Chapter 5.6, Problem 2QQ

To determine

To find: the value of $\frac{dx}{dt}$ .

Answer to Problem 2QQ

The value of $\frac{dx}{dt}$ is B − 1.

Explanation of Solution

Given information:

  $\frac{dz}{dt}=1$ and $\frac{dx}{dt}=3\frac{dy}{dt}$ .

  $x=4$ and $y=3$

We have to calculate the value of $\frac{dx}{dt}$ .

From Pythagoras theorem,

  $x^2+y^2=z^2$

Differentiating with respect to $t$ , we get

  $2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}\mathrm{.....}\left(1\right)$

From right angle triangle,

  $y=3$ $z$ 

  $x=4$

We know that,

  $\begin{array}{l}z=\sqrt{x^2+y^2}\\ z=\sqrt{4^2+3^2}\\ z=\sqrt{16+9}\\ z=\sqrt{25}\\ z=5\end{array}$

Put the value of $z=5$ , $x=4$ , $y=3$ , $\frac{dz}{dt}=1$ and $\frac{dx}{dt}=3\frac{dy}{dt}$ in equation (1).

  $\begin{array}{l}2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}\\ 2\left(4\right)\left(3\frac{dy}{dt}\right)+2\left(3\right)\frac{dy}{dt}=2\left(5\right)\left(1\right)\\ 24\frac{dy}{dt}+6\frac{dy}{dt}=10\\ 30\frac{dy}{dt}=10\\ \frac{dy}{dt}=\frac{1}{3}\end{array}$

Put the value of $\frac{dy}{dt}=\frac{1}{3}$ in $\frac{dx}{dt}=3\frac{dy}{dt}$ .

Therefore,

  $\begin{array}{l}\frac{dx}{dt}=3\left(\frac{1}{3}\right)\\ \frac{dx}{dt}=1\end{array}$

Hence, the value of $\frac{dx}{dt}$ is 1.