A particle is travelling an elliptical path described by #(3sin(2t)), 4cos(2t))#. Find the points at which it is travelling fastest?

Question

(parametric equations)

Ezra Adams · Accepted Answer

We need to start by deriving a speed function.

Recall that #s = sqrt((x'(t))^2 + (y'(t))^2)#

Therefore we need to find the derivative of the position functions.

#x'(t) = 6cos(2t)#
#y'(t) = -8sin(2t)#

Therefore

#s = sqrt(36cos^2(2t) + 64sin^2(2t))#

We need to differentiate this in order to find the maximum.

#s' = (28sin(4x))/sqrt(64sin^2(2x) + 36cos^2(2x))#

Critical points occur when the derivative equals #0#, in other words when #28sin(4x) = 0#.

#4x = pi or 0 or 2pi#
#x= pi/4 or 0 or pi/2, (3pi)/4#

At #pi/3# the derivative is negative. Therefore #x = pi/4 +pi/2n# will always be a maximum.

We can confirm graphically

Hopefully this helps!

Paisley Johnson · Answer

#t =(2k+1) pi/4 qquad k = 0,1,2,...#

#bbr = (3sin(2t)), 4cos(2t))#

#bbv = (6cos(2t)), -8sin(2t))#

Speed #s#:

#s^2 = abs( bbv)^2 = 36 cos^2(2t) + 64 sin^2(2t) #

#= 36 + 28 sin^2(2t) qquad square#

You don't need calculus to solve this, #s^2 in [36,64]#, but that's the section it is in.

#(d(s^2))/(dt) = 112 sin (2t) cos (2t) #

#= 56 sin (4t) #

#:. (d(s^2))/(dt) = 0# for:

#4t = 0, pi , 2pi , ...#

#t = 0, pi/4, pi/2 , ...#

Given the symmetry of the orbit, you only really need to look at the first 3 solutions, as they will repeat:

So the particle is travelling fastest at:

#t = pi/4 + k pi/2 = (2k+1) pi/4 qquad k = 0,1,2,...#

Charles Arocha · Answer

undefined To find the points at which the particle is traveling fastest, we need to calculate the magnitude of its velocity vector and then determine where this magnitude is maximized. The velocity vector is given by the derivative of the position vector with respect to time.

Given the position vector $ \mathbf{r}(t) = (3\sin(2t), 4\cos(2t)) $, we find the velocity vector $ \mathbf{v}(t) $ by taking the derivative of $ \mathbf{r}(t) $ with respect to time $ t $.

$ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = (6\cos(2t), -8\sin(2t)) $

Now, we need to find the magnitude of the velocity vector:

$ |\mathbf{v}(t)| = \sqrt{(6\cos(2t))^2 + (-8\sin(2t))^2} $

Simplify:

$ |\mathbf{v}(t)| = \sqrt{36\cos^2(2t) + 64\sin^2(2t)} $

$ |\mathbf{v}(t)| = \sqrt{36(\cos^2(2t) + \sin^2(2t))} $

Since $ \cos^2(2t) + \sin^2(2t) = 1 $ (due to the Pythagorean identity), we have:

$ |\mathbf{v}(t)| = \sqrt{36} = 6 $

So, the magnitude of the velocity vector is constant and equal to 6. Therefore, the particle is traveling at its fastest speed of 6 units/s at all points along its elliptical path.