For $0 \leq t \leq 5$, a particle is moving along a curve so that its position at time $t$ is $( x ( t ) , y ( t ) )$. At time $t = 1$, the particle is at position $( 2 , - 7 )$. It is known that $\frac { d x } { d t } = \sin \left( \frac { t } { t + 3 } \right)$ and $\frac { d y } { d t } = e ^ { \cos t }$.
(a) Write an equation for the line tangent to the curve at the point $( 2 , - 7 )$.
(b) Find the $y$-coordinate of the position of the particle at time $t = 4$.
(c) Find the total distance traveled by the particle from time $t = 1$ to time $t = 4$.
(d) Find the time at which the speed of the particle is 2.5. Find the acceleration vector of the particle at this time.

15. For any time $t \geq 0$, if the position of a particle in the $x y$-plane is given by $x = t ^ { 2 } + 1$ and $y = \ln ( 2 t + 3 )$, then the acceleration vector is
(A) $\left( 2 t , \frac { 2 } { ( 2 t + 3 ) } \right)$
(B) $\quad \left( 2 t , \frac { - 4 } { ( 2 t + 3 ) ^ { 2 } } \right)$
(C) $\quad \left( 2 , \frac { 4 } { ( 2 t + 3 ) ^ { 2 } } \right)$
(D) $\left( 2 , \frac { 2 } { ( 2 t + 3 ) ^ { 2 } } \right)$
(E) $\quad \left( 2 , \frac { - 4 } { ( 2 t + 3 ) ^ { 2 } } \right)$

A particle moves along the curve defined by the equation $y = x^{3} - 3x$. The $x$-coordinate of the particle, $x(t)$, satisfies the equation $\dfrac{dx}{dt} = \dfrac{1}{\sqrt{2t+1}}$, for $t \geq 0$ with initial condition $x(0) = -4$.
(a) Find $x(t)$ in terms of $t$.
(b) Find $\dfrac{dy}{dt}$ in terms of $t$.
(c) Find the location and speed of the particle at time $t = 4$.

A particle moves in the $xy$-plane so that its position at any time $t$, $0 \leq t \leq \pi$, is given by $x(t) = \frac{t^2}{2} - \ln(1+t)$ and $y(t) = 3\sin t$.
(a) Sketch the path of the particle in the $xy$-plane below. Indicate the direction of motion along the path.
(b) At what time $t$, $0 \leq t \leq \pi$, does $x(t)$ attain its minimum value? What is the position $(x(t), y(t))$ of the particle at this time?
(c) At what time $t$, $0 < t < \pi$, is the particle on the $y$-axis? Find the speed and the acceleration vector of the particle at this time.

4. A moving particle has position $( x ( t ) , y ( t ) )$ at time $t$. The position of the particle at time $t = 1$ is $( 2,6 )$, and the velocity vector at any time $t > 0$ is given by $\left( 1 - \frac { 1 } { t ^ { 2 } } , 2 + \frac { 1 } { t ^ { 2 } } \right)$.
(a) Find the acceleration vector at time $t = 3$.
(b) Find the position of the particle at time $t = 3$.
(c) For what time $t > 0$ does the line tangent to the path of the particle at $( x ( t ) , y ( t ) )$ have a slope of 8 ?
(d) The particle approaches a line as $t \rightarrow \infty$. Find the slope of this line. Show the work that leads to your conclusion.

An object moving along a curve in the $xy$-plane has position $(x(t), y(t))$ at time $t$ with $$\frac{dx}{dt} = \cos\left(t^3\right) \text{ and } \frac{dy}{dt} = 3\sin\left(t^2\right)$$ for $0 \leq t \leq 3$. At time $t = 2$, the object is at position $(4,5)$.
(a) Write an equation for the line tangent to the curve at $(4,5)$.
(b) Find the speed of the object at time $t = 2$.
(c) Find the total distance traveled by the object over the time interval $0 \leq t \leq 1$.
(d) Find the position of the object at time $t = 3$.

A particle moves in the $xy$-plane so that its position at any time $t$, for $-\pi \leq t \leq \pi$, is given by $x(t) = \sin(3t)$ and $y(t) = 2t$.
(a) Sketch the path of the particle in the $xy$-plane provided. Indicate the direction of motion along the path.
(b) Find the range of $x(t)$ and the range of $y(t)$.
(c) Find the smallest positive value of $t$ for which the $x$-coordinate of the particle is a local maximum. What is the speed of the particle at this time?
(d) Is the distance traveled by the particle from $t = -\pi$ to $t = \pi$ greater than $5\pi$? Justify your answer.

