Find rate of change of the area of a triangle at a particular moment

Question

The side lengths of a right-angled triangle are $a, b$ and $c$ centimeters where $c$ is the length of the hypotenuse and $a, b$ and $c$ are changing as differentiable functions of time. It is known that the perimeter $a+b+c$ is always constant and at the moment $a=3$ and $b=4$ the rate of change of $a$ is $108$ cm/sec. What is the rate of change of the area of the triangle (in cm$^2$/sec units) at this moment?

This is a related rates question (the answer is $72$). What I tried and first of all I calculated the derivative of the area of a right triangle ($ab/2$) and everything is known except the derivative of side $b$. I need to find some relation between $b$ and $a$ but I am not sure how.

Answer 1

$\frac{d a}{dt} + \frac{d b}{dt} + \frac{d c}{dt} = 0$ ...(i)

$c^2 = a^2 + b^2$ and that gives you

$c \frac{d c}{dt} = a \frac{d a}{dt} + b \frac{d b}{dt}$

Now when $a = 3, b = 4$, $c = 5$

$5 \frac{d c}{dt} = 3 \frac{d a}{dt} + 4 \frac{d b}{dt}$ ...(ii)

From (i) and (ii), you can find $\frac{d b}{dt} = -96$ as you know $\frac{d a}{dt} = 108$.

Now $\frac{d A}{dt} = \frac{1}{2} \big(a \frac{d b}{dt} + b \frac{d a}{dt} \big)$

Answer 2

Hint

$$c^2(t)=a^2(t)+b^2(t)\to c(t)c'(t)=a(t)a'(t)+b(t)b'(t)\quad (1)$$ and $$a(t)+b(t)+c(t)=k=cte\to a'(t)+b'(t)+c'(t)=0.\quad (2)$$

The area is given by

$$A(t)=\frac{a(t)b(t)}{2}\to A'(t)=\frac{a'(t)b(t)+a(t)b'(t)}{2}.\quad (3)$$

You have $a'(t)=108$ when $a(t)=3$ and $b(t)=4$ (it gives us that $c(t)=5$). Use that at $(1)$ and $(2)$ and find $b'(t)$. Then find your answer at $(3)$.