Day 6 - Prep

Task 6.1

We can generalize what we've done to find the work done by any force (doesn't have to be constant), acting on any object, along any path. Recall that work is a transfer of energy. Consider the following examples:

  • A tornado picks up a couch and applies forces to the couch as it swirls around the center. Work is the transfer of the energy from the tornado to the couch, giving the couch its kinetic energy.
  • When an object falls, gravity does work on the object. The work done by gravity converts potential energy to kinetic energy.
  • If we consider the flow of water down a river, it is gravity that gives the water its kinetic energy. We can place a hydroelectric dam next to a river to capture a lot of this kinetic energy. Work transfers the kinetic energy of the river to rotational energy of the turbine, which eventually ends up as electrical energy available in our homes.

When we study work, we are really studying how energy is transferred. This is one of the key components of modern life. Recall that the work done by a vector field $\vec F$ through a displacement $\vec d$ is the dot product $\vec F\cdot \vec d$.

  • When object moves from $A=(6,0)$ to $B=(0,3)$ encountering the constant force $\vec F = (2,5)$, the work is done by $\vec F$ as the object moves from $A$ to $B$ is simply the displacement $B-A=(-6,3)$ dotted by the force, so we have $W = \vec F\cdot \vec d = (2,5)\cdot(-6,3) = -12+15=3$.

An object moves from $A=(6,0)$ to $B=(0,4)$. A parametrization of the object's path is $\vec r(t) = (-3,2)t+(6,0)$ for $0\leq t\leq 2$.

  1. For $0\leq t\leq 1$, the force encountered is $\vec F = (2,5)$. For $1\leq t\leq 2$, the force encountered is $(2,7)$. How much work is done in the first second? How much work is done in the last second? How much total work is done?
  2. If we encounter a constant force $\vec F$ over a little displacement $d\vec r$, explain why the little work done is $\ds dW = \vec F\cdot d\vec r =\vec F\cdot \frac{d\vec r}{dt}dt $.
  3. Suppose that the force constantly changes as we move along the curve. At $t$, we encountered the force $\vec F(t) = (2,5+2t)$, which we could think of as the wind blowing stronger and stronger to the north. Explain why the total work done by this force along the path is $$\ds W=\int \vec F\cdot d\vec r = \int_0^2 (2,5+2t)\cdot (-3,2)dt.$$ Then compute this integral and show you get 16.
  4. If you are familiar with the units of energy, complete the following. What are the units of $\vec F$, $d\vec r$, and $dW$.

If a force of magnitude $F$ acts through a displacement with magnitude $d$, then the most basic definition of work is $W=Fd$, the product of the force and the displacement. Recall that this basic definition has a few assumptions.

  • The force $F$ must act in the same direction as the displacement.
  • The force $F$ must be constant throughout the displacement.
  • The displacement must be in a straight line.

The dot product let's us remove the first assumption as work is $W=\vec F\cdot \vec r,$ where $\vec F$ is a force acting through a displacement $\vec r$. We just saw we can remove the assumption that $\vec F$ is constant to obtain $$W=\int \vec F \cdot d\vec r = \int_a^b F\cdot \frac{d\vec r}{dt}dt, $$ provided we have a parametrization of $\vec r$ with $a\leq t\leq b$. We now get rid of the assumption that $\vec r$ is a straight line.

  1. Suppose that we move along the circle $C$ parametrized by $\vec r(t) = (3\cos t,3\sin t)$. As we move along $C$, we encounter a rotational force $\vec F(x,y) = (-2y,2x)$.
    1. Draw $C$. Then at several points on the curve, draw the vector field $\vec F(x,y)$. For example, at the point $(3,0)$ you should have the vector $\vec F(3,0)=(-2(0),2(3))=(0,6)$, a vector sticking straight up 6 units. Are we moving with the vector field, or against the vector field?
    2. Explain why we can state that a little bit of work done by a force $\vec F$ over a small displacement $d\vec r$ is $dW = \vec F\cdot d\vec r$. Why does it not matter that $\vec r$ does not move in a straight line?
    3. Since a little work done by $\vec F$ along $d\vec r$ (a small bit of $C$) is $dW = \vec F\cdot d\vec r$, we know that the total work done is $\int dW = \int \vec F\cdot d\vec r$. This gives us $$W = \int_C\left(-2y,2x\right)\cdot d\vec r = \int_0^{2\pi}\left(-2(3\sin t),2(3\cos t)\right)\cdot(-3\sin t, 3\cos t)dt.$$ Complete the integral, showing that the work done by $\vec F$ along $C$ is $36\pi$.
We now have that the work done by a vector field $\vec F$, along a curve $C$ with parametrization $\vec r(t)$ for $a\leq t\leq b$, is

