Integrating Vector-Valued Functions - Calculus 2

1. The velocity vector of a particle moving in a plane is $\mathbf{v}(t)$. Which of the following expressions represents the total distance traveled by the particle from $t=a$ to $t=b$?
   Answer: $\int_a^b \|\mathbf{v}(t)\| \,dt$ The total distance traveled by a particle is the integral of its speed over the time interval. The speed is the magnitude of the velocity vector, $\|\mathbf{v}(t)\|$. Therefore, the total distance is given by the arc length formula, $\int_a^b \|\mathbf{v}(t)\| \,dt$.\ Choice B, $\| \int_a^b \mathbf{v}(t) \,dt \|$, represents the magnitude of the displacement vector, which is the straight-line distance between the start and end points, not the total distance traveled along the path. \ Choice C is the displacement vector itself. \ Choice D is the average velocity vector over the interval.
2. The velocity of a particle is $\mathbf{v}(t)$. It is known that $\int_1^3 \mathbf{v}(t) \,dt = \langle 4, -2 \rangle$. If $\mathbf{r}(1) = \langle 1, 5 \rangle$, what is $\mathbf{r}(3)$?
   Answer: $\langle 5, 3 \rangle$ The Fundamental Theorem of Calculus for vector-valued functions states that $\int_a^b \mathbf{v}(t) \,dt = \mathbf{r}(b) - \mathbf{r}(a)$.\ In this case, $a=1$ and $b=3$. So, $\int_1^3 \mathbf{v}(t) \,dt = \mathbf{r}(3) - \mathbf{r}(1)$.\ We are given $\int_1^3 \mathbf{v}(t) \,dt = \langle 4, -2 \rangle$ and $\mathbf{r}(1) = \langle 1, 5 \rangle$.\ We can rearrange the formula to solve for $\mathbf{r}(3)$: $\mathbf{r}(3) = \mathbf{r}(1) + \int_1^3 \mathbf{v}(t) \,dt$.\ $\mathbf{r}(3) = \langle 1, 5 \rangle + \langle 4, -2 \rangle = \langle 1+4, 5-2 \rangle = \langle 5, 3 \rangle$.
3. The velocity of a particle is $\mathbf{v}(t) = \langle \sec^2(t), 2\sin(t)\cos(t) \rangle$ on the interval $-\pi/2 < t < \pi/2$. If $\mathbf{r}(0) = \langle 1, 0 \rangle$, find $\mathbf{r}(\pi/4)$.
   Answer: $\langle 2, 1/2 \rangle$ To find the position, we integrate the velocity. We can use the identity $2\sin(t)\cos(t) = \sin(2t)$ to simplify the y-component. So, $\mathbf{v}(t) = \langle \sec^2(t), \sin(2t) \rangle$.\ $\mathbf{r}(t) = \int \langle \sec^2(t), \sin(2t) \rangle \,dt = \langle \tan(t) + C_1, -\frac{1}{2}\cos(2t) + C_2 \rangle$.\ Use the initial condition $\mathbf{r}(0) = \langle 1, 0 \rangle$ to find the constants.\ x-comp: $\tan(0) + C_1 = 1 \Rightarrow 0 + C_1 = 1 \Rightarrow C_1 = 1$.\ y-comp: $-\frac{1}{2}\cos(0) + C_2 = 0 \Rightarrow -\frac{1}{2}(1) + C_2 = 0 \Rightarrow C_2 = 1/2$.\ So, $\mathbf{r}(t) = \langle \tan(t) + 1, -\frac{1}{2}\cos(2t) + \frac{1}{2} \rangle$.\ Now, evaluate at $t = \pi/4$:\ $\mathbf{r}(\pi/4) = \langle \tan(\pi/4) + 1, -\frac{1}{2}\cos(2 \cdot \pi/4) + \frac{1}{2} \rangle$.\ $= \langle 1 + 1, -\frac{1}{2}\cos(\pi/2) + \frac{1}{2} \rangle = \langle 2, -\frac{1}{2}(0) + \frac{1}{2} \rangle = \langle 2, 1/2 \rangle$.
