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.