$$W = \int_C \vec F\cdot d\vec r= \int_a^b \vec F(\vec r(t))\cdot \frac{d\vec r}{dt}dt.$$ Note that we put the $C$ under the integral $\int_C$ to remind us that we are integrating along the curve $C$. This means we need to get a parametrization of the curve $C$, and give bounds before we can integrate with respect to $t$.

If we let $\vec F = (M,N)$ and we let $\vec r(t)=(x,y)$, so that $d\vec r = (dx,dy)$, then we can write work in the differential form $$W = \int_C \vec F\cdot d\vec r= \int_C (M,N)\cdot (dx,dy) = \int_C Mdx+Ndy.$$

Task 6.2

Suppose an object travels along the path given by $\vec r(t) = (3t,-2t^2)$. The velocity is $\vec v(t) = (3,-4t)$ and the acceleration is $\vec a(t)=(0,-4)$.

  1. Is there a time $t$ at which the velocity and acceleration vectors are parallel? Explain.
  2. Compute the vector component of the acceleration vector that is parallel to the velocity vector. In other words, compute $\text{proj}_{\vec v}\vec a$. We'll call this vector $\vec a_{\parallel \vec v}$.
  3. What is the vector component of the acceleration vector that is orthogonal to the velocity vector? We'll call this vector $\vec a_{\perp \vec v}$.
  4. Draw a picture that shows the relationship among $\vec v$, $\vec a$, $\vec a_{\parallel \vec v}$, and $\vec a_{\perp \vec v}$.

Task 6.3

As the semester goes, we'll be learning to use Mathematica. You can install Mathematica for free as BYU-I student (see here to obtain Mathematica if you do not already have it). If you've never used Mathematica before, no worries. Here is Fast Introduction to Mathematica for Math Students. After reading the "Entering Input" section, feel free to click on the tabs on the left for more information about any topic needed.

  1. Let's write a block of code in Mathematica to compute the arc length of any parameterized curve.
    1. First, define a vector function in Mathematica to represent the parameterized curve $\ds \vec r(t) = \left(t^3,\frac{3t^2}{2}\right)$. Something like r = {t^3,3t^2/2}.
    2. Define some variables to hold the upper and lower limits for the parameter $t$ (something like a=1 and b=3).
    3. Add a line to your block of code that uses ParametricPlot[] to create a graph of the function. This verifies that the function is defined correctly.
    4. Compute the derivative or $\vec r$ using the derivative command ( D[]).
    5. Compute the length of the derivative using the Norm[] command.
    6. Compute the arc length of the curve using the Integrate[] command. You can use the N[] command to get a decimal approximation.
    7. Try putting all of the commands above into a single block of code, so that you can run it all with one execution.
  2. Copy the block of code that you created, then change the interval of integration to $2\leq t\leq 5$, and see if the plot as well as arc length update.
  3. Let's now use the work above to examine arc length for a few other curves. For each curve below, set up an integral formula which would give the length. Then sketch the curve. Try using them in the program you wrote above. Do not worry about integrating them by hand (they will get ugly really fast, and some are impossible).
    1. The parabola $\vec r(t) = (t,t^2)$ for $t\in[0,3]$.
    2. The ellipse $\vec r(t) = (4\cos t,5\sin t)$ for $t\in[0,2\pi]$.
    3. The hyperbola $\vec r(t) = (\tan t,\sec t)$ for $t\in[-\pi/ 4,\pi/4]$.

Task 6.4

The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.

  • Work on paths (line integrals): section 2.3, checkpoint 6.18, example 6.23 and exercises 49-54
  • Return to any of the previous day's OpenStax problems to locate extra practice.

Day 6 - In class

Brain Gains (Rapid Recall, Jivin' Generation)

  • For the vector field $\vec F(x,y) = (x+y, x^2)$ state $M$ and $N$.

