11.2.1How fast am I going in component form?

Applied Calculus in Component Form

11-40.

A tiny marble is rolling around an $xy$ -plane such that at any time, $t > 0$ , in seconds, its position is modeled by the vector-valued function $\langle \frac { 1 } { 25 } t ^ { 3 } - 50 , \frac { 1 } { 2 } t ^ { 2 } + 30 \rangle$ .

Roberta is wondering where the marble will be located $10$ seconds after it starts rolling. She evaluates the vector at $t = 10$ and concludes that the marble will be at the point $\left(–10, 80\right)$ .

However, her teammate Eleanor is not convinced. “You are assuming that the initial position of the marble when $t = 0$ is at $\left(–50, 30\right)$ . But we are not given that information.”

Who is correct? Be prepared to share your reasoning with the class.
As it turns out, the marble is located at coordinate point $\left(25, 12\right)$ at $t = 0$ . Write a parametric function that represents the position of the marble at any time $t$ . Then use it to determine the actual position of the marble when $t = 10$ .
In the context of this problem, compare and contrast a vector-valued function that represents position with a parametric function that represents position. What information does a parametric function give that a vector does not? Likewise, what information does a vector give us that a parametric equation does not?

11-41.

Julio is having a great time at Speedland, especially because it is a school day and his Calculus teacher, Ms. Koolnerd, has brought the whole class to the park for a fieldtrip.

Julio is on the roller coaster shown at right. Just as the roller coaster starts descending, Ms. Koolnerd taps him on the shoulder from the cart behind him and hands him a card. On it was written as a parametric function that models the path of the coaster.

$x(t) = 6t^2 - 0.4t^3$ , $y(t) = 90 + 80 \cos( \frac { \pi } { 100 } ( 6 t ^ { 2 } - 0.4 t ^ { 3 } ) )$
where $x$ and $y$ are in feet above the ground and $t$ is time in seconds.

First quadrant curve, starting almost at top of y axis, running fairly level, with car on curve as the curve starts to fall, then changing from concave down to concave up, turning & rising continuing up & right.

Julio quickly whips out his graphing calculator and sketches a graph of the parametric curve. You should do the same.

Determine the height of the coaster when $t = 0$ . Is that the highest it ever goes? How low does it go? Be prepared to share your strategy to answer these questions.
Julio knows that the velocity of the coaster will change as it descends. “The lower we go, the faster we will fall!” But what about the horizontal component? Will the horizontal velocity change or stay constant? With your team, use $x^\prime (t)$ to answer Julio’s question.
Julio wants to program his camera to take a picture of himself while the coaster is descending. He estimates that $t = 3$ will be a good time to take the shot. Write a parametric function that represents the roller coaster’s velocity at any time $t$ . Then use the equations to calculate the velocity at $t = 3$ in vector form. Is the coaster descending at this time? How do you know?
Julio is curious if he will be moving fast at $t = 3$ , but he finds it difficult to use the velocity vector to answer that question. “What I need,” thinks Julio, “is a scalar measurement.”
Julio decides to evaluate $\| v ( t ) \| = \sqrt { ( \frac { d x } { d t } ) ^ { 2 } + ( \frac { d y } { d t } ) ^ { 2 } }$ at $t = 3$ to calculate the magnitude of the velocity vector at that time. Determine this measurement. Be sure to include appropriate units and explain its significance in the context of the problem.

11-42.

Georgia drops her gum while gazing in awe at a bike rider racing down the street. At the moment that the gum hits the concrete, the bicycle wheel runs over the gum. The curve at right, called a cycloid, models the path her gum travels as the bicycle proceeds down the street.

The vector-valued function $\langle t - \operatorname { sin } ( t ) , 1 - \operatorname { cos } ( t ) \rangle$ can be used to identify the position of the gum in feet above the ground where $t$ is measured in seconds.

First quadrant, top part of circle, starting at the origin, ending at the x axis, repeating with another top part of circle, continuing to the right & up, point on left portion of first semicircle.

Georgia is curious about the shape of the path of her lost gum on an $xy$ -plane. At what time(s) will it be rising? At what time(s) will it be falling? She decides to examine the slopes of the lines tangent to the curve by writing a ratio:

$\frac { y ^ { \prime } ( t ) } { x ^ { \prime } ( t ) } = \frac { \frac { d y } { d t } } { \frac { d x } { d t } } = \frac { d y } { d x }$

With your team, try Georgia’s method. Calculate the slope of the line tangent to the cycloid’s path, $\frac { d y } { d x }$ , when $t =\frac { 2 \pi } { 3 }$ seconds. Is the gum rising or falling at that moment? Explain how you know.
Does the slope you found in part (a) represent the gum’s velocity at $t =\frac { 2 \pi } { 3 }$ ? If not, what does $\frac { d y } { d x }$ represent in the context of this problem?
Use component form to find an expression for the velocity of the gum at time $t =\frac { 2 \pi } { 3 }$ . Then write a note to Georgia about why the velocity vector must be expressed with two components.
If possible, calculate the speed of the gum when $t =\frac { 2 \pi } { 3 }$ . If it is not possible, explain
why not.

