Finals week plans
- Our last class is Monday Dec 11 next week.
- Final Exam information:
- Any score above 70% on the final exam will boost your grade a half letter step, while a score less than 30% will drop your grade a half letter step.
- The final exam will not be graded in mastery fashion, rather will be graded in a more traditional way (to make comparisons with other courses).
- The final exam has a time limit of 4 hours. Once you click start, the exam will appear for download. You then have 4 hours to complete the exam and upload your work.
- The final will be available in I-Learn. You can take the final at any time. No submissions will be accepted after Wed Dec 13 at midnight.
- For those who have not yet reached 80% mastery on the quizzes, there will be three more attempts for you to reach this benchmark.
- Quiz 13 will have objectives from units 3, 4, 5, and 6.
- Quiz 14A will only have objectives from units 5 and 6. This quiz is due Tues Dec 12 at 5pm, with solutions being released at 5pm. This quiz opens Mon Dec 13 after class.
- Quiz 14B will have all objectives on it. It is due on Wed Dec 13 at midnight. This quiz opens Monday Dec 11 after class.
Brain Gains (Recall/Generation)
When I see a problem involving work, I always go through this three step process.
- Is the curve a simple closed curve (does it form the boundary of some region R)? If so, use Green's Theorem $\int_CMdx+Ndy = \iint_R N_x-M_ydA$.
- Does the vector field have a potential? Use the fundamental theorem of line integral $\int_C\vec\nabla f \cdot d\vec r = f(B)-f(A)$, work done is the difference in potential.
- If the curve is not a simple closed curve, and there is no potential, then return to unit 2 and directly use the long formula $\int_CMdx+Ndy$ (and cry a little and know I'll probably make lots of mistakes).
- For the vector field $\vec F = (2x+3y^2, 6xy+4y)$, compute the work done along the curve $\vec r(t) = (3\cos t, 3\sin t)$ for $t\in [0,\pi] $.
Solution
The vector field has a potential, namely $f = x^2+3xy^2+2y^2$. The start point is $\vec r(0) = (3,0)$. The end point is $\vec r(0) = (-3,0)$. The work done is the difference in potential, so $$f(-3,0)-f(3,0) = (9-0-0)-(9-0-0)=0.$$
- For the vector field $\vec F = (-4y, 4x)$, compute the work done along the curve $\vec r(t) = (3\cos t, 3\sin t)$ for $t\in [0,\pi] $.
Solution
This vector field has no potential. The curve is not a simple closed curve either. We have to resort back to our formula from unit 2.
- For the vector field $\vec F = (-4y, 6x)$, compute the work done along the curve $\vec r(t) = (3\cos t, 3\sin t)$ for $t\in [0,2\pi] $.
Solution
Green's Theorem will make quick work of this one, because the curve is a closed curve.
- Consider the parametric surface $\vec r(u,v) = (2u, 3v, u^2+v^2)$ for $u\in [-3,3]$ and $v\in [0,3]$. Start by computing the normal vector $\vec n = \vec r_u\times \vec r_v$, and then give an equation of the tangent plane to this surface at $\vec r(-1,1)$.
Solution
We'll do this on the board.
Group problems
- Consider the surface parametrized by $\vec r(u,v) = (u\cos v, u\sin v, u^2)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$.
- Draw the surface.
- Compute $d\sigma = \left |\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial v}\right|dudv$.
- Set up an integral formula to compute $\bar z$ for this surface.
- Consider the parametric surface $\vec r(u,v) = (u, u\cos v,u\sin v)$ for $0\leq v\leq \pi$ and $0\leq u\leq 4$.
- Draw the surface.
- Compute $\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial v}$.
- Give an equation of the tangent plane to the surface at $(u,v) = (1,\pi/2)$.
- Set up an integral formula to compute the surface area $\sigma$.
- Set up an integral formula to compute $\bar y$ for this surface.
- Let $\vec F = (5y,5x)$.
- Find a potential for $\vec F$, or explain why none exists.
- Compute the work done by $\vec F$ to get from $(3,0)$ to $(0,5)$ along a straight line path.
- Compute the work done by $\vec F$ to go once counter-clockwise along the circle $\vec r(t) = (3\cos t, 3\sin t)$.
- Let $\vec F = (-5y,5x)$.
- Find a potential for $\vec F$, or explain why none exists.
- Compute the work done by $\vec F$ to get from $(3,0)$ to $(0,5)$ along a straight line path.
- Compute the work done by $\vec F$ to go once counter-clockwise along the circle $\vec r(t) = (3\cos t, 3\sin t)$. [Use Green's Theorem.]
- Compute the derivative of each vector field $\vec F$ below (obtaining a square matrix). Then find a potential for $\vec F$ or explain why the vector field has no potential.
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- Compute the divergence and curl of each vector field.
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- If a vector field has a potential, then what is the curl of that vector field?
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