Solution

The notation we'll be using is $\vec F(x,y)=(M,N)$ or $\vec F(x,y,z) = (M,N,P)$. This gives

  • $M=x+y$
  • $N=x^2$.

With Mathematica, when we've already defined $\vec F$, we can use the following to get the components. Note that because N[] is already defined as a Mathematica function, we have to use different notation for the name of the function. 

F[x_, y_] := {x + y, x^2}
M[x_, y_] := F[x, y][[1]]
Nn[x_, y_] := F[x, y][[2]]
F[x, y]
M[x, y]
Nn[x, y]
  • For the curve $\vec r(t) = (t^2, t^3)$, state $x$, $y$, $dx$, $dy$, and $ds$ in terms of $t$.

Solution

We have

  • $x = t^2$
  • $y = t^3$
  • $dx=2t dt$ (the differential of $x$ equals $2t$ multiplied by the differential of $t$. )
  • $dy=3t^2 dt$ (the $dt$ here matters. We have $\frac{dy}{dt} = 3t^2$, and we have $dy=3t^2 dt$.)
  • $ds = \sqrt{(2t)^2+(3t^2)^2}dt$ - This is the quantity we use for arc length.

Note that $dx$ is the differential of $x$, whereas $\frac{dx}{dt}$ is the derivative of $x$ with respect to $t$.

Here's an option for defining all these values in Mathematica. The ClearAll[] command is used to make sure that any previous definitions given for any of these variables are removed from memory. We'll start doing this from here on out, as it's good programming practice. 

ClearAll[x, y, dx, dy, dt, ds]
r[t_] := {t^2, t^3}
x[t_] := r[t][[1]]
y[t_] := r[t][[2]]
dx[t_, dt_] := x'[t] dt
dy[t_, dt_] := y'[t] dt
ds[t_, dt_] := Norm[r'[t]] dt
x[t]
y[t]
dx[t, dt]
dy[t, dt]
ds[t, dt]
ds[t, dt] // ComplexExpand
  • Given $\vec F(x,y) = (x+y, x^2)$ and $\vec r(t) = (t^2, t^3)$ state $\vec F(\vec r(t))$.

Solution

Recall that $\vec r(t) = (t^2, t^3)$ means $x=t^2$ and $y=t^3$, so $$\vec F(\vec r(t)) = \vec F(x=t^2, y=t^3) = (t^2+t^3, (t^2)^2).$$

Function composition in Mathematica can be done with the At and Apply commands (using @ and ). The Apply[] command (using ) is needed with the output of a function is a vector (which is true for the vector-valued function $\vec r(t)$. 

ClearAll[F, r]
F[x_, y_] := {x + y, x^2}
r[t_] := {t^2, t^3}
F @@ r@t
F @@ r[t]
Apply[F, r[t]]
F[r[t]] (*Does not work because the output of r[t] is a vector*)
  • Set up the work integral $\int_C Mdx+Ndy$ for $\vec F(x,y) = (x+y, x^2)$ and $\vec r(t) = (t^2, t^3)$ for $t\in [-1,3] $ (so replace each term with what it equals in terms of $t$).

Solution

We have the following:

  • $M=x+y = t^2+t^3$,
  • $N=x^2 = (t^2)^2$,
  • $dx = 2tdt$,
  • $dy = 3t^2dt$.

This gives $$\int_C Mdx+Ndy =\int_{-1}^3 \underbrace{(t^2+t^3)}_{M}\underbrace{(2tdt)}_{dx}+\underbrace{(t^2)^2}_{N}\underbrace{(3t^2dt)}_{dy} .$$ 

ClearAll[F, r]
F[x_, y_] := {x + y, x^2}
r[t_] := {t^2, t^3}
F @@ r[t]
r'[t]
(F @@ r[t]) . r'[t]
Integrate[(F @@ r[t]) . r'[t], {t, -1, 3}]
  • Set up the work integral $\int_C \vec F\cdot d\vec r$ for $\vec F(x,y) = (3xy, x+2y)$ and $\vec r(t) = (t^2+1, 2t)$ for $t\in [0,2] $.

Solution

We have the following:

  • $F(\vec r(t)) = (3(t^2+1)(2t), (t^2+1)+2(2t))$,
  • $\frac{d\vec r}{dt}=(2t,2)$,
  • $d\vec r=(2t,2)dt$.

