In the beginning of a multivariable calculus class, I'm trying to calculate the following problem.
A particle moves in the plane along an ellipse curve so that at time t ≥ 0 it is at the point $ r(t) = (3cos(2πt), 2sin(2πt)). $
(a) At what points on the curve is the speed greatest?
(b) Find the greatest speed.
I've derived r, to get an expression for the derivative at t, that is, the speed at t: $ r'(t) = (-6 \pi sin(2πt), 4 \pi cos(2πt)) $.
Regarding a, now I'm not sure how to use r'(t) to find the maximum. We don't have tools like gradients at this point. Are we instead supposed to logically think about the shape of the ellips and what that means for its speed?
(The answer for a is $(0,\pm 2)$.)
Regarding b, is I was able to establish $(0,\pm2)$ in a, I assume I'd now like to know what t in r'(t) gives $(0,\pm2)$ so that I may plug that t into r'(t) and find the speed? How do I go about finding what t in r'(t) gives $(0,\pm2)$?
You almost got it. What you currently have, $\vec{r'}$ is the velocity vector at any point in time (with both the x and y components). You need to use the Pythagorean Theorem to find the speed at any point in time by doing the following:
$$s(t) = \sqrt{(-6 \pi \sin(2 \pi t))^2) + (4 \pi \cos(2 \pi t))^2}$$
which simplifies to:
$$s(t) = 2 \pi \sqrt{9 \sin^2(2 \pi t) + 4 \cos^2(2 \pi t)}$$
Now, we can find the derivative of $s(t)$ in order to find its maximum: $$s'(t)= \frac{20{\pi}^2\cos\left(2{\pi}t\right)\sin\left(2{\pi}t\right)}{\sqrt{9\sin^2\left(2{\pi}t\right)+4\cos^2\left(2{\pi}t\right)}}$$
Now, we can use some reasoning to determine where $s'(t)=0$ (acceleration is zero when speed is at max/min). Since we know a fraction is only equal to zero when the numerator is 0, and by dividing everything by $20 \pi^2$, we can obtain the following: $$\cos(2 \pi t) \cdot \sin(2 \pi t) = 0$$ which gives the solutions: $$t = \frac{n}{2}-\frac{1}{4},\, t=\frac{n}{2} \text{ where } n \in \mathbb{Z}$$ From here, a plugging in $t$ into $s(t)$ will quickly determine its value, and whether it is a min/max.