Day 7

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 $W=-9$.

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.