Finals week plans
- Our last class is Monday next week.
- Final Exam information:
- The final is optional. If you chose to not take it, then none of the info below applies.
- 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 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 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 at 5pm, with solutions being released at 5pm. This quiz opens Mon after class.
- Quiz 14B will have all objectives on it. It is due on Wed at midnight. This quiz opens Monday Dec after class.
Brain Gains
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 Discussion
- 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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