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 11 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)
Click for some code snippets to help work with surfaces in Mathematica.
Module[{r, u, v, ru, rv, n, uBounds, vBounds},
r[u_, v_] := {u Cos[v], u Sin[v], v};
uBounds = {u, 2, 4};
vBounds = {v, 0, 2 Pi};
ru[u_, v_] := Derivative[1, 0][r][u, v];
rv[u_, v_] := Derivative[0, 1][r][u, v];
n[u_, v_] := Cross[ru[u, v], rv[u, v]];
Manipulate[Show[
ParametricPlot3D[r[u, v], uBounds, vBounds, PlotStyle -> Opacity[0.5]],
Graphics3D[{Thick,
Red, Arrow[{r[u1, v1], r[u1, v1] + ru[u1, v1]}],
Blue, Arrow[{r[u1, v1], r[u1, v1] + rv[u1, v1]}],
Green, Arrow[{r[u1, v1], r[u1, v1] + n[u1, v1]}]}],
PlotRange -> All],
{{u1, (uBounds[[2]] + uBounds[[3]])/2}, uBounds[[2]], uBounds[[3]]},
{{v1, (vBounds[[2]] + vBounds[[3]])/2}, vBounds[[2]], vBounds[[3]]}]]
Module[{r, u, v, ru, rv, n, uBounds, vBounds},
r[u_, v_] := {(2 - Cos[u]) Cos[v], (2 - Cos[u]) Sin[v], Sin[u]};
uBounds = {u, 0, 2 Pi};
vBounds = {v, 0, 2 Pi};
ru[u_, v_] := Derivative[1, 0][r][u, v];
rv[u_, v_] := Derivative[0, 1][r][u, v];
n[u_, v_] := Cross[ru[u, v], rv[u, v]];
Manipulate[Show[
ParametricPlot3D[r[u, v], uBounds, vBounds, PlotStyle -> Opacity[0.5]],
Graphics3D[{Thick,
Red, Arrow[{r[u1, v1], r[u1, v1] + ru[u1, v1]}],
Blue, Arrow[{r[u1, v1], r[u1, v1] + rv[u1, v1]}],
Opacity[0.5], Green, Polygon[{r[u1, v1], r[u1, v1] + ru[u1, v1], r[u1, v1] + ru[u1, v1] + rv[u1, v1], r[u1, v1] + rv[u1, v1]}],
Opacity[1], Black, Polygon[{r[u1, v1], r[u1, v1] + ru[u1, v1] du, r[u1, v1] + ru[u1, v1] du + rv[u1, v1] dv, r[u1, v1] + rv[u1, v1] dv}]}],
PlotRange -> All, ImageSize -> Large],
{{u1, (uBounds[[2]] + uBounds[[3]])/2}, uBounds[[2]], uBounds[[3]]},
{{v1, (vBounds[[2]] + vBounds[[3]])/2}, vBounds[[2]], vBounds[[3]]},
{{du, 0.5}, 0, 1},
{{dv, 0.5}, 0, 1}
]]
Here are some other surfaces to work with.
r[u_, v_] := {(2 - Cos[u]) Cos[v], (2 - Cos[u]) Sin[v], Sin[u]};
uBounds = {u, 0, 2 Pi};
vBounds = {v, 0, 2 Pi};
r[u_, v_] := {u, v, 4 - u^2 - v^2};
uBounds = {u, -2, 2};
vBounds = {v, -2, 2};
r[u_, v_] := {u Cos[v], u Sin[v], 4 - u^2};
uBounds = {u, 0, 2};
vBounds = {v, 0, 2 Pi};
r[u_, v_] := {u Cos[v], u, u Sin[v]};
uBounds = {u, 0, 2};
vBounds = {v, 0, 2 Pi};
r[u_, v_] := {2 Sin[u] Cos[v], 3 Sin[u] Sin[v], 5 Cos[u]};
uBounds = {u, 0, Pi};
vBounds = {v, 0, 2 Pi};
r[u_, v_] := {2 Cos[v], u, 2 Sin[v]};
uBounds = {u, -2, 5};
vBounds = {v, 0, 2 Pi};
- For the surface $\vec r(u,v) = (u\cos v,u\sin v,v)$ for $0\leq v\leq 6\pi$ and $2\leq u\leq 4$, set up an integral formula to compute the surface area (recall that $d\sigma = \left|\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v}\right|dudv$).
Solution
The partial derivatives are $$\frac{\partial \vec r}{\partial u} = (\cos v,\sin v,0),\quad \frac{\partial \vec r}{\partial v} = (-u\sin v,u\cos v,1), $$ A normal vector $\vec n$ is the cross product of these two, so $$\vec n = (\sin v, - \cos v,u\cos^2v+u\sin^2v) = (\sin v, -\cos v,u).$$ The surface area is hence $$\sigma =\iint_S d\sigma = \int_{2}^{4}\int_{0}^{6\pi}\sqrt{\sin^2v+\cos^2v+u^2}dvdu = \int_{2}^{4}\int_{0}^{6\pi}\sqrt{1+u^2}dvdu .$$
- For the surface above, set up an integral formula to compute $\bar x$, $\bar y$, or $\bar z$.