4. The velocity of a particle is given by $\mathbf{v}(t) = \langle 4t, 3t^2 \rangle$. What is the average velocity of the particle from $t=0$ to $t=2$?
   Answer: $\langle 4, 4 \rangle$ The average value of a vector function $\mathbf{f}(t)$ over an interval $[a, b]$ is given by $\frac{1}{b-a} \int_a^b \mathbf{f}(t) \,dt$.\ For this problem, the average velocity is $\frac{1}{2-0} \int_0^2 \langle 4t, 3t^2 \rangle \,dt$.\ First, compute the definite integral: \ $\int_0^2 \langle 4t, 3t^2 \rangle \,dt = \left\langle [2t^2]_0^2, [t^3]_0^2 \right\rangle = \langle 2(2)^2 - 0, 2^3 - 0 \rangle = \langle 8, 8 \rangle$.\ Then, multiply by $\frac{1}{b-a} = \frac{1}{2}$: \ Average velocity = $\frac{1}{2} \langle 8, 8 \rangle = \langle 4, 4 \rangle$.\ Choice A is the average of the initial and final velocities. Choice B is the total displacement, not the average velocity. Choice C is the velocity at the midpoint of the interval.
5. A particle has velocity $\mathbf{v}(t) = \langle 2t, 1 \rangle$. If its initial position is $\mathbf{r}(0) = \langle 0, 0 \rangle$, what is the magnitude of its position vector at $t=2$?
   Answer: $2\sqrt{5}$ First, find the position vector $\mathbf{r}(t)$ by integrating the velocity vector $\mathbf{v}(t)$.\ $\mathbf{r}(t) = \int \langle 2t, 1 \rangle \,dt = \langle t^2 + C_1, t + C_2 \rangle$.\ Given the initial position $\mathbf{r}(0) = \langle 0, 0 \rangle$, we find the constants: \ $0^2 + C_1 = 0 \Rightarrow C_1 = 0$.\ $0 + C_2 = 0 \Rightarrow C_2 = 0$.\ So, the position vector is $\mathbf{r}(t) = \langle t^2, t \rangle$.\ Next, find the position vector at $t=2$:\ $\mathbf{r}(2) = \langle 2^2, 2 \rangle = \langle 4, 2 \rangle$.\ Finally, calculate the magnitude of this vector:\ $\|\mathbf{r}(2)\| = \|\langle 4, 2 \rangle\| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$.
6. A particle's velocity is $\mathbf{v}(t) = \langle 3t^2 - 1, 2e^{2t} \rangle$. If the particle is at $\mathbf{r}(0) = \langle 2, 0 \rangle$, what is its position at $t=1$?
   Answer: $\langle 2, e^2 - 1 \rangle$ The final position $\mathbf{r}(1)$ can be found using the relationship $\mathbf{r}(b) = \mathbf{r}(a) + \int_a^b \mathbf{v}(t) \,dt$.\ Here, $a=0, b=1$. First, calculate the displacement by integrating the velocity from 0 to 1:\ $\int_0^1 \langle 3t^2 - 1, 2e^{2t} \rangle \,dt = \left\langle [t^3 - t]_0^1, [e^{2t}]_0^1 \right\rangle$.\ Evaluating the definite integrals:\ For the x-component: $(1^3 - 1) - (0^3 - 0) = 0$.\ For the y-component: $e^{2(1)} - e^{2(0)} = e^2 - 1$.\ The displacement is $\langle 0, e^2 - 1 \rangle$.\ Now, add the initial position to the displacement:\ $\mathbf{r}(1) = \mathbf{r}(0) + \text{displacement} = \langle 2, 0 \rangle + \langle 0, e^2 - 1 \rangle = \langle 2, e^2 - 1 \rangle$.