11-43.

Leslie wants to practice using derivatives to find the slopes of lines tangent to graphs of parametric equations, so her friend John makes up this example:

What is the slope of the line tangent to parametric curve given by $x(t) = 4 \cos(\frac { \pi } { 4 }− t^3)$ , $y(t) = 2 \sin(\frac { \pi } { 4 }− t^3)$ , at the point where $t = 0$ ?

“Piece of cake!” says Leslie.

What are $\frac { d x } { d t }$ and $\frac { d y } { d t }$ when $t = 0$ ?
“You sneak!” exclaims Leslie. “I can’t use $\frac { \frac { d y } { d t } } { \frac { d x } { d t } }$ to find $\frac { d y } { d t }$ .” Why not?
“You can’t stump me that easily,” says Leslie. “I’ll find a way to use rectilinear equations to do this.” With your team, help Leslie find another method to determine the slope at $t = 0$ , explicitly.
Write the equation of the tangent line for $t = 0$ .
In this situation, both $\frac { d x } { d t }$ and $\frac { d y } { d t }$ were zero. What would have been the slope if only one of them was zero?

11-44.

A point travels along the curve $y = xe^{x}$ with a horizontal velocity of $\frac { d x } { d t } = 4$ units per second. How fast is the point’s distance from the origin changing when $x = -1$ ? Homework Help ✎

11-45.

For any three points $A$ , $B$ , and $C$ , express $\vec { B C }$ in terms of $\vec { A B }$ and $\vec { A C }$ . Homework Help ✎

11-46.

What percent of the area inside the limaçon $r = 3 + 6\cos\left(θ\right)$ lies within the inner loop? 11-46 HW eTool (Desmos). Homework Help ✎

11-47.

Let $g(h(t)) = \cos\sqrt { t }$ , $g\left(t\right)= \cos\left(t\right)$ , and $h(t) = \sqrt { t }$ . Homework Help ✎

What are $\frac { d g } { d t }$ and $\frac { d h } { d t }$ ?

Verify that $\frac { d g } { d h } = \frac { \frac { d g } { d t } } { \frac { d h } { d t } }$ .

11-48.

Multiple Choice: If $⟨x\left(t\right), y\left(t\right)⟩$ is the position vector of an object at time $t$ , then the velocity vector is: Homework Help ✎

$\langle x ^ { \prime } ( t ) , y ^ { \prime } ( t ) \rangle$

$\langle \frac { d x } { d t } , \frac { d y } { d t } \rangle$

(a) and (b)

$\frac { d y } { d x }$

$\langle 1 , \frac { d y } { d x } \rangle$

11-49.

Multiple Choice: The arc length of $y = x^{3/2}$ from $x = 0$ to $x = 4$ is closest to: Homework Help ✎

$6.874$

$6.874$

$8.944$

$9.073$

11-50.

Multiple Choice: The interval of convergence of the power series $\displaystyle\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n - 1 } \frac { x ^ { n } } { n }$ is: Homework Help ✎

$(–∞, ∞)$

$[–1, 1]$

$[–1, 1)$

(–1, 1]

$\left(–1, 1\right)$

11-51.

Multiple Choice: No calculator! Determine which of the values below is the result when four right endpoint rectangles are used to calculate the area under $f\left(x\right) = 2x^{2} - 3x + 1$ over $1 ≤ x ≤ 5$ . Homework Help ✎

$32 \text{ un}^{2}$

$34 \text{ un}^{2}$

$50 \text{ un}^{2}$

$67 \text{ un}^{2}$

$70 \text{ un}^{2}$

11-52.

Let $x =\sqrt { 4 t ^ { 2 } + 1 }$ , $y = 2t$ , and $t > 0$ . An equivalent rectangular equation is: Homework Help ✎

$x =\sqrt { y ^ { 2 } + 1 }$

$x = y^{2} + 1$

$x = y + 1$

$y =\sqrt { x ^ { 2 } + 1 }$

$y = x + 1$

Calculus Third Edition

11.2.1How fast am I going in component form?

Position, Velocity, and Acceleration Vectors