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.
