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

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!

#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:

• #{(s^2(0) = 36 ),(bb(s^2(pi/4) = 64)),(s^2(pi/2) = 36) :}#

So the particle is travelling fastest at:

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