Solution
We already know the surface area is given by $$\sigma =\iint_S d\sigma = \int_{2}^{4}\int_{0}^{6\pi}\sqrt{\sin^2v+\cos^2v+u^2}dvdu = \int_{2}^{4}\int_{0}^{6\pi}\sqrt{1+u^2}dvdu .$$ This gives $$\ds \bar x = \frac{\iint_S x d\sigma}{\iint_S d\sigma} = \frac{ \int_{2}^{4}\int_{0}^{6\pi}(u\cos v)\sqrt{1+u^2}dvdu }{ \int_{2}^{4}\int_{0}^{6\pi}\sqrt{1+u^2}dvdu }. $$ $$\ds \bar y = \frac{\iint_S y d\sigma}{\iint_S d\sigma} = \frac{ \int_{2}^{4}\int_{0}^{6\pi}(u\sin v)\sqrt{1+u^2}dvdu }{ \int_{2}^{4}\int_{0}^{6\pi}\sqrt{1+u^2}dvdu }. $$ $$\ds \bar z = \frac{\iint_S z d\sigma}{\iint_S d\sigma} = \frac{ \int_{2}^{4}\int_{0}^{6\pi}(v)\sqrt{1+u^2}dvdu }{ \int_{2}^{4}\int_{0}^{6\pi}\sqrt{1+u^2}dvdu }. $$
- Give an equation of the tangent plane to the surface above at $(u,v) = (3,\pi)$. The cross product of the partial derivatives of will get you a normal vector to the surface.
Solution
The partial derivatives $$\frac{\partial \vec r}{\partial u} = (\cos v,-\sin v,0),\quad \frac{\partial \vec r}{\partial v} = (-u\sin v,u\cos v,1) $$ are tangent to the surface. Their cross product will be a normal vector to the plane. The cross product of these two is $$\vec n(u,v) = (\sin v, - \cos v,u\cos^2v+u\sin^2v) = (\sin v, -\cos v,u).$$ We now use $(3,\pi)$ to find $(x,y,z) = \vec r(3,\pi) = (-3,0,\pi)$ and $\vec n(3,\pi) = (0,1,3)$. A plane with normal vector $(0,1,3)$ that passes through $(-3,0,\pi)$ has an equation $$0(x+3)+(1)(y-0)+3(z-\pi)=0.$$
- Recall for a two dimensional region that $\bar x = \dfrac{\iint_R xdA}{\iint_R dA}$. If you know that the area of a region is $A = 3$ and the centroid is $(5,7)$, compute $\iint_R 2xdA$.
Solution
Since $\bar x = \dfrac{\iint_R xdA}{\iint_R dA}$, we know that $\iint_R xdA = \bar x \iint_R dA = \bar x A$. In other words, we can replace $\iint_R xdA$ with $\bar x A$. We then compute $$\iint_R 2xdA = 2\iint_R xdA = 2\bar x A = 2(5)(3) = 30.$$
- A region $R$ has area $A = 3$ and centroid $(\bar x, \bar y) = (5,7)$. Compute the integral $\ds \iint_R (2-3y+4x) dA$.
Solution
Remember that $\iint_R dA = A$, as well as $\iint_R xdA = \bar x A$ and $\iint_R ydA = \bar y A$. We then compute $$\begin{align*} \iint_R (2-3y+4x) dA &=2\iint_R dA-3\iint_R y dA+4\iint_R x dA\\ &=2A - 3\bar y A + 4\bar x A \\ &=(2-3\bar y+4\bar x)A\\ &=(2-3(7)+4(5))(3)\\ &=3. \end{align*} $$
- A curve $C$ traverses around a region $R$ with $A = 3$ and the centroid is $(5,7)$. Compute the work done by $\vec F = (2x+3y,x^2+y^2)$ along $C$ (Recall Green's theorem: $\oint Mdx+Ndy = \iint_R N_x-M_y dA$).
Solution
Green's theorem will apply here, because we have a simple closed curve. This gives $$\iint_R N_x-M_y dA =\iint_R 2x-3 dA =2\iint_R xdA - \iint_R 3 dA = 2\bar xA - 3A = 2(5)(3)-3(3) = 21.$$
Group problems
- Compute the work done by $\vec F = (-3y,3x)$ to move an object counterclockwise once along the circle $\vec r(t) = (5\cos t, 5\sin t),$ using Green's theorem $\iint_R N_x-M_y dA$.
- Compute the work done by $\vec F = (y^2,3x)$ to move an object counterclockwise once along the circle $\vec r(t) = (5\cos t, 5\sin t),$ using Green's theorem $\iint_R N_x-M_y dA$.
- Compute the work done by each vector field below to move an object counterclockwise once along the triangle with corners $(0,0)$, $(2,0)$, and $(0,3)$.
- $\vec F = (2x-y,2x+4y)$
- $\vec F = (x^2+y^2,x+y)$
- 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 the surface area $\sigma$.
- 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 $d\sigma = \left|\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial v}\right|dudv$.
- Set up an integral formula to compute $\bar x$ for this surface.
- Give an equation of the tangent plane to the surface at $(u,v) = (1,\pi/2)$.
- Let $\vec F = (5y,5x)$.
- Find a potential for $\vec F$, or explain why none exists.
- 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 go once counter-clockwise along the circle $\vec r(t) = (3\cos t, 3\sin t)$.
- 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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