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

  • 1
    $\begingroup$ Constant velocity of $10 cm/s$ means that the particle travels $10 cm$ (on the parabola) in one second. $\endgroup$ – PierreCarre May 3 at 18:22
  • $\begingroup$ For a particle parametrized by $t \mapsto (x(t),y(t))$, its velocity at time $t$ is the length of the tangent vector to the path at time $t$. So the velocity at time $t$ equals $\sqrt{x'(t)^2 + y'(t)^2}$. Since we are given that the particle moves along a parabola, we have $y(t) = x(t)^2$, so $y'(t) = 2x(t)x'(t)$ and $x'(t)^2 + y'(t)^2 = x'(t)^2 + 4x^2(t)x'(t)^2 = x'(t)^2 (1 + 4x(t)^2)$. On the other hand, $\theta(t) = \arctan(y(t)/x(t)) = \arctan(x^2(t)/x(t)) = \arctan(x(t))$. Now the easy part is doing calculus... $\endgroup$ – Jane Doé May 3 at 18:23

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 $$\theta'=\frac{\sqrt{37}}{37}.$$

  • $\begingroup$ It says that the particle is moving along the curve $y=x^2$ with a constant velocity of $10$ cm/s. Not along the x-axis. x as a function of times just can't be changing constantly like that. In fact, as the particle keeps going father and farther along the curve, there should be a smaller and smaller change in x (it's going to grow at a decreasing rate) and that can't be described as constant growth. $\endgroup$ – Michael Rybkin May 3 at 20:51
  • $\begingroup$ The relationship we get is $\frac{d\theta}{dt}=\cos^2\theta\frac{dx}{dt}$. The cosine of the angle theta, $\cos{\theta}$, when $x=3$ can certainly be found, but the real problem is that we don't really know what $\frac{dx}{dt}$ is when $x=3$. $\endgroup$ – Michael Rybkin May 3 at 20:51
  • $\begingroup$ @MichaelRybkin Check the adjustment I have now made. What we know is sufficient to determine the requested rate. $\endgroup$ – Allawonder May 3 at 21:14
  • $\begingroup$ Your answer makes sense to me. I think it is correct. It follows that the $x'(t)$ is given by $\frac{10}{\sqrt{1+4t^2}}$ and $x(t)$ must be $5\sinh^{-1}(2t)$ (it's the answer to the differential equation $x'(t)=\frac{10}{\sqrt{1+4t^2}}$ with the initial condition $x(0)=0$): see on Wolfram Alpha. $\endgroup$ – Michael Rybkin May 3 at 22:10

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