7. A projectile is launched from the origin with an initial velocity vector $\mathbf{v}(0) = \langle 10, 50 \rangle$. The acceleration due to gravity is $\mathbf{a}(t) = \langle 0, -10 \rangle$. What is the position vector $\mathbf{r}(t)$ of the projectile?
   Answer: $\langle 10t, -5t^2 + 50t \rangle$ We are given acceleration and need to find position, which requires two integrations.\ First, integrate $\mathbf{a}(t)$ to find $\mathbf{v}(t)$:\ $\mathbf{v}(t) = \int \langle 0, -10 \rangle \,dt = \langle C_1, -10t + C_2 \rangle$.\ Use the initial velocity $\mathbf{v}(0) = \langle 10, 50 \rangle$ to find the constants: $C_1 = 10$ and $C_2 = 50$.\ So, $\mathbf{v}(t) = \langle 10, -10t + 50 \rangle$.\ Next, integrate $\mathbf{v}(t)$ to find $\mathbf{r}(t)$:\ $\mathbf{r}(t) = \int \langle 10, -10t + 50 \rangle \,dt = \langle 10t + D_1, -5t^2 + 50t + D_2 \rangle$.\ Since the projectile is launched from the origin, $\mathbf{r}(0) = \langle 0, 0 \rangle$. Use this to find the constants: $D_1 = 0$ and $D_2 = 0$.\ Therefore, the position vector is $\mathbf{r}(t) = \langle 10t, -5t^2 + 50t \rangle$.
8. A particle's acceleration is $\mathbf{a}(t) = \langle 6t, -t^{-2} \rangle$. At $t=1$, its position is $\mathbf{r}(1) = \langle -2, 0 \rangle$ and at $t=2$, its position is $\mathbf{r}(2) = \langle 2, \ln(2) \rangle$. Find the velocity $\mathbf{v}(1)$.
   Answer: $\langle 0, 1 \rangle$ This problem requires two integrations and solving a system of equations for the constants. \ Integrate $\mathbf{a}(t)$ to get $\mathbf{v}(t) = \langle 3t^2 + C_1, t^{-1} + C_2 \rangle = \langle 3t^2 + C_1, 1/t + C_2 \rangle$. \ Integrate $\mathbf{v}(t)$ to get $\mathbf{r}(t) = \langle t^3 + C_1 t + D_1, \ln(t) + C_2 t + D_2 \rangle$. \ Use the two position data points: \ $\mathbf{r}(1) = \langle 1+C_1+D_1, \ln(1)+C_2+D_2 \rangle = \langle 1+C_1+D_1, C_2+D_2 \rangle = \langle -2, 0 \rangle$. \ $\mathbf{r}(2) = \langle 8+2C_1+D_1, \ln(2)+2C_2+D_2 \rangle = \langle 2, \ln(2) \rangle$. \ From the y-components: $C_2+D_2=0$ and $\ln(2)+2C_2+D_2=\ln(2) \Rightarrow 2C_2+D_2=0$. Subtracting these gives $C_2=0$, and so $D_2=0$. \ From the x-components: $1+C_1+D_1=-2 \Rightarrow C_1+D_1=-3$ and $8+2C_1+D_1=2 \Rightarrow 2C_1+D_1=-6$. Subtracting the first from the second gives $C_1=-3$, and so $D_1=0$. \ We need the constants for velocity, which are $C_1=-3$ and $C_2=0$. \ So, $\mathbf{v}(t) = \langle 3t^2 - 3, 1/t \rangle$. \ Finally, $\mathbf{v}(1) = \langle 3(1)^2 - 3, 1/1 \rangle = \langle 0, 1 \rangle$.
9. The velocity of a particle is $\mathbf{v}(t) = \langle 2t, 3t^2 \rangle$. Its position vector $\mathbf{r}(t)$ is such that $\mathbf{r}(1)$ is orthogonal to $\mathbf{v}(1)$. If the x-component of $\mathbf{r}(1)$ is 3, find the position vector $\mathbf{r}(t)$.
