Question

The base of an isosceles triangle 10 feet long and the base angles are decreasing at a rate of 2 degress per second. Find the rate of change of the area when the base angles are 45 degrees?

Answer 1

Here in the figure BC=10ft. And we assume $\angle B=\angle C =\theta$ (cause isosceles triangle).

Also BD=DC=5ft (cause AD is the perpendicular bisector of BC).

Now, in $\triangle ABD$ , using trigonometry:-

$$\tan(\angle ABD)=\frac{AD}{BD}\\~\\ \Rightarrow\tan\theta=\frac{AD}{5}\\~\\ \Rightarrow AD=5\tan\theta$$

So, now we calculate the area of the $\triangle ABC$ :-

$$\triangle=\frac{1}{2}\cdot5\tan\theta\cdot10\\~\\ \Rightarrow\triangle=25\tan\theta\\$$

Now, as we're dealing with rate of change of different stuff we differentiate both sides wrt $t$ :-

$$\frac{d\triangle}{dt}=25\sec^2\theta\frac{d\theta}{dt}\\~\\ \Rightarrow \frac{d\triangle}{dt}=25\cdot(\sqrt{2})^2\cdot2\\~\\ \Rightarrow \frac{d\triangle}{dt}=100$$

We, put $\theta=45^\degree$ , and use the rate of change of $\theta$ given in the question.

So, rate of change of area is $100$ sq ft per second.