An object's two dimensional velocity is given by #v(t) = ( tsin(pi/3t) , 2cos(pi/2t )- 3t )#. What is the object's rate and direction of acceleration at #t=1 #?

1 Answer
Dec 27, 2017

Answer:

Using calculus, take the derivative of #vec v(t)# with respect to time #t# to get the acceleration function, then solve for acceleration at #t = 1# to get #vec a(1) = (pi/6 + sqrt(3)/2, - pi - 3) ~~ (2.25564958316718, -6.141592653589793)#
Telling you that it accelerates around #2.25564958316718 m/s^2# in the #x#-direction and #-6.141592653589793 m/s^2# in the #y#-direction.

Explanation:

The calculus definition for acceleration is the derivative of velocity with respect to time:

#vec a = (d vec v)/(dt)#

To take the derivative of the two-dimensional velocity, which is in this case the following function:

#vec v(t) = (t sin(pi/3 t), 2 cos(pi/2 t) - 3t)#,

Take the derivative of each dimension, as such:

#(d vec v)/(dt) = (d/dt (t sin(pi/3 t)), d/dt (2 cos(pi/2 t) - 3t)#

Let's focus on the #x#-value first:

#d/dt (t sin(pi/3 t))#

We'll call that a function of #x#:

#x(t) = t sin(pi/3 t)#

So we're taking:

#(dx)/(dt) = d/dt (t sin(pi/3 t))#

First, let's use the product rule:

#dx = t * d(sin(pi/3 t)) + sin(pi/3 t) dt#

To get the differential #d(sin(pi/3 t))#, we need the chain rule...

Let's call this other function xi (a Greek letter, #xi#):

#xi(t) = sin(pi/3 t)#

And break it down into a composition of simpler functions:

#xi_1(t) = pi/3 t#

#xi_2(t) = sin(t)#

So #xi(t) = xi_2(xi_1(t))#.

To get #(d xi)/(dt)#, we first need to take #(d xi_1)/(dt)#:

#(d xi_1)/(dt) = d/dt (pi/3 t) = pi/3#

Then "multiply" by #dt# to obtain the differential #d xi_1#:

#(d xi_1)/(dt) = pi/3 rarr d xi_1 = pi/3 dt#

Next, take #(d xi_2)/(d xi_1)#, which means replacing all input with #xi_1#:

#(d xi_2)/(d xi_1) = d/(d xi_1) sin(xi_1) = cos(xi_1)#

Obtaining the differential #d xi_2#:

#d xi_2 = cos(xi_1) d xi_1#

And replacing #xi_1# and #d xi_1# with what we had solved for earlier, which is in terms of #t# and #dt#:

#d xi_2 = cos(pi/3 t) pi/3 dt#

Therefore, "dividing" by #dt#:

#(d xi_2)/(dt) = cos(pi/3 t) pi/3#

#(d xi_2)/(dt) = pi/3 cos(pi/3 t)#

That is not only #(d xi_2)/(dt)#, but #(d xi)/(dt)# as a whole:

#(d xi)/(dt) = pi/3 cos(pi/3 t)#

Solving for the differential:

#d xi = pi/3 cos(pi/3 t) dt#

Substituting back into #dx#:

#dx = t * d xi + sin(pi/3 t) dt#

#dx = t * pi/3 cos(pi/3 t) dt + sin(pi/3 t) dt#

#dx = pi/3 t cos(pi/3 t) dt + sin(pi/3 t) dt#

Finally, solving for the derivative:

#(dx)/(dt) = pi/3 t cos(pi/3 t) + sin(pi/3 t)#

Substituting back into our function:

#(d vec v)/(dt) = ((dx)/(dt), d/dt (2 cos(pi/2 t) - 3t)#

#(d vec v)/(dt) = (pi/3 t cos(pi/3 t) + sin(pi/3 t), d/dt (2 cos(pi/2 t) - 3t)#

Next, we'll look at the #y#-value, as a function #y(t)#:

#y(t) = 2 cos(pi/2 t) - 3t#

So we're solving for the derivative:

#(dy)/(dt) = d/dt (2 cos(pi/2 t) - 3t)#

First, by the sum rule:

#(dy)/(dt) = d/dt (2 cos(pi/2 t)) + d/dt (-3t)#

Taking the constatns out by linearity:

#(dy)/(dt) = 2 d/dt (cos(pi/2 t)) - 3 d/dt (t)#

#d/dt (t)# is just #1#:

#(dy)/(dt) = 2 d/dt (cos(pi/2 t)) - 3#

As for #d/dt (cos(pi/2 t))#, we could use the chain rule again, defining a function of ypsilon (a Greek letter #upsilon#):

#upsilon(t) = cos(pi/2 t)#

Again splitting it:

#upsilon_1(t) = pi/2 t#

#upsilon_2(t) = cos(t)#

#upsilon(t) = upsilon_2(upsilon_1(t))#

Then taking #(d upsilon_1)/(dt)#:

#(d upsilon_1)/(dt) = d/dt (pi/2 t) = pi/2#

#d upsilon_1 = pi/2 dt#

Next, #(d upsilon_2)/(d upsilon 1)#:

#(d upsilon_2)/(d upsilon 1) = d/(d upsilon_1) cos(upsilon_1) = -sin(upsilon_1)#

#d upsilon_2 = -sin(upsilon_1) d upsilon_1#

"Unrolling" things:

#d upsilon_2 = -sin(pi/2 t) pi/2 dt#

#d upsilon_2 = - pi/2 sin(pi/2 t) dt#

Taking the derivative:

#(d upsilon_2)/(dt) = - pi/2 sin(pi/2 t)#

#(d upsilon)/(dt) = - pi/2 sin(pi/2 t)#

Back to #(dy)/(dt)#:

#(dy)/(dt) = 2 (d upsilon)/(dt) - 3#

#(dy)/(dt) = 2 * - pi/2 sin(pi/2 t) - 3#

#(dy)/(dt) = - pi sin(pi/2 t) - 3#

Substituting back into the derivative of the velocity function:

#(d vec v)/(dt) = (pi/3 t cos(pi/3 t) + sin(pi/3 t), (dy)/(dt))#

#(d vec v)/(dt) = (pi/3 t cos(pi/3 t) + sin(pi/3 t), - pi sin(pi/2 t) - 3)#

And that should get us the acceleration function:

#vec a(t) = (pi/3 t cos(pi/3 t) + sin(pi/3 t), - pi sin(pi/2 t) - 3)#

Now, we just need to substitute for #t = 1#:

#vec a(1) = (pi/3 (1) cos(pi/3 (1)) + sin(pi/3 (1)), - pi sin(pi/2 (1)) - 3)#

And solve:

#vec a(1) = (pi/3 cos(pi/3) + sin(pi/3), - pi sin(pi/2) - 3)#

#vec a(1) = (pi/3 (1/2) + sqrt(3)/2, - pi (1) - 3)#

#vec a(1) = (pi/6 + sqrt(3)/2, - pi - 3)#

This approximates to:

#vec a(1) ~~ (2.25564958316718, -6.141592653589793)#

We can see that it accelerates around #2.25564958316718 m/s^2# in the #x#-direction and #-6.141592653589793 m/s^2# in the #y#-direction.