Question

Let $x$ be the length of one of the equal sides of an isosceles triangle, and let $\theta$ be the angle between them. If $x$ is increasing at the rate $(1 / 12) m / hr$, and $\theta$ is increasing at the rate of $\pi / 180$ radians/hr then the rate in $m ^2 / hr$ at which the area of the triangle is increasing when $x =12 m$ and $\theta=\pi / 4$

• A. $2^{1 / 2}\left(1+\frac{2 \pi}{5}\right)$
• B. $\frac{73}{2} \cdot 2^{1 / 2}$
• C. $\frac{3^{1 / 2}}{2}+\frac{\pi}{5}$
• D. $2^{1 / 2}\left(\frac{1}{2}+\frac{\pi}{5}\right)$

Answer 1

Answer: (D)

$A=\frac{1}{2} x^2 \sin \theta \quad \Rightarrow \quad 2 A=x^2 \sin \theta$
$2 \frac{ dA }{ dt }= x ^2 \cos \theta \frac{ d \theta}{ dt }+\sin \theta 2 x \frac{ dx }{ dt }$
$2 \frac{ dA }{ dt }=(144)\left(\frac{1}{\sqrt{2}}\right) \frac{\pi}{180}+\frac{1}{\sqrt{2}} \cdot 2 \cdot 12 \cdot \frac{1}{12}=\frac{12 \pi}{15 \sqrt{2}}+\frac{2}{\sqrt{2}}$

$\frac{ dA }{ dt }=\frac{2 \pi}{5 \sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{\sqrt{2} \pi}{5}+\frac{\sqrt{2}}{2}=\sqrt{2}\left(\frac{\pi}{5}+\frac{1}{2}\right)$