Find velocity and acceleration vectors when \(t = 0\) for \( \vec{r}(t) = e^{2t}\hat{i} + e^{-t}\hat{j} \)

Find velocity and acceleration at \( t=3/4 \) for \( \vec{r}(t) = 3\cos(2\pi t)\hat{i} + \) \( 3\sin(2\pi t)\hat{j} \)

As time \(t>0\) increases, a particle travels along the curve \( \mathcal{C} \) with position parameterized by \( \vec{r}(t) = 2(\cos t+t\sin t)\hat{i} + \) \( 2(\sin t-t\cos t)\hat{j} \). For each \(t>0\), compute the tangent (velocity) vector and the speed. What is the length of the curve which the particle travels over between \(t=0\) and \(t=4\pi\)?

For the length of the curve, use the equation \( L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 } ~dt } \). See the vector functions arc length page for details.

In the middle of the video solution for this problem is a part that you were not asked to do in the problem statement, from about 8 minutes to about 10 and a half minutes in the video.

Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i} + t^2\hat{j} + t^3\hat{k}\).

Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i}+3e^t\hat{j} + 4e^t\hat{k}\).

Find velocity, acceleration and speed for the position function \( \vec{r}(t) = (3\cos t)\hat{i} + \) \( (3\sin t)\hat{j} - 4t\hat{k} \).

For the position function \(\vec{r}(t)=\langle t-\sin t,1-\cos t\rangle\), find the velocity and acceleration vectors at the point \((\pi,2)\).

The equations we need are \(\vec{v}(t)=\vec{r}'(t)\) and \(\vec{a}(t)=\vec{v}'(t)\). So the velocity vector is \(\vec{v}(t)=\langle 1-\cos t, \sin t\rangle\).
Now we need to find t. We are given the point \((\pi,2)\) on the position vector. So we have two equations we can use to find t,
\(t-\sin t=\pi\) and \(1-\cos t=2\). The second one is easier to solve for t, so we have
\(1-\cos t=2 \to \) \(\cos t=-1 \to \) \(t=\pi\)
Just to double-check, if we let \(t=\pi\) in the equation \(t-\sin t=\pi\), the answer checks.
So \(\vec{v}(\pi)=\langle 1-\cos \pi, \sin\pi\rangle\) giving us \(\vec{v}(\pi)=\langle 2,0 \rangle\).

To get the acceleration vector, we take the derivative of the velocity vector, i.e. \(\vec{a}(t)=\vec{v}'(t)\). So \(\vec{a}(t)=\langle \sin t, \cos t\rangle\) and at the point \((\pi,2)\) the acceleration vector is \(\vec{a}(\pi)=\langle 0,-1\rangle\).
Although the problem statement did not ask for a graph, here is a graph of the position function, the velocity vector and the acceleration vector.

\( \vec{v}(\pi) = \langle 2,0\rangle \); \( \vec{a}(\pi) = \langle 0,-1 \rangle \)

Find the position vector function \(\vec{r}(t)\) for a particle with acceleration \(\vec{a}(t) = \langle 2t,2\sin(t),\cos(4t)\rangle\), initial velocity \(\vec{v}(0)=\langle 1,-3,2\rangle\) and initial position \(\vec{r}(0)=\langle 2,4,-1\rangle\).

\( \vec{r}(t) = \langle (1/3)t^3+t+2, \) \( -2\sin(t)-t+4, \) \( (-1/16)\cos(4t)+2t-15/16\rangle \)

Determine the velocity vector, speed and acceleration vector of an object when \(t=1\) given by the position vector \( \vec{r}(t) = t\vhat{i} + (-0.5t^2+4)\vhat{j} \).

\( \vec{v}(1)=\langle 1,-1 \rangle\), speed=\(\sqrt{2} \), \( \vec{a}(1)=\langle 0,-1 \rangle \)

Determine the velocity vector, speed and acceleration vector of an object when \(t=2\) for the position vector \( \vec{r}(t) = \cos(\pi t)\vhat{i} + \sin(\pi t)\vhat{j} + \) \( (t^2/2)\vhat{k} \)

\( \vec{v}(2) = \langle 0,\pi,2 \rangle \), speed = \( \sqrt{\pi^2+4} \), \( \vec{a}(2) = \langle -\pi^2,0,1 \rangle \)

Find the velocity and acceleration vectors and speed of the vector function \( \vec{r}(t) = \langle -5t,-3t^2,4t^4+3 \rangle \) at \(t=1\).

\( \vec{v}(t) = \langle -5,-6,16\rangle\), \( \vec{a}(t) = \langle 0,-6,48 \rangle \), speed = \( \sqrt{317} \)

A car travels with a velocity vector given by \( \vec{v}(t) = \langle t^2,e^t+1 \rangle \), where t is measured in seconds and the vector components are measured in feet. If the initial position of the car is \( \vec{r}(0) = \langle 1,3 \rangle \), find the position of the car after one second.

\( \vec{r}(1) = \langle 4/3, e+3 \rangle \) feet

A particle moves along a curve whose parametric equations are \( x(t) = e^{-t}\), \( y(t) = 2\cos(3t) \), \( z(t) = 2\sin(3t) \). (a) Determine the velocity and acceleration vectors.
(b) Find the magnitude of the velocity (speed) and acceleration at \(t=0\).

(a) \( \vec{v}(t) = -e^{-t}\vhat{i} - 6\sin(3t)\vhat{j} + 6\cos(3t)\vhat{k} \); \( \vec{a}(t) = e^{-t}\vhat{i} - 18\cos(3t)\vhat{j} - 18\sin(3t)\vhat{k} \)
(b) \( \|\vec{v}(0)\| = \sqrt{37} \); \( \|\vec{a}(0)\| = 5\sqrt{13} \)

Given the velocity of a particle is given by \(\langle e^t + 2t, 3t^2+1, 2\cos(2t) \rangle\) and at time \(t=0\) the particle is at position \(\langle 2,0,-1\rangle\), then determine the position of the particle at time \(t=2\).