Answer to Question #155192 in Calculus for Phyroe

Question

The base of an isosceles triangle is 8 feet kong. If the altitude is 6 feet long and increasing 3 inches per minute, at what rate are the base angles changing?

Answer 1

Consider the isosceles triangle as shown in the figure below:

In the triangle,

$tan(\theta)=\frac{h}{4}$

Differentiate both sides with respect to $t$ as,

$\frac{d}{dt}tan(\theta)=\frac{1}{4} \frac{d}{dt}(h)$

$sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{4} \frac{dh}{dt}$

When $h=6,\frac{dh}{dt}=3$ and $sec^2(\theta)=1+tan^2(\theta)=1+(\frac{6}{4})^2=\frac{13}{4}$

Therefore,

$\frac{13}{4}\frac{d\theta}{dt}=\frac{1}{4}(3)$

$13\frac{d\theta}{dt}=3$

$\frac{d\theta}{dt}=\frac{3}{13}$

Hence, the base angles are changing at rate $\frac{d\theta}{dt}=\frac{3}{13}$ radian per minutes.