   Answer: $\langle t^2 + 2, t^3 - 3 \rangle$ First, integrate $\mathbf{v}(t)$ to find the general form of $\mathbf{r}(t)$:\ $\mathbf{r}(t) = \int \langle 2t, 3t^2 \rangle \,dt = \langle t^2 + C_1, t^3 + C_2 \rangle$.\ Next, use the given information to find the specific vector $\mathbf{r}(1)$. We have $\mathbf{v}(1) = \langle 2(1), 3(1)^2 \rangle = \langle 2, 3 \rangle$.\ Let $\mathbf{r}(1) = \langle 3, y_1 \rangle$. The condition that $\mathbf{r}(1)$ is orthogonal to $\mathbf{v}(1)$ means their dot product is zero: \ $\mathbf{r}(1) \cdot \mathbf{v}(1) = \langle 3, y_1 \rangle \cdot \langle 2, 3 \rangle = 0$.\ $3(2) + y_1(3) = 0 \Rightarrow 6 + 3y_1 = 0 \Rightarrow y_1 = -2$.\ So, we have the specific position $\mathbf{r}(1) = \langle 3, -2 \rangle$.\ Now use this to find the constants of integration:\ From $\mathbf{r}(t) = \langle t^2 + C_1, t^3 + C_2 \rangle$, we have $\mathbf{r}(1) = \langle 1^2 + C_1, 1^3 + C_2 \rangle = \langle 1+C_1, 1+C_2 \rangle$.\ Set this equal to $\langle 3, -2 \rangle$:\ $1 + C_1 = 3 \Rightarrow C_1 = 2$.\ $1 + C_2 = -2 \Rightarrow C_2 = -3$.\ Therefore, the position vector is $\mathbf{r}(t) = \langle t^2 + 2, t^3 - 3 \rangle$.
10. Consider $\mathbf{p}(t) = \langle t^2 \sin(t^3), e^{t^2} \cos(t), \frac{t}{\sqrt{1-t^2}} \rangle$. If $\mathbf{Q}(x) = \int_0^x \mathbf{p}(t) \, dt$, what is $\frac{d^2}{dx^2}[\mathbf{Q}(x) \cdot \mathbf{Q}(x)]$ at $x = 0$?
    Answer: $2$ By the Fundamental Theorem of Calculus, $\mathbf{Q}'(x) = \mathbf{p}(x)$ and $\mathbf{Q}(0) = \mathbf{0}$. Let $f(x) = \mathbf{Q}(x) \cdot \mathbf{Q}(x) = ||\mathbf{Q}(x)||^2$. Then $f'(x) = 2\mathbf{Q}(x) \cdot \mathbf{Q}'(x) = 2\mathbf{Q}(x) \cdot \mathbf{p}(x)$. At $x = 0$: $f'(0) = 2\mathbf{Q}(0) \cdot \mathbf{p}(0) = 2\mathbf{0} \cdot \mathbf{p}(0) = 0$. Taking the second derivative: $f''(x) = 2\mathbf{Q}'(x) \cdot \mathbf{p}(x) + 2\mathbf{Q}(x) \cdot \mathbf{p}'(x) = 2\mathbf{p}(x) \cdot \mathbf{p}(x) + 2\mathbf{Q}(x) \cdot \mathbf{p}'(x) = 2||\mathbf{p}(x)||^2 + 2\mathbf{Q}(x) \cdot \mathbf{p}'(x)$. At $x = 0$: $f''(0) = 2||\mathbf{p}(0)||^2 + 2\mathbf{0} \cdot \mathbf{p}'(0) = 2||\mathbf{p}(0)||^2$. Since $\mathbf{p}(0) = \langle 0, 1, 0 \rangle$, we have $||\mathbf{p}(0)||^2 = 1$, so $f''(0) = 2$.
