Two sides of a triangle are $2 \mathrm {~m}$ and $3 \mathrm {~m}$ in length and the angle between them is increasing at a rate of $0.06 \mathrm { rad } / \mathrm { s }$. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is $\frac { \pi } { 3 }$.
A) $1.145 \mathrm {~m} ^ { 2 } / \mathrm { s }$
B) $- 0.955 \mathrm {~m} ^ { 2 } / \mathrm { s }$
C) $0.090 \mathrm {~m} ^ { 2 } / \mathrm { s }$
D) $5.045 \mathrm {~m} ^ { 2 } / \mathrm { s }$
E) $- 1.955 \mathrm {~m} ^ { 2 } / \mathrm { s }$

Question

Quizplus · Accepted Answer

Let's label the triangle. 
[asy]
pair A,B,C;
A=(0,0);
B=(3,0);
C=(1,3);
draw(A--B--C--cycle);
draw(rightanglemark(C,A,B,4));
label("$2$",(A+B)/2,dir(-90));
label("$3$",(C+B)/2,dir(60));
[/asy]
And let's use the law of cosines to find the length of the third side $a$. $$a^2=2^2+3^2-2\cdot2\cdot3\cdot \cos	heta$$
$$a^2=13-12\cos	heta$$
Now the area is given by the formula $\frac{1}{2}ab\sin	heta$. (Note that this formula is only true when the angle is between the two given sides, which is the case here.) Taking the derivative of this with respect to time gives: $$\frac{dA}{dt}=\frac{1}{2}\left(\frac{da}{dt}\cdot b+a\cdot\frac{db}{dt}ight)\sin	heta+\frac{1}{2}ab\cos	heta\cdot\frac{d	heta}{dt}$$
Plugging in $a=\sqrt{13-12\cos	heta}$, $b=2$, $db/dt=0$, $d	heta/dt=0.06$, and $	heta=\pi/3$, we get: $$\frac{dA}{dt}=\frac{1}{2}\left(-3\sqrt{3}\sin\frac{\pi}{3}ight)(0)+\frac{1}{2}\left(\sqrt{13-12\cos(\pi/3)}ight)(0.06)\left(2ight)+\frac{1}{2}(2)\left(\sqrt{13-12\cos(\pi/3)}ight)\left(\frac{1}{2}ight)$$
Evaluating this gives $\frac{dA}{dt}=0.09	ext{ m}^2	ext{/s}$, which corresponds to choice $\boxed{	extbf{(C)}}$.

Two Sides of a Triangle Are 2 m2 \mathrm {~m}2 m And 3 m3 \mathrm {~m}3 m

Two Sides of a Triangle Are $2 \mathrm {~m}$ And $3 \mathrm {~m}$