# 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 ?

Dec 27, 2017

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

#### Explanation:

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

$\vec{a} = \frac{d \vec{v}}{\mathrm{dt}}$

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

$\vec{v} \left(t\right) = \left(t \sin \left(\frac{\pi}{3} t\right) , 2 \cos \left(\frac{\pi}{2} t\right) - 3 t\right)$,

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:

$\frac{d}{\mathrm{dt}} \left(t \sin \left(\frac{\pi}{3} t\right)\right)$

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

$x \left(t\right) = t \sin \left(\frac{\pi}{3} t\right)$

So we're taking:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(t \sin \left(\frac{\pi}{3} t\right)\right)$

First, let's use the product rule:

$\mathrm{dx} = t \cdot d \left(\sin \left(\frac{\pi}{3} t\right)\right) + \sin \left(\frac{\pi}{3} t\right) \mathrm{dt}$

To get the differential $d \left(\sin \left(\frac{\pi}{3} t\right)\right)$, we need the chain rule...

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

$\xi \left(t\right) = \sin \left(\frac{\pi}{3} t\right)$

And break it down into a composition of simpler functions:

${\xi}_{1} \left(t\right) = \frac{\pi}{3} t$

${\xi}_{2} \left(t\right) = \sin \left(t\right)$

So $\xi \left(t\right) = {\xi}_{2} \left({\xi}_{1} \left(t\right)\right)$.

To get $\frac{d \xi}{\mathrm{dt}}$, we first need to take $\frac{d {\xi}_{1}}{\mathrm{dt}}$:

$\frac{d {\xi}_{1}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(\frac{\pi}{3} t\right) = \frac{\pi}{3}$

Then "multiply" by $\mathrm{dt}$ to obtain the differential $d {\xi}_{1}$:

$\frac{d {\xi}_{1}}{\mathrm{dt}} = \frac{\pi}{3} \rightarrow d {\xi}_{1} = \frac{\pi}{3} \mathrm{dt}$

Next, take $\frac{d {\xi}_{2}}{d {\xi}_{1}}$, which means replacing all input with ${\xi}_{1}$:

$\frac{d {\xi}_{2}}{d {\xi}_{1}} = \frac{d}{d {\xi}_{1}} \sin \left({\xi}_{1}\right) = \cos \left({\xi}_{1}\right)$

Obtaining the differential $d {\xi}_{2}$:

$d {\xi}_{2} = \cos \left({\xi}_{1}\right) d {\xi}_{1}$

And replacing ${\xi}_{1}$ and $d {\xi}_{1}$ with what we had solved for earlier, which is in terms of $t$ and $\mathrm{dt}$:

$d {\xi}_{2} = \cos \left(\frac{\pi}{3} t\right) \frac{\pi}{3} \mathrm{dt}$

Therefore, "dividing" by $\mathrm{dt}$:

$\frac{d {\xi}_{2}}{\mathrm{dt}} = \cos \left(\frac{\pi}{3} t\right) \frac{\pi}{3}$

$\frac{d {\xi}_{2}}{\mathrm{dt}} = \frac{\pi}{3} \cos \left(\frac{\pi}{3} t\right)$

That is not only $\frac{d {\xi}_{2}}{\mathrm{dt}}$, but $\frac{d \xi}{\mathrm{dt}}$ as a whole:

$\frac{d \xi}{\mathrm{dt}} = \frac{\pi}{3} \cos \left(\frac{\pi}{3} t\right)$

Solving for the differential:

$d \xi = \frac{\pi}{3} \cos \left(\frac{\pi}{3} t\right) \mathrm{dt}$

Substituting back into $\mathrm{dx}$:

$\mathrm{dx} = t \cdot d \xi + \sin \left(\frac{\pi}{3} t\right) \mathrm{dt}$

$\mathrm{dx} = t \cdot \frac{\pi}{3} \cos \left(\frac{\pi}{3} t\right) \mathrm{dt} + \sin \left(\frac{\pi}{3} t\right) \mathrm{dt}$

$\mathrm{dx} = \frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) \mathrm{dt} + \sin \left(\frac{\pi}{3} t\right) \mathrm{dt}$

Finally, solving for the derivative:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) + \sin \left(\frac{\pi}{3} t\right)$

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 $y \left(t\right)$:

$y \left(t\right) = 2 \cos \left(\frac{\pi}{2} t\right) - 3 t$

So we're solving for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(2 \cos \left(\frac{\pi}{2} t\right) - 3 t\right)$