11. The velocity of a particle is given by $\mathbf{v}(t) = \langle \frac{1}{t+1}, e^{-t} \rangle$. What is the displacement of the particle from $t=0$ to $t=1$?
    Answer: $\langle \ln(2), 1 - e^{-1} \rangle$ Displacement is the definite integral of the velocity vector over the given interval. We need to calculate $\int_0^1 \mathbf{v}(t) \,dt$.\ $$\int_0^1 \langle \frac{1}{t+1}, e^{-t} \rangle \,dt = \left\langle \int_0^1 \frac{1}{t+1} \,dt, \int_0^1 e^{-t} \,dt \right\rangle$$ For the x-component: $\int_0^1 \frac{1}{t+1} \,dt = [\ln|t+1|]_0^1 = \ln(2) - \ln(1) = \ln(2)$.\ For the y-component: $\int_0^1 e^{-t} \,dt = [-e^{-t}]_0^1 = (-e^{-1}) - (-e^0) = -e^{-1} + 1 = 1 - e^{-1}$.\ Therefore, the displacement vector is $\langle \ln(2), 1 - e^{-1} \rangle$.
12. The rate of change of the velocity of a particle is given by $\mathbf{a}(t) = \langle 6, 6t \rangle$. What is the net change in velocity from $t=1$ to $t=3$?
    Answer: $\langle 12, 24 \rangle$ The net change in a quantity is the definite integral of its rate of change. The net change in velocity from $t=1$ to $t=3$ is given by $\Delta \mathbf{v} = \mathbf{v}(3) - \mathbf{v}(1) = \int_1^3 \mathbf{a}(t) \,dt$.\ $\int_1^3 \langle 6, 6t \rangle \,dt = \left\langle [6t]_1^3, [3t^2]_1^3 \right\rangle$.\ Evaluating the x-component: $6(3) - 6(1) = 18 - 6 = 12$.\ Evaluating the y-component: $3(3)^2 - 3(1)^2 = 3(9) - 3(1) = 27 - 3 = 24$.\ Therefore, the net change in velocity is $\langle 12, 24 \rangle$. The constants of integration for velocity cancel out when computing a definite integral, so the initial velocity is not needed.
13. A particle's velocity is $\mathbf{v}(t) = \langle \frac{1}{1+t^2}, \frac{1}{\sqrt{1-t^2}} \rangle$ for $-1 < t < 1$. If $\mathbf{r}(0) = \langle 1, -1 \rangle$, find $\mathbf{r}(t)$.
    Answer: $\langle \arctan(t) + 1, \arcsin(t) - 1 \rangle$ To find the position vector $\mathbf{r}(t)$, we must integrate the velocity vector $\mathbf{v}(t)$.\ $\mathbf{r}(t) = \int \langle \frac{1}{1+t^2}, \frac{1}{\sqrt{1-t^2}} \rangle \,dt$.\ The antiderivatives are standard inverse trigonometric functions: \ $\mathbf{r}(t) = \langle \arctan(t) + C_1, \arcsin(t) + C_2 \rangle$.\ Now, use the initial condition $\mathbf{r}(0) = \langle 1, -1 \rangle$ to find the constants.\ For the x-component: $\arctan(0) + C_1 = 1 \Rightarrow 0 + C_1 = 1 \Rightarrow C_1 = 1$.\ For the y-component: $\arcsin(0) + C_2 = -1 \Rightarrow 0 + C_2 = -1 \Rightarrow C_2 = -1$.\ Therefore, the position vector is $\mathbf{r}(t) = \langle \arctan(t) + 1, \arcsin(t) - 1 \rangle$.
14. Let $\mathbf{R}(t)$ be a vector-valued function representing position. Its rate of change is given by $\mathbf{R}'(t) = \langle \frac{1}{t+1}, \frac{2t}{t^2+1} \rangle$. If $\mathbf{R}(0) = \langle 1, 1 \rangle$, find the vector representing the net change in position from $t=0$ to $t=1$, which is $\mathbf{R}(1) - \mathbf{R}(0)$.
