- 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 1 of Unit 4.
Brain Gains (Rapid Recall, Jivin' Generation)
- Recall the derivative of a vector field $\vec F(x,y,z)=(M,N,P)$ is the 3 by 3 matrix $$\ds D\vec F(x,y,z) = \begin{bmatrix}\dfrac{\partial \vec F}{\partial x}&\dfrac{\partial \vec F}{\partial y}&\dfrac{\partial \vec F}{\partial z}\end{bmatrix} = \begin{bmatrix}M_x&M_y&M_z\\N_x&N_y&N_z\\P_x&P_y&P_z\end{bmatrix}$$ Compute the derivative $D\vec F(x,y,z)$ of the vector field $\vec F(x,y,z) = (3x+4yz, 5x^2+2z, xyz)$.
Solution
We can use Mathematica to check.
F[x_, y_, z_] := {3 x + 4 y z, 5 x^2 + 2 z, x y z}
D[F[x, y, z], {{x, y, z}}] // MatrixForm
- The divergence of a vector field $\vec F(x,y,z) = (M,N,P)$ is $$\begin{align*} \text{div}(\vec F) &= \vec \nabla \cdot \vec F \\ &= \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)\cdot \left(M,N,P\right) \\ &= \frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}\\ &= M_x+N_y+P_z . \end{align*}$$ Compute the divergence of the vector field $\vec F(x,y,z) = (3x+4yz, 5x^2+2z, xyz)$.
Solution
We can use Mathematica to check our work.
F[x_, y_, z_] := {3 x + 4 y z, 5 x^2 + 2 z, x y z}
Div[F[x, y, z], {x, y, z}]
- The curl of a vector field $\vec F(x,y,z) = (M,N,P)$ is $$\begin{align*} \text{curl}(\vec F) &= \vec \nabla \times \vec F \\ &= \left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z} \right)\times \left(M,N,P\right) \\ &= \left(\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z},\frac{\partial M}{\partial z}-\frac{\partial P}{\partial x},\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) \\ &= \left(P_y-N_z,M_z-P_x,N_x-M_y\right) . \end{align*}$$ Compute the curl of the vector field $\vec F(x,y,z) = (3x+4yz, 5x^2+2z, xyz)$.
Solution
We can use Mathematica to check our work.
F[x_, y_, z_] = {3 x + 4 y z, 5 x^2 + 2 z, x y z}
Curl[F[x, y, z], {x, y, z}]
- Consider the vector field $\vec F(x,y) = (6x+y^2,2xy-4y^3)$.
- Compute the derivative $D\vec F(x,y)$.
- Find a function $f(x,y)$ so that $\vec \nabla f(x,y) = \vec F$ (recall that this means $f$ is a potential for $\vec F$).
Solution
ClearAll[F, f, x, y]
F[x_, y_] := {6 x + y^2, 2 x y - 4 y^3}
D[F[x, y], {{x, y}}] // MatrixForm
potential = DSolve[D[f[x, y], {{x, y}}] == F[x,y], f[x, y], {x, y}]
D[f[x, y] /. potential, {{x, y}}] // Flatten
- Consider the surface $\vec r(u,v) = (2u+3v,4u+5v,6u+7v)$. The differential is $$d\vec r=\begin{pmatrix}2\\4\\6\end{pmatrix}du+\begin{pmatrix}3\\5\\7\end{pmatrix}dv.$$ Find the area of the parallelogram with edges $(2,4,6)du$ and $(3,5,7)dv$.
Solution
The magnitude of the cross product of these two vectors is $$\begin{align*} |(2,4,6)du\times(3,5,7)dv| &=\left|(2,4,6)\times(3,5,7)\right|dudv\\ &=\left|(4(7)-5(6),3(6)-2(7),2(5)-3(4))\right|dudv\\ &=\left|(-2,4,-2)\right|dudv\\ &=\sqrt{4+16+4}dudv\\ &=\sqrt{24}dudv \end{align*}$$ We could do this with Mathematica, to avoid making a simple error on the cross product.
Cross[{2, 4, 6} du, {3, 5, 7} dv] // Norm
- Draw the surface $S$ with parametrization $\vec r(u,v) = (4\cos u, v, 4\sin u)$ for $0\leq u\leq \pi$ and $2\leq v\leq 5$.
Solution
The surfaces involves circles of radius 4 when only examining the $xz$-plane. The $y$ axis is independent of the others, so this is half a cylinder (above $xy$plane) for $2\leq y\leq 5$.
The code for Mathematica is below.
ParametricPlot3D[{4 Cos[u], v, 4 Sin[u]}, {u, 0, Pi}, {v, 2, 5}]
We'll be using a surface with given bounds to do quite a bit in things at once, so let's get in the habit of putting the variable definitions up front.
r[u_, v_] := {4 Cos[u], v, 4 Sin[u]}
uB = {u, 0, Pi}
vB = {v, 2, 5}
ParametricPlot3D[r[u, v], uB, vB]
- Consider the vector field $\vec F(x,y,z) = (2x+3z, 4y,3x-3z^2)$ and the helical curve $C$ parametrized by $\vec r(t) = (t, \cos (\pi t), \sin(\pi t))$ for $0\leq t\leq 2$.
- Find a potential for $\vec F$, so find a function $f$ so that $\vec F = \vec \nabla f$.
- Find work done by $\vec F$ to move an object along $C$. You can use the fundamental theorem of line integrals which states $$\int_C \vec F \cdot d\vec r = f(B)-f(A),$$ provided $f$ is a potential for $\vec F$ with $A$ and $B$ being the starting and endpoint of the curve $C$.
Soluiton
A potential is $f(x,y,z) = x^2 + 3xz +2y^2-z^3$. The start point on the curve is $A = \vec r(0)=(0,1,0)$ and the end point is $B = (2,1,0)$, The work done along the curve is $$\int_C \vec F \cdot d\vec r = f(B)-f(A) = (4+2)-(2)=4.$$
ClearAll[F, f, r, potential]
F[x_, y_, z_] := {2 x + 3 z, 4 y, 3 x - 3 z^2}
sol = DSolve[D[f[x, y, z], {{x, y, z}}] == F[x, y, z], f, {x, y, z}] //Flatten
potential = f /. sol;
(*Manually type in the end points*)
potential[2, 1, 0] - potential[0, 1, 0]
(*Use function notation to evaluate the potential at the end points.*)
r[t_] := {t, Cos[Pi t], Sin[Pi t]}
potential @@ r@2 - potential @@ r@0
(*The apply function (@@) is required for function composition when the output is a vector.*)
(*Here we compute work using the formula from earlier in the semester*)
Integrate[Dot[Apply[F, r[t]], D[r[t], t]], {t, 0, 2}]
Integrate[Dot[F @@ r@t, r'@t], {t, 0, 2}]
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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