- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
This is day 2 of Unit 4.
Brain Gains (Rapid Recall, Jivin' Generation)
- For the vector field $\vec F(x,y,z)=(x+yz,xz+z,xy+y)$, start by computing the derivative (to check if there is a potential), and then find a potential.
Solution
The second derivative is $$\begin{bmatrix} 1 & z & y \\ z & 0 & x+1 \\ y & x+1 & 0 \\ \end{bmatrix}.$$ Because the derivative is symmetric, we have reason to believe a potential may exist. We integrate each component, with respect to the corresponding variable.
- $\int x+yz dx = x^2/2+xyz +C(y,z)$
- $\int xz+z dy = xyz +yz +C(x,z)$
- $\int xy+y dz = xyz +yz +C(x,y)$
A potential for $\vec F(x,y,z)$ is $f(x,y,z) = x^2/2+xyz+yz$.
We can verify all this with the Mathematica code below.
ClearAll[F, f, x, y, z]
F[x_, y_, z_] := {x + y z, x z + z, x y + y}
D[F[x, y, z], {{x, y, z}}] // MatrixForm
potential = DSolve[D[f[x, y, z], {{x, y, z}}] == F[x, y, z], f[x, y, z], {x, y, z}] // Expand // Flatten
D[f[x, y, z] /. potential, {{x, y, z}}]
- Let $\vec F(x,y,z) = ( x+yz,xz+z,xy+y)$ and $C$ be the straight segment from $(2,-5,0)$ to $(1,2,3)$. Compute the work done by $\vec F$ along $C$ by first finding a potential for $\vec F$.
Solution
A potential for $\vec F$ is $f(x,y,z) = x^2/2+xyz+yz$. To compute the work done, we just need the difference in the potential. This gives $$f(1,2,3)-f(2,-5,0) = (1/2+6+6)-(2+0+0) = 10.5.$$
- Draw the surface $\vec r(u,v) = (u\cos v, u, u\sin v)$ for $0\leq u\leq 2$ and $0\leq v\leq \pi$.
Solution
The surface is half a cone, with the axis of symmetry on the $y$-axis. Notice that we draw half ($v$ bounds stop at $\pi$) a circle of radius $u$ when $y=u$. The Mathematica code below will draw this surface.
ClearAll[r, u, v]
r[u_, v_] := {u Cos[v], u, u Sin[v]}
uB = {u, 0, 2}
vB = {v, 0, Pi}
ParametricPlot3D[r[u, v], uB, vB]
- Let $\vec F(x,y,z) = (a x+by+cz,dx+ey+fz,gx+hy+iz)$. Compute $D\vec F(x,y,z)$, $\vec \nabla \times \vec F$, and $\vec \nabla \cdot \vec F$.
Solution
With Mathematica, we have the following solutions.
F[x_, y_, z_] := {a x + b y + c z, d x + e y + f z, g x + h y + i z }
D[F[x, y, z], {{x, y, z}}] // MatrixForm
Curl[F[x, y, z], {x, y, z}]
Div[F[x, y, z], {x, y, z}]
Group Problems
- For the vector field $\vec F = (xyz, 3x^2+4y, 2x+3y+4z)$, compute the derivative, the divergence, and the curl. Use Mathematica to check your work.
- 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.
- $\vec F = (2x,3y)$ [Check: $D\vec F = \begin{bmatrix}2&0\\0&3\end{bmatrix}$ and $f = x^2+\frac{3}{2}y^2$. We can quickly check that $\vec \nabla f = (2x,3y)$, as needed.]
- $\vec F = (2y,3x)$
- $\vec F = (3y,3x)$
- $\vec F = (4x,5y,6z)$
- $\vec F = (4x,5z,6y)$
- $\vec F = (4x,5z,5y)$
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- Compute the work done by the vector field $\vec F = (4x+2xy,x^2+2y)$ along the curve $C$ parametrized by $\vec r(t) = (3t-1,-5t+2)$ for $0\leq t\leq 1$. [Hint: First find a potential.]
Solution
The vector field has a potential as the derivative $D\vec F =\begin{bmatrix}- &2x \\2x &-\end{bmatrix}$ is symmetric.
- A potential for the vector field is $f(x,y) = 2x^2+x^2y+y^2$ (note $\int 4x+2xy dx = 2x^2+x^2y +C(y)$ and $\int x^2+2y dy = x^2y+y^2+D(x)$).
- The start point is $\vec r(0) = (-1,2)$ and the end point is $\vec r(1) = (2,-3)$.
The work done by $\vec F$ is the difference in potential, which gives $$\int_C\vec F\cdot d\vec r = f(2,-3) - f(-1,2)=(8-12+9)-(2+2+4) = 5-8 = - 3.$$
- Draw each curve or surface given below.
- $\vec r(t) = (3\cos t,3\sin t)$ for $0\leq t\leq 2\pi$.
- $\vec r(u,v) = (3\cos u,3\sin u,v)$ for $0\leq u\leq 2\pi$ and $0\leq v\leq 5$.
- $\vec r(u,v) = (4\cos u,v, 3\sin u)$ for $0\leq u\leq \pi$ and $0\leq v\leq 7$.
- $\vec r(t) = (3\cos t,3\sin t,4t)$ for $0\leq t\leq 6\pi$. (Check: Helix)
- $\vec r(u,v) = (u\cos v,u\sin v,u)$ for $0\leq v\leq 2\pi$ and $0\leq u\leq 4$. (Check: Cone)
- $\vec r(u,v) = (u\cos v,u\sin v,v)$ for $0\leq v\leq 6\pi$ and $2\leq u\leq 4$. (Check: Spiral stair case)
- $\vec r(t) = (0,t,9-t^2)$ for $0\leq t\leq 3$.
- $\vec r(u,v) = (u\cos v,u\sin v,9-u^2)$ for $0\leq v\leq 2\pi$ and $0\leq u\leq 3$.
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