A particle moving along a curve in the plane has position $( x ( t ) , y ( t ) )$ at time $t$, where $$\frac { d x } { d t } = \sqrt { t ^ { 4 } + 9 } \text { and } \frac { d y } { d t } = 2 e ^ { t } + 5 e ^ { - t }$$ for all real values of $t$. At time $t = 0$, the particle is at the point $( 4,1 )$.
(a) Find the speed of the particle and its acceleration vector at time $t = 0$.
(b) Find an equation of the line tangent to the path of the particle at time $t = 0$.
(c) Find the total distance traveled by the particle over the time interval $0 \leq t \leq 3$.
(d) Find the $x$-coordinate of the position of the particle at time $t = 3$.

At time $t$, a particle moving in the $x y$-plane is at position $( x ( t ) , y ( t ) )$, where $x ( t )$ and $y ( t )$ are not explicitly given. For $t \geq 0 , \frac { d x } { d t } = 4 t + 1$ and $\frac { d y } { d t } = \sin \left( t ^ { 2 } \right)$. At time $t = 0 , x ( 0 ) = 0$ and $y ( 0 ) = - 4$. (a) Find the speed of the particle at time $t = 3$, and find the acceleration vector of the particle at time $t = 3$. (b) Find the slope of the line tangent to the path of the particle at time $t = 3$. (c) Find the position of the particle at time $t = 3$. (d) Find the total distance traveled by the particle over the time interval $0 \leq t \leq 3$.

An object moving along a curve in the $xy$-plane is at position $(x(t), y(t))$ at time $t$, where $$\frac{dx}{dt} = \sin^{-1}\left(1 - 2e^{-t}\right) \text{ and } \frac{dy}{dt} = \frac{4t}{1 + t^{3}}$$ for $t \geq 0$. At time $t = 2$, the object is at the point $(6, -3)$. (Note: $\sin^{-1} x = \arcsin x$)
(a) Find the acceleration vector and the speed of the object at time $t = 2$.
(b) The curve has a vertical tangent line at one point. At what time $t$ is the object at this point?
(c) Let $m(t)$ denote the slope of the line tangent to the curve at the point $(x(t), y(t))$. Write an expression for $m(t)$ in terms of $t$ and use it to evaluate $\lim_{t \rightarrow \infty} m(t)$.
(d) The graph of the curve has a horizontal asymptote $y = c$. Write, but do not evaluate, an expression involving an improper integral that represents this value $c$.

An object moving along a curve in the $x y$-plane is at position $( x ( t ) , y ( t ) )$ at time $t$ with $$\frac { d x } { d t } = \arctan \left( \frac { t } { 1 + t } \right) \text { and } \frac { d y } { d t } = \ln \left( t ^ { 2 } + 1 \right)$$ for $t \geq 0$. At time $t = 0$, the object is at position $( - 3 , - 4 )$. (Note: $\tan ^ { - 1 } x = \arctan x$ ) (a) Find the speed of the object at time $t = 4$. (b) Find the total distance traveled by the object over the time interval $0 \leq t \leq 4$. (c) Find $x ( 4 )$. (d) For $t > 0$, there is a point on the curve where the line tangent to the curve has slope 2 . At what time $t$ is the object at this point? Find the acceleration vector at this point.

2. The velocity vector of a particle moving in the $x y$-plane has components given by
$$\frac { d x } { d t } = 14 \cos \left( t ^ { 2 } \right) \sin \left( e ^ { t } \right) \text { and } \frac { d y } { d t } = 1 + 2 \sin \left( t ^ { 2 } \right) , \text { for } 0 \leq t \leq 1.5 .$$
At time $t = 0$, the position of the particle is $( - 2,3 )$.
(a) For $0 < t < 1.5$, find all values of $t$ at which the line tangent to the path of the particle is vertical.
(b) Write an equation for the line tangent to the path of the particle at $t = 1$.
(c) Find the speed of the particle at $t = 1$.
(d) Find the acceleration vector of the particle at $t = 1$.

