# Question #1fa6b

Feb 6, 2017

Without knowing what the said angle is, we can only get a formula for answering this question.

#### Explanation:

We are interested in the Area of a right triangle, $A$ and one of the acute angles, $\theta$.

We are told that $\frac{d \theta}{\mathrm{dt}} = \frac{\pi}{16}$ rad / s and we are asked to find $\frac{\mathrm{dA}}{\mathrm{dt}}$ when $\theta$ is _ _ _ . (not given).

We need an equation relating $A$ and $\theta$.

Sketch a right triangle with hypotenuse $40$ cm and base angle $\theta$

The area of a triangle is $\frac{1}{2} \text{base" xx "height}$.

In my picture the base is the side adjacent to $\theta$, so its length is $40 \cos \theta$. The height is $40 \sin \theta$.

The area is given by $A = 800 \sin \theta \cos \theta$ ${\text{cm}}^{2}$.

We now need to differentiate with respect to $t$. Rather than using the product rule for an implicit differentiation, let's rewrite the function first using $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$

$A = 400 \sin \left(2 \theta\right)$

Now differentiate w.r.t. $t$.

$\frac{\mathrm{dA}}{\mathrm{dt}} = 800 \cos \left(2 \theta\right) \frac{d \theta}{\mathrm{dt}}$

We were given $\frac{d \theta}{\mathrm{dt}}$, so we can write

$\frac{\mathrm{dA}}{\mathrm{dt}} = 50 \pi \cos \left(2 \theta\right)$

Now, if we are given a $\theta$, we can find the rate of change of the area.