Answer
$1~rad/sec$
Work Step by Step
Step 1. We are given $y=x^2$,$\frac{dx}{dt}=10m/sec$, and $x=3m, y=3^2=9m$
Step 2. With $y=x^2$, we have $\frac{dy}{dt}=2x\frac{dx}{dt}=2(3)(10)=60 m/sec$
Step 3. At point $P(x,y)$, the angle can be found as $tan\theta=\frac{y}{x}$. Thus:
$sec^2\theta\frac{d\theta}{dt}=\frac{xy'-yx'}{x^2}=\frac{3(60)-9(10)}{3^2}=\frac{3(60)-9(10)}{3^2}=10$
Step 4. Since $cos^2\theta=(\frac{x}{\sqrt {x^2+y^2}})^2=(\frac{3}{\sqrt {3^2+9^2}})^2=\frac{1}{10}$,
We have
$\frac{d\theta}{dt}=10cos^2\theta=1 rad/sec$
