Answer
a) $ \frac{dl}{dt}= \frac {1} {\sqrt{x^2+y^2}} \left[x \frac{dx}{dt}+ y \frac{dy}{dt}\right] $
b) $ \frac{dl}{dt}= \frac {1}{10} \frac{ft}{s}$; diagonal increasing
Work Step by Step
a)
$l= \sqrt{x^2+y^2} $............. eq (1)
Differentiating equation (1) with respect to t.
$ \frac{dl}{dt}= \frac{dl}{dx} \frac{dx}{dt}+ \frac{dl}{dy}\frac{dy}{dt}$
Putting $ l=\sqrt{x^2 +y^2} $ in RHS
$ \frac{dl}{dt}= \frac{d(\sqrt{x^2+y^2)}} {dx}\frac{dx}{dt}+ \frac{d(\sqrt{x^2+y^2)}} {dy}\frac{dy}{dt}$
$ \frac{dl}{dt}= \frac{1}{2} (x^2+y^2)^{-\frac{1}{2}} \frac{d(x^2+y^2)}{dx} \frac{dx}{dt}+ \frac{1}{2} (x^2+y^2)^{-\frac{1}{2}} \frac{d(x^2+y^2)}{dy}\frac{dy}{dt}$
$ \frac{dl}{dt}= \frac {1}{2} (x^2+y^2) ^ {-\frac {1}{2}} (2x) \frac{dx}{dt}+ \frac {1}{2} (x^2+y^2) ^ {-\frac {1}{2}} (2y) \frac{dy}{dt}$
$ \frac{dl}{dt}= \frac{x}{\sqrt{x^2+y^2}} \frac{dx}{dt}+ \frac{y}{\sqrt{x^2+y^2}} \frac{dy}{dt}$
$ \frac{dl}{dt}= \frac {1} {\sqrt{x^2+y^2}} \left[x \frac{dx}{dt}+ y \frac{dy}{dt}\right] $........ eq (2)
b)
Putting $x = 3 ft$ and $y = 4$, $\frac{dx}{dt}= \frac {1}{2} $ ft/s, $ \frac{dy}{dt}= -\frac {1}{4} $ ft/s in equation (2)
$ \frac{dl}{dt}= \frac{1} {\sqrt {3^2+4^2}} \left[3 \times \frac {1}{2} + 4 \times \left(-\frac {1}{4}\right)\right] $
$ \frac{dl}{dt}= \frac {1}{10} \frac{ft}{s}$
Since $ \frac{dl}{dt}= \frac {1}{10} \frac{ft}{s} > 0$, diagonal increasing
