Related Rates Ladder Question

Solution 1:

Let, $h$ & $a$ be the height & the base of the right triangle formed at any instance then

In right triangle, using Pythagorean theorem, $$25^2=h^2+a^2$$$$\implies h=\sqrt{625-a^2}$$ Now, the area of the right triangle at any instance $$A=\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(a)(h)$$ $$A=\frac{1}{2}a\sqrt{625-a^2}$$ Hence, the rate of change of the area of right triangle, $$\frac{dA}{dt}=\frac{1}{2}\frac{d}{dt}(a\sqrt{625-a^2})$$ $$=\frac{1}{2}\left(\sqrt{625-a^2}-\frac{a^2}{\sqrt{625-a^2}}\right)\frac{da}{dt}$$ $$=\frac{1}{2}\left(\frac{625-2a^2}{\sqrt{625-a^2}}\right)\frac{da}{dt}$$ Now, setting the corresponding values, length of base, $a=11$ ft & speed of ladder base, $\frac{da}{dt}=2$ ft/sec, we get $$\frac{dA}{dt}=\frac{1}{2}\left(\frac{625-2\cdot11^2}{\sqrt{625-11^2}}\right)(2)$$ $$\color{red}{\frac{dA}{dt}}=\color{blue}{\frac{383}{2\sqrt{126}}\approx 17.06 ft^2/sec}$$