$t$	0	2	4	6	8	10	12
$P ( t )$	0	46	53	57	60	62	63

[Figure]

A particle is moving along a curve so that its position at time $t$ is $(x(t), y(t))$, where $x(t) = t^2 - 4t + 8$ and $y(t)$ is not explicitly given. Both $x$ and $y$ are measured in meters, and $t$ is measured in seconds. It is known that $\frac{dy}{dt} = te^{t-3} - 1$.
(a) Find the speed of the particle at time $t = 3$ seconds.
(b) Find the total distance traveled by the particle for $0 \leq t \leq 4$ seconds.
(c) Find the time $t$, $0 \leq t \leq 4$, when the line tangent to the path of the particle is horizontal. Is the direction of motion of the particle toward the left or toward the right at that time? Give a reason for your answer.
(d) There is a point with $x$-coordinate 5 through which the particle passes twice. Find each of the following.
(i) The two values of $t$ when that occurs
(ii) The slopes of the lines tangent to the particle's path at that point
(iii) The $y$-coordinate of that point, given $y(2) = 3 + \frac{1}{e}$

For $t \geq 0$, a particle is moving along a curve so that its position at time $t$ is $(x(t), y(t))$. At time $t = 2$, the particle is at position $(1, 5)$. It is known that $\frac{dx}{dt} = \frac{\sqrt{t+2}}{e^{t}}$ and $\frac{dy}{dt} = \sin^{2} t$.
(a) Is the horizontal movement of the particle to the left or to the right at time $t = 2$? Explain your answer. Find the slope of the path of the particle at time $t = 2$.
(b) Find the $x$-coordinate of the particle's position at time $t = 4$.
(c) Find the speed of the particle at time $t = 4$. Find the acceleration vector of the particle at time $t = 4$.
(d) Find the distance traveled by the particle from time $t = 2$ to $t = 4$.

For $t \geq 0$, a particle is moving along a curve so that its position at time $t$ is $( x ( t ) , y ( t ) )$. At time $t = 2$, the particle is at position $( 1,5 )$. It is known that $\frac { d x } { d t } = \frac { \sqrt { t + 2 } } { e ^ { t } }$ and $\frac { d y } { d t } = \sin ^ { 2 } t$.
(a) Is the horizontal movement of the particle to the left or to the right at time $t = 2$ ? Explain your answer. Find the slope of the path of the particle at time $t = 2$.
(b) Find the $x$-coordinate of the particle's position at time $t = 4$.
(c) Find the speed of the particle at time $t = 4$. Find the acceleration vector of the particle at time $t = 4$.
(d) Find the distance traveled by the particle from time $t = 2$ to $t = 4$.

At time $t \geq 0$, a particle moving along a curve in the $xy$-plane has position $( x ( t ) , y ( t ) )$ with velocity vector $v ( t ) = \left( \cos \left( t ^ { 2 } \right) , e ^ { 0.5 t } \right)$. At $t = 1$, the particle is at the point $( 3 , 5 )$.
(a) Find the $x$-coordinate of the position of the particle at time $t = 2$.
(b) For $0 < t < 1$, there is a point on the curve at which the line tangent to the curve has a slope of 2. At what time is the object at that point?
(c) Find the time at which the speed of the particle is 3.
(d) Find the total distance traveled by the particle from time $t = 0$ to time $t = 1$.

For time $t \geq 0$, a particle moves in the $x y$-plane with position $( x ( t ) , y ( t ) )$ and velocity vector $\left\langle ( t - 1 ) e ^ { t ^ { 2 } } , \sin \left( t ^ { 1.25 } \right) \right\rangle$. At time $t = 0$, the position of the particle is $( - 2,5 )$.
(a) Find the speed of the particle at time $t = 1.2$. Find the acceleration vector of the particle at time $t = 1.2$.
(b) Find the total distance traveled by the particle over the time interval $0 \leq t \leq 1.2$.
(c) Find the coordinates of the point at which the particle is farthest to the left for $t \geq 0$. Explain why there is no point at which the particle is farthest to the right for $t \geq 0$.

A particle moving along a curve in the $x y$-plane is at position $( x ( t ) , y ( t ) )$ at time $t > 0$. The particle moves in such a way that $\frac { d x } { d t } = \sqrt { 1 + t ^ { 2 } }$ and $\frac { d y } { d t } = \ln \left( 2 + t ^ { 2 } \right)$. At time $t = 4$, the particle is at the point $( 1,5 )$.
(a) Find the slope of the line tangent to the path of the particle at time $t = 4$.
(b) Find the speed of the particle at time $t = 4$, and find the acceleration vector of the particle at time $t = 4$.
(c) Find the $y$-coordinate of the particle's position at time $t = 6$.
(d) Find the total distance the particle travels along the curve from time $t = 4$ to time $t = 6$.

