Question

Find the speed along the path, where along the path is the speed greatest? and where is it the least?

Answer 1

Given Equation:\[r(t) = <\cos^{3}t,\sin ^{3}t >\]

Answer 2

Using this equation:
\[S(t) = u'(t)\sqrt{f'(u)^{2} + g'(u)^{2}}\]

Answer 3

I first took the derivative of r(t) and obtained r'(t):\[r'(t) = <-3\sin(t)\cos ^{2}(t),3\cos(t)\sin ^{2}(t) >\]

Which gives the Velocity and the Absolute value of Velocity is speed.

The question I have is how does that apply to the formula to find speed.

Answer 4

Also r(t) is \[0 \le t \le \frac{ \Pi }{ 2 }\]

Answer 5

@sarahusher can you help at all?

Answer 6

@dumbcow can you help when you see this?

Answer 7

ok looks like the formula they gave for speed S(t) can only be used if r(t) can be in terms of f(u(t)) and g(u(t)) , i dont think that will work here
however, speed is the magnitude of the velocity in component form
\[S(t) = \sqrt{r'_x(t)^{2} + r'_y(t)^{2}}\]
you already know r'(t)
\[S(t) = \sqrt{(-3\sin t \cos^{2} t)^{2} + (3\cos t \sin^{2} t)^{2}}\]
which simplifies to:
\[S(t) = 3 \sin t \cos t \]
then find max/min
\[S'(t) = 3\cos^{2}t - 3\sin^{2}t = 0\]
\[\rightarrow \tan^{2} t = 1\]
\[\rightarrow t = \frac{\pi}{4}\]
plugging this in we find max speed is 1.5 when t = pi/4
note the endpoints give a speed of 0