This gives $$\int_C \vec F\cdot d\vec r =\int_{0}^2 \left[\underbrace{(3(t^2+1)(2t))}_{M}\underbrace{(2t)}_{dx/dt}+\underbrace{((t^2+1)+2(2t))}_{N}\underbrace{(2)}_{dy/dt}\right]dt .$$ 

ClearAll[F, r]
F[x_, y_] := {3 x y, x + 2 y}
r[t_] := {t^2 + 1, 2 t}
F @@ r[t]
r'[t]
(F @@ r[t]) . r'[t]
Integrate[(F @@ r[t]) . r'[t], {t, 0, 2}]

Group Problems

  1. Use the arc length formula and the parameterization $\vec r(t)=(a\cos t,a\sin t)$ of a circle to verify that the circumference of a circle of radius $a$ is $2\pi a$. $$s=\int_?^? \sqrt{(?)^2+(?)^2}dt.$$
  2. A force given by $\vec F = (y,-x+y)$ acts on an object as it moves along the curve $\vec r(t) =(t,t^2)$ for $0\leq t\leq 2$. Note that this means $x=t$ and $y=t^2$, so $\vec F = (t^2,-t+t^2)$. Compute $\ds \int_C\vec F\cdot \frac{d\vec r}{dt}dt$ (the work done by the force along the curve).
  3. Consider the curve $C$ parametrized by $\vec r(t) = (t, t^2)$ for $0\leq t\leq 2$.
    • Give a vector equation of the tangent line to the curve at $t=1$ (fill in the blanks below): $$\begin{pmatrix}?\\?\end{pmatrix} = \begin{pmatrix}?\\?\end{pmatrix}t+\begin{pmatrix}?\\?\end{pmatrix}.$$
    • Set up an integral that gives the length of this curve. Just set it up (fill in the blanks below). $$s=\int_?^? \sqrt{(?)^2+(?)^2}dt.$$

Day 7 - Prep

Task 7.1

Consider the parametric curve $\vec r(t) = (t^2-3t, 4t-5)$ for $0\leq t\leq 3$.

  1. Draw the curve.
  2. Compute the velocity and speed at any time $t$.
  3. Give an equation of tangent line at $t=2$.
  4. Compute the acceleration at $t=2$.
  5. At $t=2$, compute the vector component of the acceleration that is parallel to the velocity, and the vector component of the acceleration that is orthogonal to the velocity.
  6. Set up an integral to compute the arc length of the curve. Then compute the integral (using software is fine).

Task 7.2

Consider the parabolic curve $y=4-x^2$ for $-1\leq x\leq 2$, and the vector field $\vec F(x,y) = (2x+y,-x)$. A parametrization $\vec r(t)$ of this parabolic curve that starts at $(-1,3)$ and ends at $(2,0)$ is $\vec r(t) = (t, 4-t^2)$.

  1. Compute $d\vec r$ and state $dx$ and $dy$. What are $M$ and $N$ in terms of $t$?
  2. Compute the work done by $\vec F$ on an object that moves along the parabola from $(-1,3)$ to $(2,0)$ (i.e. compute $\int _C Mdx+Ndy$). Check your work using the Sage code at the end of this problem.
  3. How much work is done by $\vec F$ to move an object along the same parabola from $(2,0)$ to $(-1,3)$. In other words, if you traverse along a path backwards, how much work is done?
  4. We now want to know how much work is done by the same vector field on an object that moves along a straight line from $(-1,3)$ to $(2,0)$.
    • Give a parametrization $\vec r(t)$ of the straight line curve that starts at $(-1,3)$ and ends at $(2,0)$. Make sure you give bounds for $t$.
    • Compute $d\vec r$ and state $dx$ and $dy$. What are $M$ and $N$ in terms of $t$?
    • Compute the work done by $\vec F$ to move an object along the straight line path from $(-1,3)$ to $(2,0)$. Again, check your work using the Sage code at the end of this problem. Note that you must type the times symbol in (3*t-1, ...), otherwise, you'll get an error.
  5. Optional (we'll discuss this in class if you don't have it). How much work does it take to go along the closed path that starts at $(2,0)$, follows the parabola $y=4-x^2$ to $(-1,3)$, and then returns to $(2,0)$ along a straight line. Show that this total work is $W=-9$.