    Answer: $\langle \ln(2), \ln(2) \rangle$ The question asks for the net change in position, $\mathbf{R}(1) - \mathbf{R}(0)$. By the Fundamental Theorem of Calculus for vector-valued functions, this is equal to the definite integral of the rate of change, $\int_0^1 \mathbf{R}'(t) \,dt$. The initial condition $\mathbf{R}(0) = \langle 1, 1 \rangle$ is extra information that is not needed to find the net change, but would be needed to find the final position $\mathbf{R}(1)$.\ We calculate $\int_0^1 \langle \frac{1}{t+1}, \frac{2t}{t^2+1} \rangle \,dt$.\ For the x-component: $\int_0^1 \frac{1}{t+1} \,dt = [\ln(t+1)]_0^1 = \ln(2) - \ln(1) = \ln(2)$.\ For the y-component, we use a u-substitution with $u = t^2+1$, so $du = 2t \,dt$. The integral is $\int \frac{1}{u} du = \ln(u) = \ln(t^2+1)$.\ $\int_0^1 \frac{2t}{t^2+1} \,dt = [\ln(t^2+1)]_0^1 = \ln(1^2+1) - \ln(0^2+1) = \ln(2) - \ln(1) = \ln(2)$.\ The net change vector is $\langle \ln(2), \ln(2) \rangle$.
15. The acceleration of a particle is $\mathbf{a}(t) = \langle \sin(t), -\cos(t) \rangle$. The velocity of the particle at $t=\pi$ is $\mathbf{v}(\pi) = \langle 1, 1 \rangle$. What is the velocity vector $\mathbf{v}(t)$?
    Answer: $\langle -\cos(t), 1 - \sin(t) \rangle$ To find the velocity vector $\mathbf{v}(t)$, integrate the acceleration vector $\mathbf{a}(t)$:\ $\mathbf{v}(t) = \int \langle \sin(t), -\cos(t) \rangle \,dt = \langle -\cos(t) + C_1, -\sin(t) + C_2 \rangle$.\ Use the given condition $\mathbf{v}(\pi) = \langle 1, 1 \rangle$ to solve for the constants $C_1$ and $C_2$.\ For the x-component: $-\cos(\pi) + C_1 = 1 \Rightarrow -(-1) + C_1 = 1 \Rightarrow 1 + C_1 = 1 \Rightarrow C_1 = 0$.\ For the y-component: $-\sin(\pi) + C_2 = 1 \Rightarrow 0 + C_2 = 1 \Rightarrow C_2 = 1$.\ Substituting the constants back into the expression for $\mathbf{v}(t)$ gives $\mathbf{v}(t) = \langle -\cos(t), -\sin(t) + 1 \rangle = \langle -\cos(t), 1 - \sin(t) \rangle$.
16. The velocity of a particle is $\mathbf{v}(t) = \langle 2t+1, 3 \rangle$. The particle is at the point $(5, 8)$ when $t=2$. Where is the particle at $t=0$?
    Answer: $(-1, 2)$ We can find the position at $t=0$ using the relationship $\mathbf{r}(a) = \mathbf{r}(b) - \int_a^b \mathbf{v}(t) \,dt$.\ Here, $a=0$ and $b=2$. We have $\mathbf{r}(0) = \mathbf{r}(2) - \int_0^2 \mathbf{v}(t) \,dt$.\ First, calculate the integral representing the displacement from $t=0$ to $t=2$:\ $\int_0^2 \langle 2t+1, 3 \rangle \,dt = \left\langle [t^2+t]_0^2, [3t]_0^2 \right\rangle$.\ Evaluating the definite integrals:\ For the x-component: $(2^2+2) - (0^2+0) = 6$.\ For the y-component: $3(2) - 3(0) = 6$.\ The displacement from $t=0$ to $t=2$ is $\langle 6, 6 \rangle$.\ Now, solve for $\mathbf{r}(0)$:\ $\mathbf{r}(0) = \mathbf{r}(2) - \langle 6, 6 \rangle = \langle 5, 8 \rangle - \langle 6, 6 \rangle = \langle -1, 2 \rangle$.\ So the particle is at $(-1, 2)$ at $t=0$.
