The base of an isosceles triangle is 10 feet long. If the altitude is decreasing at a rate of 2
inches per second, at what rate is the base angle changing when the height is 12 feet?
So far I am confused as to what I should be finding the derivative of and what values I should plug in afterwards.
Best Answer
Draw the picture. You'll see a right triangle with an "adjacent" side that is $5$ feet long (half of the base) and a varying height $x$, which is the "opposite" side. Let $\theta$ be the "base angle", so that $$ \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac x 5. $$ Then $$ \theta = \arctan \frac x 5. $$ You need $$ \frac{d\theta}{dt} = \frac{d\theta}{dx}\cdot\frac{dx}{dt} $$ at a time when $x=12$ and $\dfrac{dx}{dt}=\left( -\dfrac 2 {12} \text{ feet per second} \right) = (-2 \text{ inches per second}).$