9:00 AM Jamboard Links
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12:45 PM Jamboard Links
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Learning Reminders
- We are in the 13th week of the semester. If you are on track for an A, then ideally you're finishing your SDL project for the 5th unit, and proposed something for the 6th unit.
- There are only 2 weeks left, which means you can submit at most 2 more SDL projects.
- The final SDL project (6th) can be over any topic from the entire semester. You can use it to expand what we do in the 6th unit, or you may choose to revisit something from a prior unit that you would like to spend more time with.
Rapid Recall
- 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
- $M=x+y$
- $N=x^2$.
- For the curve $\vec r(t) = (t^2, t^3)$, state $dx$, $dy$, and $ds$ in terms of $t$.
Solution
We have
- $dx=2t dt$ (the differential of $x$ equals $2t$ multiplied by the differential of $y$. )
- $dy=3t^2 dt$
- $ds = \sqrt{(2t)^2+(3t^2)^2}dt$ - 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 $t$.
- 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 $$\vec F(\vec r(t)) = \vec F(x=t^2, y=t^3) = (t^2+t^3, (t^2)^2).$$
- 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 $$\int_C Mdx+Ndy =\int_{-1}^3 \underbrace{(t^2+t^3)}_{M}\underbrace{(2tdt)}_{dx}+\underbrace{(t^2)^2}_{N}\underbrace{(3t^2dt)}_{dy} .$$
- 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 $$\int_C \vec F\cdot d\vec r =\int_{0}^2 \left[\underbrace{(3(t^2+1)(2t))}_{M}\underbrace{(2t)}_{dx/dt}+\underbrace{((t^2+1)+2(2t))}_{N}\underbrace{(2)}_{dy/dt}\right]dt .$$
Group problems
- 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 $2\pi a$. $$s=\int_?^? \sqrt{(?)^2+(?)^2}dt.$$
- 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).
- 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). $$s=\int_?^? \sqrt{(?)^2+(?)^2}dt.$$
- A wire lies along the curve $C$. The density (mass per length) of the wire at a point $(x,y)$ on the curve is given by $\delta(x,y) = y+2$. This means a little bits of mass along a small length $ds$ is given by $$dm = \delta ds = (y+2)ds = (y(t)+2)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt.$$ Set up an integral formula that gives the total mass of the wire (put everything in terms of $t$.
- The wire contains charged particles. The charge density (charge per length) at a point $(x,y)$ on the curve is given by the product $q(x,y)=xy$. This means the charge along a little length $ds$ is given by $dQ = (q) ds = (xy)ds$. Set up an integral formula that gives the total charge on the wire.
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