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Section Vector Functions

Worksheet FUN-3 Reading

All of the readings for this course come from the OpenStax Calculus Volume 3 textbook, which is available for free online at https://openstax.org/details/books/calculus-volume-3. You can read the textbook online or download a PDF version. The textbook is also available in print from various retailers if you prefer a physical copy.
For this lesson, you should read sections 3.1 Vector Valued Functions and Space Curves, and 3.2 Calculus of Vector Valued Functions. You should read through the sections carefully, making sure to understand the definitions, examples, and key ideas. You should also work through the exercises at the end of each section to test your understanding and practice applying the concepts. Some of the exercises at the end of the section are assigned in the WP assignments.

Worksheet FUN-3 Videos

Vector-Valued Functions
Calculus of Vector-Valued Functions

Subsection FUN-3 Written Practice

Worksheet Worksheet

INFORMATION ABOUT ALL WRITTEN PRACTICE (WP) ASSIGNMENTS:
THE WP ASSIGNMENTS ARE DESIGNED TO PREPARE YOU FOR THE QUIZZES AND ASSESSMENTS
  • Work on every problem on every assignment.
  • If you get stuck or do not understand a solution, ask questions.
  • For each problem, try it once or twice before looking at solutions or asking for help.
  • Mark the problems you did correctly on the first try.
  • Highlight problems where you got help (using solutions, videos, a tutor, etc) -These are the problem types that you will need more practice on to be ready for the assessments.
  • Understand that just copying the solutions instead of working through the problems will greatly reduce your chances of success in this class.
Vector-Valued Functions: Complete problems 11, 15, 19, 23, 25, 33
Calculus of Vector-Valued Functions: Complete problems 41, 43, 47, 55, 63, 67, 71, 75
In addition to these problems, on separate paper complete the following exploration. This link Velocity and Acceleration will take you to the exploration. On the right side of the screen, you will be given some instructions and questions to answer. Follow the instructions and write out your answers to each of the questions.

Worksheet FUN-3 Sample Outcomes

For the quiz on this outcome you would be given 30 minutes to complete 2 problems. Here are samples of the types of problems you will encounter:

1.

Find the equation for the line tangent to \(\vec{r}(t)=e^{2t}\hat{i}+e^t\hat{j}\) at \(t=0\text{.}\)Leave the answer in vector form.
Hint.
To find the equation of the tangent line, you need to compute the derivative of the vector-valued function and evaluate it at the given point.
Answer.
Tangent line: \(\lt 1+2t,1+t \gt\)
Solution.
First we need to find \(\vec{r}(0)\text{:}\) \(\vec{r}(0)=1\hat{i}+1\hat{j}\text{.}\)
Next, we compute the derivative: \(\vec{r}^{\prime}(t)=2e^{2t}\hat{i}+e^t\hat{j}\text{.}\)
Then we evaluate at \(t=0\text{:}\) \(\vec{r}^{\prime}(0)=2\hat{i}+1\hat{j}\text{.}\) The tangent line is \(\lt 1+2t,1+t \gt\text{.}\)

2.

Find the distance traveled along the helix given by \(\vec{r}(t)=\lt 5\cos(2t),5\sin(2t),2t \gt\) as \(t\) varies from 0 to \(\frac{3\pi}{2}\text{.}\)
Hint.
To find the distance traveled, you need to compute the arc length integral over the interval in terms of \(t\text{.}\)
Answer.
Distance traveled: \(3\pi\sqrt{26}\) units
Solution.
To set up the arc length integral, we first compute the derivative of \(\vec{r}(t)\text{:}\)
\(\vec{r}^{\prime}(t)=\lt -10\sin(2t),10\cos(2t),2 \gt\text{.}\)
The magnitude of this vector is:
\(|\vec{r}^{\prime}(t)|=\sqrt{100\sin^2(2t)+100\cos^2(2t)+4}=\sqrt{104}=2\sqrt{26}\text{.}\)
The arc length is then:
\(S=\int_0^{3\pi/2} 2\sqrt{26} dt\)
\(S=2\sqrt{26}t | _0^{3\pi/2}=3\pi\sqrt{26}\) units.

3.

A particle is moving along a curve in space with velocity (measured in m/s) of \(\vec{v}(t)=\lt 2t,3t^2,4 \gt\text{.}\)
(a)
Find the position vector of the particle, \(\vec{r}(t)\text{,}\) if \(\vec{r}(3)=\lt 2,-2,6 \gt\text{.}\)
Hint.
To find the position vector, integrate the velocity vector and use the initial condition.
Answer.
\(\vec{r}(t)=\lt t^2-7, t^3-29, 4t-6 \gt\)
Solution.
Integrating the velocity vector:
\(\vec{r}(t)=\lt \int 2t \, dt, \int 3t^2 \, dt, \int 4 \, dt \gt\)
\(=\lt t^2, t^3, 4t \gt + \vec{c}\text{.}\)
Using the initial condition \(\vec{r}(3)=\lt 2,-2,6 \gt\text{:}\)
\(\lt 9, 27, 12 \gt + \vec{c}=\lt 2,-2,6 \gt\)
\(\vec{c}=\lt -7, -29, -6 \gt\text{.}\)
Therefore: \(\vec{r}(t)=\lt t^2-7, t^3-29, 4t-6 \gt\text{.}\)
(b)
Find the unit tangent vector at \(t=3\text{.}\)
Hint.
The unit tangent vector is the derivative of the position vector, normalized.
Answer.
\(\vec{T}(3)=\lt \frac{6\sqrt{781}}{781}, \frac{27\sqrt{781}}{781}, \frac{4\sqrt{781}}{781} \gt\)
Solution.
First, we find the derivative of the position vector. But notice that \(\vec{r}^{\prime}(t)=\vec{v}(t)\text{:}\)
\(\vec{v}(t)=\lt 2t,3t^2,4 \gt\text{.}\)
Then we evaluate at \(t=3\text{:}\)
\(\vec{v}(3)=\lt 6,27,4 \gt\text{.}\)
The magnitude is:
\(|\vec{v}(3)|=\sqrt{36+729+16}=\sqrt{781}\text{.}\)
The unit tangent vector is:
\(\vec{T}(3)=\frac{\lt 6,27,4 \gt}{\sqrt{781}} = \lt \frac{6\sqrt{781}}{781}, \frac{27\sqrt{781}}{781}, \frac{4\sqrt{781}}{781} \gt\text{.}\)