d y}{d t}=a\left(1+t^{-2}\right) \end{aligned}$ (2)

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Put the values of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively

$\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1} \\$

$\begin{aligned} & &=\frac{t\left(t+\frac{1}{t}\right)}{t\left(t-\frac{1}{t}\right)} \end{aligned}$

$=\frac{\left(t+\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)} \\$

$\begin{aligned} & &=\frac{a\left(t+\frac{1}{t}\right)}{a\left(t-\frac{1}{t}\right)} \end{aligned}$ [Multiply and divide by a]

$=\frac{x}{y}$ $\left[\because x=a\left(t+\frac{1}{t}\right) \& y=a\left(t-\frac{1}{t}\right)\right]$

Hence proved

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