17. Let $\mathbf{w}(t) = \langle \cos(t^2), \sin(t^2), t \rangle$. If the arc length of $\mathbf{w}(t)$ from $t = 0$ to $t = a$ is given by $L = \int_0^a \sqrt{4t^2 \sin^2(t^2) + 4t^2 \cos^2(t^2) + 1} \, dt$, what can you conclude about the relationship between this curve and the standard helix?
    Answer: The arc length simplifies to $L = \int_0^a \sqrt{4t^2 + 1} \, dt$ To find the arc length, we need $||\mathbf{w}'(t)||$. Computing: $\mathbf{w}'(t) = \langle -2t\sin(t^2), 2t\cos(t^2), 1 \rangle$. Then $||\mathbf{w}'(t)||^2 = (-2t\sin(t^2))^2 + (2t\cos(t^2))^2 + 1^2 = 4t^2\sin^2(t^2) + 4t^2\cos^2(t^2) + 1 = 4t^2(\sin^2(t^2) + \cos^2(t^2)) + 1 = 4t^2 + 1$. Therefore $||\mathbf{w}'(t)|| = \sqrt{4t^2 + 1}$, and the arc length is $L = \int_0^a \sqrt{4t^2 + 1} \, dt$. The given expression in the problem statement correctly simplifies to this form using the trigonometric identity $\sin^2(t^2) + \cos^2(t^2) = 1$. Options A, C, and D make claims about curvature, geometry, and torsion that are not directly supported by the arc length calculation.
18. A particle's acceleration is $\mathbf{a}(t) = \langle 6t, 2 \rangle$. Given the initial conditions $\mathbf{v}(1) = \langle 3, 0 \rangle$ and $\mathbf{r}(1) = \langle 0, 1 \rangle$, find the particle's position $\mathbf{r}(2)$.
    Answer: $\langle 7, 2 \rangle$ First, integrate the acceleration vector $\mathbf{a}(t)$ to find the velocity vector $\mathbf{v}(t)$.\ $\mathbf{v}(t) = \int \langle 6t, 2 \rangle \,dt = \langle 3t^2 + C_1, 2t + C_2 \rangle$.\ Use the condition $\mathbf{v}(1) = \langle 3, 0 \rangle$ to find the constants.\ For the x-component: $3(1)^2 + C_1 = 3 \Rightarrow C_1 = 0$.\ For the y-component: $2(1) + C_2 = 0 \Rightarrow C_2 = -2$.\ So, $\mathbf{v}(t) = \langle 3t^2, 2t - 2 \rangle$.\ Next, integrate the velocity vector $\mathbf{v}(t)$ to find the position vector $\mathbf{r}(t)$.\ $\mathbf{r}(t) = \int \langle 3t^2, 2t - 2 \rangle \,dt = \langle t^3 + D_1, t^2 - 2t + D_2 \rangle$.\ Use the condition $\mathbf{r}(1) = \langle 0, 1 \rangle$ to find the constants.\ For the x-component: $(1)^3 + D_1 = 0 \Rightarrow D_1 = -1$.\ For the y-component: $(1)^2 - 2(1) + D_2 = 1 \Rightarrow -1 + D_2 = 1 \Rightarrow D_2 = 2$.\ So, $\mathbf{r}(t) = \langle t^3 - 1, t^2 - 2t + 2 \rangle$.\ Finally, evaluate $\mathbf{r}(2)$:\ $\mathbf{r}(2) = \langle (2)^3 - 1, (2)^2 - 2(2) + 2 \rangle = \langle 8 - 1, 4 - 4 + 2 \rangle = \langle 7, 2 \rangle$.