First, by the sum rule:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(2 \cos \left(\frac{\pi}{2} t\right)\right) + \frac{d}{\mathrm{dt}} \left(- 3 t\right)$

Taking the constatns out by linearity:

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \frac{d}{\mathrm{dt}} \left(\cos \left(\frac{\pi}{2} t\right)\right) - 3 \frac{d}{\mathrm{dt}} \left(t\right)$

$\frac{d}{\mathrm{dt}} \left(t\right)$ is just $1$:

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \frac{d}{\mathrm{dt}} \left(\cos \left(\frac{\pi}{2} t\right)\right) - 3$

As for $\frac{d}{\mathrm{dt}} \left(\cos \left(\frac{\pi}{2} t\right)\right)$, we could use the chain rule again, defining a function of ypsilon (a Greek letter $\upsilon$):

$\upsilon \left(t\right) = \cos \left(\frac{\pi}{2} t\right)$

Again splitting it:

${\upsilon}_{1} \left(t\right) = \frac{\pi}{2} t$

${\upsilon}_{2} \left(t\right) = \cos \left(t\right)$

$\upsilon \left(t\right) = {\upsilon}_{2} \left({\upsilon}_{1} \left(t\right)\right)$

Then taking $\frac{d {\upsilon}_{1}}{\mathrm{dt}}$:

$\frac{d {\upsilon}_{1}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(\frac{\pi}{2} t\right) = \frac{\pi}{2}$

$d {\upsilon}_{1} = \frac{\pi}{2} \mathrm{dt}$

Next, $\frac{d {\upsilon}_{2}}{d \upsilon 1}$:

$\frac{d {\upsilon}_{2}}{d \upsilon 1} = \frac{d}{d {\upsilon}_{1}} \cos \left({\upsilon}_{1}\right) = - \sin \left({\upsilon}_{1}\right)$

$d {\upsilon}_{2} = - \sin \left({\upsilon}_{1}\right) d {\upsilon}_{1}$

"Unrolling" things:

$d {\upsilon}_{2} = - \sin \left(\frac{\pi}{2} t\right) \frac{\pi}{2} \mathrm{dt}$

$d {\upsilon}_{2} = - \frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right) \mathrm{dt}$

Taking the derivative:

$\frac{d {\upsilon}_{2}}{\mathrm{dt}} = - \frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right)$

$\frac{d \upsilon}{\mathrm{dt}} = - \frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right)$

Back to $\frac{\mathrm{dy}}{\mathrm{dt}}$:

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \frac{d \upsilon}{\mathrm{dt}} - 3$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \cdot - \frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right) - 3$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \pi \sin \left(\frac{\pi}{2} t\right) - 3$

Substituting back into the derivative of the velocity function:

$\frac{d \vec{v}}{\mathrm{dt}} = \left(\frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) + \sin \left(\frac{\pi}{3} t\right) , \frac{\mathrm{dy}}{\mathrm{dt}}\right)$

$\frac{d \vec{v}}{\mathrm{dt}} = \left(\frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) + \sin \left(\frac{\pi}{3} t\right) , - \pi \sin \left(\frac{\pi}{2} t\right) - 3\right)$

And that should get us the acceleration function:

$\vec{a} \left(t\right) = \left(\frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) + \sin \left(\frac{\pi}{3} t\right) , - \pi \sin \left(\frac{\pi}{2} t\right) - 3\right)$

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

$\vec{a} \left(1\right) = \left(\frac{\pi}{3} \left(1\right) \cos \left(\frac{\pi}{3} \left(1\right)\right) + \sin \left(\frac{\pi}{3} \left(1\right)\right) , - \pi \sin \left(\frac{\pi}{2} \left(1\right)\right) - 3\right)$

And solve:

$\vec{a} \left(1\right) = \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3}\right) + \sin \left(\frac{\pi}{3}\right) , - \pi \sin \left(\frac{\pi}{2}\right) - 3\right)$

$\vec{a} \left(1\right) = \left(\frac{\pi}{3} \left(\frac{1}{2}\right) + \frac{\sqrt{3}}{2} , - \pi \left(1\right) - 3\right)$

$\vec{a} \left(1\right) = \left(\frac{\pi}{6} + \frac{\sqrt{3}}{2} , - \pi - 3\right)$

This approximates to:

$\vec{a} \left(1\right) \approx \left(2.25564958316718 , - 6.141592653589793\right)$

We can see that it accelerates around $2.25564958316718 \frac{m}{s} ^ 2$ in the $x$-direction and $- 6.141592653589793 \frac{m}{s} ^ 2$ in the $y$-direction.