Two sides of a triangle are 15cm and 20cm long respectively.
$A)$ How fast is the third side increasing if the angle between the given sidesis 60 degrees and is increasing at the rate of $2^\circ/sec$? $B)$ How fast is the area increasing?
$A)$ I used $c^2=a^2+b^2-2ab\cos(\theta)$ so I got the missing side $c=28.72$ is this right? and then I get confused with the implicit differentiation where, $2c$$\frac{dc}{dt}$ $= $2a$\frac{da}{dt}$ $+$ $2b$$\frac{db}{dt}$ $-$ $2\cos(\theta)$ ($a$$\frac{db}{dt}$+$b$$\frac{da}{db})$ I know that 60degrees is $\frac{Pi}{3}$ to radiance but I keep on getting the wrong answer, they said it's supposed to be $\frac{dc}{dt}=0.5$ I don't know what to substitute with $\frac{da}{dt}$ and $\frac{db}{dt}$ is it the $2^\circ/sec$ ? I'm so confused.
$B)$ I also get confuse here on what formula to use.
Best Answer
In part a), $\theta$ is the only variable mentioned as changing.
So $\dfrac{d\theta}{dt} = 2^{\circ}/\text{sec} = \dfrac{\pi}{90}\text{rad/sec}$ and $\dfrac{da}{dt} = \dfrac{db}{dt} = 0 \text{cm/sec}$.
Can you show what you plugged into the formula to get $c \approx 28.72$? I got $c = 5\sqrt{13} \approx 18.03$.
Also, when you differentiated $c^2 = a^2+b^2-2ab\cos\theta$ you should get $2c\dfrac{dc}{dt} = 2ab\sin\theta \dfrac{d\theta}{dt}$
In part b), note that the area of a triangle is $A = \dfrac{1}{2}ab\sin\theta$.