Chapter 5.6, Problem 6E

To determine

To relate: $\frac{d\text{A}}{dt}$ to $\frac{da}{dt}$ , $\frac{db}{dt}$ and $\frac{d\theta }{dt}$ .

Answer to Problem 6E

The value of $\frac{d\text{A}}{dt}$ is $\frac{1}{2}\left[b\sin \theta \frac{da}{dt}+a\sin \theta \frac{db}{dt}+ab\cos \theta \frac{d\theta }{dt}\right]$ .

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

  $\text{A}=\left(\frac{1}{2}\right)ab\sin \theta$

A = Area

Calculation :

We have to relate $\frac{d\text{A}}{dt}$ to $\frac{da}{dt}$ , $\frac{db}{dt}$ and $\frac{d\theta }{dt}$ .

Therefore,

  $\text{A}=\left(\frac{1}{2}\right)ab\sin \theta$

Since all variables are differentiable function of $t$

Differentiate with respect to t .

  $\frac{d\text{A}}{dt}=\frac{1}{2}\left[b\sin \theta \frac{da}{dt}+a\sin \theta \frac{db}{dt}+ab\cos \theta \frac{d\theta }{dt}\right]\left[\begin{array}{l}\text{Since }\frac{d}{dt}\sin \theta \\ =\cos \theta \frac{d\theta }{dt}\end{array}\right]$

Hence, the value of $\frac{d\text{A}}{dt}$ is $\frac{1}{2}\left[b\sin \theta \frac{da}{dt}+a\sin \theta \frac{db}{dt}+ab\cos \theta \frac{d\theta }{dt}\right]$ .