Speed of a particle given parametric equations of x and y.

Question

So my $x(t)=\pi t +\cos(2\pi t -(\pi/2))$, and my $y(t)=\pi t +\sin(2\pi t -(\pi/2))$. I implicitly derived and got $$\frac{dy}{dt}=\frac{(\pi+\cos(2\pi t -(\pi/2))*2\pi)}{(\pi-\sin(2\pi t -(\pi/2))*2\pi)}$$

From here, how do I find the tangent line when t=0 and also the speed when t=0? Also, how do I determine the total amount of time the object has a non-negative vertical velocity during the time interval $0\le t \le1$ ?

Thanks!

Answer 1

Hint: Tangent line is $y-y_0=\dfrac{dy}{dt}\left(\dfrac{dx}{dt}\right)^{-1}(x-x_0)$. Also, speed is simply $\sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac {dy}{dt} \right)^2} $. For a derivation, see here. Vertical velocity is when $\dfrac{dy}{dt}$ is $\geq0$.

Answer 2

For this sort of problem, it's probably not a good idea to calculate $dy/dx$ and try to write the tangent line in the form $y=f(x)$. The problem is that curves described by these sorts of parametric equations will often have a vertical tangent somewhere, and this will cause problems.

A better approach is to write the tangent line in the form $$ (y-y_0)\frac{dx}{dt} = (x-x_0)\frac{dy}{dt} $$

This form doesn't suffer from any problems with vertical tangents.