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 $x=12m$ 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)

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

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