Chapter 5.6, Problem 40E

To determine

To find: the vertical velocity $\frac{dy}{dt}$ at that point.

Answer to Problem 40E

The vertical velocity is C- $-\text{2}\text{.25}$

Explanation of Solution

Given information :

  $\frac{dx}{dt}=3$

  $x=0.6$

  $y=0.8$

Assume all variables are differentiable function of t .

Calculation:

We have to find the vertical velocity $\frac{dy}{dt}$ at that point.

The equation which defines circle is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where the centre point is $\left(a,b\right)$ and the radius of the circle is r .

The unit circle is the circle which has a centre point $\left(0,0\right)$ and the radius is 1.

Therefore,

  $x^2+y^2=1$

Differentiate with respect to t .

  $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

Putting the value of $x=0.6$ , $y=0.8$ and $\frac{dx}{dt}=3$ in above equation,

  $\begin{array}{l}2\left(0.6\right)\left(3\right)+2\left(0.8\right)\frac{dy}{dt}=0\\ 3.6+1.6\frac{dy}{dt}=0\\ 1.6\frac{dy}{dt}=-3.6\\ \frac{dy}{dt}=\frac{-3.6}{1.6}\\ \frac{dy}{dt}=-2.25\end{array}$

Hence, the vertical velocity is C- $-\text{2}\text{.25}$