Answer
(a) $3$
(b) $-\frac{5}{2\sqrt 3}$
Work Step by Step
Coordinate of the point $P$ are $ (x, \sqrt x) $ moving along the curve $y=\sqrt x$
Let $L$ be the given point with coordinate $ L (2,0) $.
Let $z$ be the distance from the point $P$ to the point $L$
$z=\sqrt {(x-2) ^2+(\sqrt x-0)^2} $ Or
$z=\sqrt {x^2-4x+4+x} $
$z=\sqrt {x^2-3x+4} $
Taking derivatives with respect to t
$\frac{dz}{dt}=\frac{d( \sqrt { x^2-3x+4})}{dt}$
$\frac{dz}{dt}=[\frac{d( \sqrt { x^2-3x+4})}{dx}] \frac{dx}{dt}$
$\frac{dz}{dt}=[\frac{1}{2} (x^2-3x+4)^{\frac{1}{2}-1} \frac{d(x^2-3x+4)}{dx}] \frac{dx}{dt}$
$\frac{dz}{dt}=[\frac{1}{2} (x^2-3x+4)^{-\frac{1}{2}} \frac{d(x^2-3x+4)}{dx} ]\frac{dx}{dt}$
$\frac{dz}{dt}=\frac{1}{2}\frac{2x-3}{\sqrt { x^2-3x+4}} \frac{dx}{dt}$
Put $x=3$ ,$\frac{dx}{dt}=4$ units /s
(a)$\frac{dz}{dt}=\frac{1} {2} \frac {2 \times 3-3}{\sqrt { 3^2-3 \times 3+4}} \times 4=3$
(b) From coordinates of points $ P (x, \sqrt x) $ and $L(2,0) $
slope of $PL=\frac{\sqrt x-0}{x-2}=\frac{\sqrt x}{x-2}$
If $\alpha$ is the inclination of $LP$, then
$\tan\alpha=\frac{\sqrt x}{x-2}$
Taking derivative with respect to t
$ \frac{d(\tan\alpha)}{dt}= \frac{ d(\frac{\sqrt x}{x-2})}{dt}$
$ \frac{d(\tan\alpha)}{d\alpha} \frac{ d\alpha}{dt}= \frac{ d(\frac{\sqrt x}{x-2})}{dx} \frac{dx}{dt}$
$\sec^2\alpha \frac{d\alpha}{dt}= \frac{(x-2) d\frac{( \sqrt x)}{dx}- \sqrt x \frac{d(x-2)}{dx}}{(x-2)^2} \frac{dx}{dt}$
$(1+\tan^2\alpha) \frac{d\alpha}{dt}= \frac{(x-2) \frac{1}{2} x^{- \frac{1}{2}}- \sqrt x }{(x-2)^2} \frac{dx}{dt}$
$(1+\tan^2\alpha) \frac{d\alpha}{dt}= \frac{ \frac{x-2}{2\sqrt x }-\sqrt x }{(x-2)^2} \frac{dx}{dt}$
$[1+ (\frac{\sqrt x}{x-2})^2] \frac{d\alpha}{dt}= \frac{ \frac{x-2}{2\sqrt x }-\sqrt x }{(x-2)^2} \frac{dx}{dt}$ ............................... eq(1)
Putting $x=3,\frac{dx}{dt}=4$ units/s in equation (1)
$[1+ (\frac{\sqrt 3}{3-2})^2] \frac{d\alpha}{dt}= \frac{ \frac{3-2}{2\sqrt 3 }-\sqrt 3 }{(3-2)^2} \times 4 $
$ \frac{d\alpha}{dt} =-\frac{5}{2\sqrt 3}$