Use SageMath to check your work above. Click to see the Sage code.

Hit evaluate at the bottom. Feel free to modify the code below to fit your needs. 

var('t','x','y')      #Define your variables
r(t) = (t,4-t^2)      #State your parametrization  
bounds = (t,-1,2)     #Give bounds for the parametrization
F(x,y) = (2*x+y,-x)   #State the vector field
xbounds = (x,-1,2)    #These bounds are useful if you want to make a good plot.
ybounds = (y,0,4)    #These bounds are useful if you want to make a good plot.

dr = r.diff(t)        #Compute the derivative
dW=F(*r(t)).dot_product(dr(t)) #Find a little bit of work.  The code r[0] gives the first component, and r[1] gives the second. 
W=integrate(dW,bounds)

pretty_print(html("""The work done by $F=%s$ along the curve r=$%s$ 
     over the bounds $%s$ is $%s$"""%tuple(map(latex, [F(x,y), r(t), bounds, W]))))
show(table([
[r"$\vec r(t)$", "$(x,y)$", r(t)],
[r"$d\vec r$", "$(dx,dy)$", dr(t)],
["$\vec F(x,y)$", F(x,y), F(*r(t))],
["$M$", F(x,y)[0], F(*r(t))[0]],
["$N$", F(x,y)[1], F(*r(t))[1]],
[r"$dW=\vec F\cdot d\vec r$", "$Mdx+Ndy$", dW],
[r"$W=\int_C \vec F\cdot d\vec r$",W,W]
]
))

p=parametric_plot(r(t),bounds)
p+=plot_vector_field(F,xbounds,ybounds)
show(p)

Task 7.3

Density is generally a mass per unit volume. However, when talking about a wire, it's simpler to let density be the mass per unit length. We can make objects out of composite materials, where the density is different at different places in the object. For example, we might have a straight wire where one end is copper and the other end is gold. In the middle, the wire slowly transitions from being all copper to all gold. Such composite materials are engineered all the time (though probably not our example wire). The density at point $(x,y,z)$ is given by the quantity $\delta (x,y,z)$.

  1. Suppose a wire $C$ has the parameterization $\vec r(t)$ for $t\in[a,b]$. Suppose the wire's density (mass per unit length) at a point $(x,y,z)$ on the wire is given by the function $\delta(x,y,z)$. Since density is a mass per length, multiplying density by a small length $ds$ gives us the mass of a small portion of the curve. We represent this symbolically using $dm=\delta(\vec r(t_0)) ds$. Explain why the mass $m$ of the wire is given by the formulas below (explain why each equal sign is true): $$m=\int_C dm = \int_C \delta ds = \int_a^b \delta(\vec r(t)) \left|\frac{d\vec r}{dt}\right|dt.$$
  2. Now suppose a wire lies along the straight segment from $(0,2,0)$ to $(1,1,3)$. A parametrization of this line is $\vec r(t) = (t,-t+2,3t)$ for $t\in[0,1]$. The wire's density (mass per unit length) at a point $(x,y,z)$ is $\delta(x,y,z)=x+y+z$.
    1. Is the wire heavier at $(0,2,0)$ or at $(1,1,3)$?
    2. What is the total mass of the wire? [Replace $x$, $y$, $z$, and $ds$ with what they equal in terms of $t$ and then integrate.]
  3. Now consider an insulated wire that lies along the curve $\vec r(t) = (7\cos t, 7\sin t)$ for $0\leq t\leq \pi$. The wire contains charged particles where the charge per unit length at location $(x,y)$ is given by $q(x,y)=y$. We'll compute the total charge on the wire.
    1. Draw the curve. Then at several points on the curve write the value of $q(x,y)$ at that point. (Optional: Should the total charge be positive or negative?)
    2. Why is the little charge $dQ$ over a little distance $ds$ approximately given by $dQ = q(x,y)ds$?
    3. The total charge is the sum of the charges over all the little pieces on the rod. This gives us the total charge as $$Q_{\text{total}}=\int_CdQ=\int_C q(x,y)ds = \int_a^b y \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$ Replace $x$ and $y$ with what they are in terms of $t$ and then finish by computing the integral above.

Task 7.4

The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.

  • Return to any of the previous day's OpenStax problems to locate extra practice.


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