For $0 \leq t \leq \pi$, a particle is moving along the curve shown so that its position at time $t$ is $(x(t), y(t))$, where $x(t)$ is not explicitly given and $y(t) = 2\sin t$. It is known that $\frac{dx}{dt} = e^{\cos t}$. At time $t = 0$, the particle is at position $(1, 0)$.
(a) Find the acceleration vector of the particle at time $t = 1$. Show the setup for your calculations.
(b) For $0 \leq t \leq \pi$, find the first time $t$ at which the speed of the particle is 1.5. Show the work that leads to your answer.
(c) Find the slope of the line tangent to the path of the particle at time $t = 1$. Find the $x$-coordinate of the position of the particle at time $t = 1$. Show the work that leads to your answers.
(d) Find the total distance traveled by the particle over the time interval $0 \leq t \leq \pi$. Show the setup for your calculations.

A particle moving along a curve in the $xy$-plane has position $(x(t), y(t))$ at time $t$ seconds, where $x(t)$ and $y(t)$ are measured in centimeters. It is known that $x'(t) = 8t - t^2$ and $y'(t) = -t + \sqrt{t^{1.2} + 20}$. At time $t = 2$ seconds, the particle is at the point $(3, 6)$.
(a) Find the speed of the particle at time $t = 2$ seconds. Show the setup for your calculations.
(b) Find the total distance traveled by the particle over the time interval $0 \leq t \leq 2$. Show the setup for your calculations.
(c) Find the $y$-coordinate of the position of the particle at the time $t = 0$. Show the setup for your calculations.
(d) For $2 \leq t \leq 8$, the particle remains in the first quadrant. Find all times $t$ in the interval $2 \leq t \leq 8$ when the particle is moving toward the $x$-axis. Give a reason for your answer.

[Calculus] The position $( x , y )$ of a point P moving on the coordinate plane at time $t$ is given by $$\left\{ \begin{array} { l } x = 4 ( \cos t + \sin t ) \\ y = \cos 2 t \end{array} \quad ( 0 \leqq t \leqq 2 \pi ) \right.$$ When the distance traveled by point P from $t = 0$ to $t = 2 \pi$ is $a \pi$, find the value of $a ^ { 2 }$. [4 points]

A point P moving on the coordinate plane has position $( x , y )$ at time $t$ $(t > 0)$ given by $$x = t - \frac { 2 } { t } , \quad y = 2 t + \frac { 1 } { t }$$ What is the speed of point P at time $t = 1$? [3 points]
(1) $2 \sqrt { 2 }$
(2) 3
(3) $\sqrt { 10 }$
(4) $\sqrt { 11 }$
(5) $2 \sqrt { 3 }$

The position $\mathrm { P } ( x , y )$ of a point P moving on the coordinate plane at time $t ( 0 < t < \pi )$ is given by $$x = \sqrt { 3 } \sin t , \quad y = 2 \cos t - 5$$ At time $t = \alpha ( 0 < \alpha < \pi )$, the velocity $\vec { v }$ of point P and $\overrightarrow { \mathrm { OP } }$ are parallel. What is the value of $\cos \alpha$? (Here, O is the origin.) [4 points]
(1) $\frac { 1 } { 10 }$
(2) $\frac { 1 } { 5 }$
(3) $\frac { 3 } { 10 }$
(4) $\frac { 2 } { 5 }$
(5) $\frac { 1 } { 2 }$

The position $( x , y )$ of a point P moving on the coordinate plane at time $t ( t \geq 0 )$ is $$x = 1 - \cos 4 t , y = \frac { 1 } { 4 } \sin 4 t.$$ When the speed of point P is maximum, find the magnitude of the acceleration of point P. [3 points]

A point P moving on the coordinate plane has position $( x , y )$ at time $t \left( 0 < t < \frac { \pi } { 2 } \right)$ given by
$$x = t + \sin t \cos t , \quad y = \tan t$$
What is the minimum speed of point P for $0 < t < \frac { \pi } { 2 }$? [3 points]
(1) 1
(2) $\sqrt { 3 }$
(3) 2
(4) $2 \sqrt { 2 }$
(5) $2 \sqrt { 3 }$