A particle moves along the graph of $y=x^2$ over the plane xy at a constant velocity of 10cm/s. Let θ denote the angle between the x-axis and the line that goes from the origin to $P(x,x^2)$. find the rate of change of θ with respect to the time in which x=3.
what I tried so far is when x=3, y=9, so the point p is at $P(3,9)$ then $tan(θ) =9/3, θ=1.25 rad$ derivating implicitly with respect to time we get:
$$\frac{dy}{dt}=2x\frac{dx}{dt}$$ but I don't what "constant velocity of 10cm/s" refers to, so I'm stuck
Since the particle moves along the curve, then its position at a certain point in time is given by $(x,x^2),$ where we think of $x$ as a function of time. Thus, the velocity of the particle at each point is given by $(x',2xx').$ Hence, its speed (the magnitude of the velocity) is given by $\sqrt{x'^2+4(xx')^2}.$ Since we are told that this is invariably $10,$ it follows that $$x'^2+4(xx')^2=100.$$
Now when $x=3,$ this simplifies to give $$x'=\frac{10}{37}\sqrt{37}.$$
But note that at all times, the relationship $$\tan \theta=\frac {x^2}{x}=x$$ holds between $x$ and $\theta.$ Thus, we have that $$\sec^2\theta\cdot\theta'=x'.$$ When $x=3,$ we have that then $$\cos \theta = \frac{\sqrt{10}}{10}$$ and therefore $$\cos^2\theta=\frac{1}{10}.$$ Hence, we have $$x'=\frac{\theta'}{\cos^2\theta}=\frac{10}{37}\sqrt{37}.$$ It now follows that $$\theta'=\frac{\sqrt{37}}{37}.$$