19. A particle's acceleration is $\mathbf{a}(t) = \langle 1/t, 0 \rangle$ for $t > 0$. Given $\mathbf{v}(1) = \langle 1, 2 \rangle$ and $\mathbf{r}(1) = \langle 1, 1 \rangle$, find $\mathbf{r}(e)$.
    Answer: $\langle e+1, 2e-1 \rangle$ First, integrate acceleration to find velocity: \ $\mathbf{v}(t) = \int \langle 1/t, 0 \rangle \,dt = \langle \ln(t) + C_1, C_2 \rangle$ for $t>0$.\ Use $\mathbf{v}(1) = \langle 1, 2 \rangle$ to find the constants: \ $\ln(1) + C_1 = 1 \Rightarrow 0 + C_1 = 1 \Rightarrow C_1 = 1$. \ $C_2 = 2$. \ So, $\mathbf{v}(t) = \langle \ln(t) + 1, 2 \rangle$.\ Next, integrate velocity to find position: \ $\mathbf{r}(t) = \int \langle \ln(t) + 1, 2 \rangle \,dt$. The integral of $\ln(t)$ is $t\ln(t) - t$.\ So, $\mathbf{r}(t) = \langle t\ln(t) - t + t + D_1, 2t + D_2 \rangle = \langle t\ln(t) + D_1, 2t + D_2 \rangle$.\ Use $\mathbf{r}(1) = \langle 1, 1 \rangle$ to find the constants: \ $1\ln(1) + D_1 = 1 \Rightarrow 0 + D_1 = 1 \Rightarrow D_1 = 1$. \ $2(1) + D_2 = 1 \Rightarrow D_2 = -1$.\ So, $\mathbf{r}(t) = \langle t\ln(t) + 1, 2t - 1 \rangle$.\ Finally, evaluate at $t=e$: \ $\mathbf{r}(e) = \langle e\ln(e) + 1, 2e - 1 \rangle = \langle e(1) + 1, 2e - 1 \rangle = \langle e+1, 2e-1 \rangle$.
20. A particle has velocity $\mathbf{v}(t) = \langle t\cos(t), t\sin(t) \rangle$. If the particle is at the origin at $t=0$, what is its position at $t = \pi/2$?
    Answer: $\langle \frac{\pi}{2} - 1, 1 \rangle$ The position $\mathbf{r}(t)$ is found by integrating $\mathbf{v}(t)$. Since $\mathbf{r}(0) = \langle 0,0 \rangle$, the position at $t=\pi/2$ is given by the definite integral $\mathbf{r}(\pi/2) = \int_0^{\pi/2} \mathbf{v}(t) \,dt$.\ Both components require integration by parts, $\int u \,dv = uv - \int v \,du$.\ For the x-component, $\int t\cos(t) \,dt$: Let $u=t, dv=\cos(t)dt$. Then $du=dt, v=\sin(t)$. The integral is $t\sin(t) - \int \sin(t) \,dt = t\sin(t) + \cos(t)$.\ Evaluating from 0 to $\pi/2$: $[t\sin(t) + \cos(t)]_0^{\pi/2} = (\frac{\pi}{2}\sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2})) - (0 + \cos(0)) = (\frac{\pi}{2} + 0) - (1) = \frac{\pi}{2} - 1$.\ For the y-component, $\int t\sin(t) \,dt$: Let $u=t, dv=\sin(t)dt$. Then $du=dt, v=-\cos(t)$. The integral is $-t\cos(t) - \int -\cos(t) \,dt = -t\cos(t) + \sin(t)$.\ Evaluating from 0 to $\pi/2$: $[-t\cos(t) + \sin(t)]_0^{\pi/2} = (-\frac{\pi}{2}\cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})) - (0 + \sin(0)) = (0 + 1) - (0) = 1$.\ Thus, the position is $\langle \frac{\pi}{2} - 1, 1 \rangle$.

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