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 $\left(\frac{1}{12}\right)m/hr,$ and $\theta$ is increasing at the rate of $\frac{\pi }{180}rad/hr,$ then the rate in$\frac{m^2}{hr}$ at which the area of the triangle is increasing when $x=12m$ and $\theta =\frac{\pi }{4},$ is

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

Answer 1

The correct option is D $2^{1/2}(\frac{1}{2}+\frac{\pi }{5})$


$A=\frac{1}{2}{x}^2\sin \theta \Rightarrow 2A=x^2\sin \theta$
$\Rightarrow 2\frac{dA}{dt}=x^2\cos \theta \frac{d\theta }{dt}+\sin \theta 2x\frac{dx}{dt}$
At $\theta =\frac{\pi }{4},x=12$
$\Rightarrow 2\frac{dA}{dt}=\left(144\right)\left(\frac{1}{\sqrt{2}}\right)\frac{\pi }{180}+\frac{1}{\sqrt{2}}\cdot 2\cdot 12\cdot \frac{1}{12}=\frac{12\pi }{15\sqrt{2}}+\frac{2}{\sqrt{2}}$
$\frac{dA}{dt}=\frac{2\pi }{5\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}\pi }{5}+\frac{\sqrt{2}}{2}$
$=\sqrt{2}(\frac{\pi }{5}+\frac{1}{2})$