Exercise 11 solutions | Ludek Cizinsky

Exercises from the book

🔹 13.2.4b

We are given a function \(r\) such that \(r(t) = \langle t^2, t^3 \rangle\).

Solution

a) (bonus) First, we are supposed to sketch the curve. You can look at the sketch here.

b) Second, we are supposed find \(r'(t)\). This can be done as follows:

\[r'(t) = \langle f'(t), g'(t) \rangle = \langle 2t, 3t^2 \rangle\]

c) (bonus) Third, we are supposed to sketch the position vector \(r(t)\) and tangent vector \(r'(t)\) at \(t = 1\). See this visulization.

🔹 13.2.11

Find the derivative of the vector function \(r(t) = t^2i + cos(t^2)j + sin^2(t)k\).

Solution

First, we know that \(i, j, k\) represent basis unit vectors for the \(3D\) space. For example, \(i = (1, 0, 0)\) etc. Thus, we know that \(r\) can be written as follows: \(r(t) = \langle t^2, cos(t^2), sin^2(t) \rangle\). We can use the same rule, as above (refer to the book page 898 and theorem 2), therefore:

\[r'(t) = \langle 2t, -2tsin(t^2), 2sin(t)cos(t)\rangle\]

You can also see visualization here.

🔹 13.3.1

We are given a curve \(r(t) = \langle 3 - t, 2t, 4t + 1\rangle \text{ where } 1 \leq t \leq 3\).

Solution

a) First, compute arc length using the formula provided in the book:

\[L = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2} dt\]

Therefore, we can write:

\[L = \int_1^3 \sqrt{(-1)^2 + (2)^2 + (4)^2} dt = \int_1^3 \sqrt{21} = 3\sqrt{21} - \sqrt{21} = 2\sqrt{21}\]

b) Second, let’s compute the length using a distance formula:

\[L2 = |v_1 - v_2|\]

where \(v_1 = r(1)\) and \(v_2 = r(3)\). Therefore, we can write:

\[L2 = \sqrt{(2 - 0)^2 + (2 - 6)^2 + (5 - 15)^2} = \sqrt{84} = 2\sqrt{21}\]

This is strange, we would certainly expect to get at least slightly different result since the b) is just an approxmation. You can see the reason why here.

🔹 13.4.3

Find the velocity, acceleration and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of \(t\):

\[r(t) = \langle -\frac{1}{2}t^2, t\rangle \text{, t = 2}\]

Solution

We know that velocity corresponds to \(r'(t) = \langle -t, 1 \rangle\), acceleration to \(\langle -1, 0 \rangle\) and speed is given as \(\|r'(t)\| = \sqrt{t^2 + 1}\). Find the visualization here.

🔹 13.4.36

Answer the following questions:

a) If a particle moves along a straight line, what can you say about its acceleration vector?

b) If a particle moves with constant speed along a curve, what can you about its acceleration vector?

Solution

To answer the first question, velocity is a vector, thus having both magnitude and direction. Change in either one of these or both results in acceleration. An object moving along a straight path would have an constant direction but its speed might vary. As a consequence, the acceleration vector will have the same direction, but its magnitude might vary.

To answer the second question, here the speed is constant, but direction of the velocity vector might vary, therefore the acceleration vector will have the same magnitude, but its direction will vary.

Exercises from the Rasmus’ notes

🔫 Exercise 1

The formula for a circular motion is:

\[R(t) = (rcos(ωt), rsin(ωt))\]

where \(r\) is the radius of the circle, and \(ω\) is a real number. The orbit time, i.e. the time it takes to complete an orbit is \(\frac{2π}{ω}\).

Suppose we want a satellite to orbit the earth at radius \(r\) from the center of the earth (we will assume that \(r\) is greater than the radius of the earth). The physical equation governing the movement states that the size of the acceleration should correspond to the force of the earths gravitational force at radius \(r\). The equation is

\[| R''(t)| = \frac{k}{r^2}\]

where \(k\) is a constant. Compute \(ω\) as a function of \(r\). Compute the corresponding speed. In words, you should compute how fast the satellite needs to travel to stay in the circular motion at a given height, and the orbit time at that height.

Answer the following questions:

As the radius goes up, does the orbit time go up or down?
As the radius goes up, does the speed go up or down?

Solution

Let’s start with computing \(ω\) as a function of \(r\). We need to compute \(\|R''(t)\|\), if you do the math correctly, you should obtain \(\|R''(t)\| = rω^2\). Now we can write:

\[rω^2 = \frac{k}{r^2} \Rightarrow ω = \pm \sqrt{\frac{k}{r^3}}\]

Second, we are supposed to compute corresponding speed \(\|R'(t)\|\). Again, if you do the math correctly, you should arrive at \(\|R'(t)\| = rω\).

We can then define speed as (ignore minus indeed since speed can not be negative by defintion):

\[speed = rω = r\sqrt{\frac{k}{r^3}} = \frac{\sqrt{k}}{\sqrt{r}}\]

Finally, we can answer the questions as follows:

As the radius goes up, we can see that the orbit time would go up according to the first obtained equation
As the radius goes up, speed goes down as we can see from the second equation

🔍 About the page

Last updated: 16/11/2022
Unless othewise stated, exercises come from the book: James Stewart, Daniel K. Clegg and Saleem Watson: Calculus: Early Transcendentals, Metric Edition, 9th edition
Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by Rasmus Ejlers Møgelberg. Please note that the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.