Brain Gains (Rapid Recall, Jivin' Generation)

  • Consider the change-of-coordinates given by $x = 2u+v$ and $y=u-2v$. Draw the region in the $xy$-plane given by $1\leq u\leq 2$ and $0\leq v\leq 1$. Suggestion: first complete the table below.

$$\begin{array}{c|c} (u,v)&(x,y)\\\hline (1,0)&(2+0,1-0) = (2,1)\\ (1,1)&(2+1,1-2) = (3,-1)\\ (2,0)&\\ (2,1)& \end{array}$$

Solution

Here's the region, with Mathematica code. 

R = ParametricRegion[{2 u + v, u - 2 v}, {{u, 1, 2}, {v, 0, 1}}];
Region[R, Axes -> True, AxesLabel -> {x, y}, AxesOrigin -> {0, 0}]
  • Compute the differential $(dx,dy)$ for the change-of-coordinates given by $x = 2u+v$ and $y=u-2v$. Write your answer as a linear combination.

Solution

We have $dx = 2du+1dv$ and $dy = du-2dv$. This gives $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix}2\\1\end{pmatrix}du + \begin{pmatrix}1\\-2\end{pmatrix}dv.$$

  • State the area of a parallelogram whose edges are the vectors $(a,b)$ and $(c,d)$.

Solution

The prep had you show that the area is $A = |ad-bc|$.

  • State the area of a parallelogram whose edges are the vectors $(2,1)$ and $(1,-2)$, so the vectors from the differential of the change-of-coordinates $x = 2u+v$, $y=u-2v$. We call this the Jacobian of the change-of-coordinates given by $x = 2u+v$ and $y=u-2v$, written $\ds\frac{\partial(x,y)}{\partial(u,v)}$.

Solution

We have $$\frac{\partial(x,y)}{\partial(u,v)}= A = |ad-bc| = |(2)(-2)-(1)(1)|=|-5|=5.$$

  • 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 = {3 x + 4 y z, 5 x^2 + 2 z, x y z}
D[F, {{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 = {3 x + 4 y z, 5 x^2 + 2 z, x y z}
Div[F, {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 = {3 x + 4 y z, 5 x^2 + 2 z, x y z}
Curl[F, {x, y, z}]

Group Problems

  1. Find the area of the parallelogram whose vertices are $(0,0)$, $(2,4)$, $(-3,1)$, and $(-1,5)$. [Check: 14]
    • Then find the area of the triangle who vertices are $(0,0)$, $(2,4)$, and $(-3,1)$. (How is the triangle related to the parallelogram?)
    • Now find the area of the triangle who vertices are $(2,4)$, $(-3,1)$, and $(-1,5)$.
    • Now find the area of the triangle who vertices are $(0,0)$, $(-3,1)$, and $(-1,5)$. (If you got the same answer for the last 3 questions, then hooray!)
  2. Find the area of the triangle whose vertices are $(-2,1)$, $(3,4)$, and $(1,7)$. [Check: 21/2]
  3. 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.
  4. Consider the change of coordinates given by $x = 3u+2v$ and $y=u-4v$.
    • Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$. You can use the Mathematica code from above to check your answer.
    • Compute the area of the region you just drew (the region should be a parallelogram).
    • Compute the differential of the change-of-coordinates, and write it in the form $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$
    • State the Jacobian $\ds\frac{\partial(x,y)}{\partial(u,v)}$.
  5. Pick values for $a,b,c,d$ in the change-of-coordinates $x = au+bv$, $y=cu+dv$.
    • Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$. You can use the Mathematica code from above to check your answer.
    • Compute the area of the region you just drew (the region should be a parallelogram).
    • Compute the differential of the change-of-coordinates, and write it in the form $$\begin{pmatrix}dx\\dy\end{pmatrix} = \begin{pmatrix} ?\\ ?\end{pmatrix}du+\begin{pmatrix}?\\?\end{pmatrix}dv.$$
    • State the Jacobian $\ds\frac{\partial(x,y)}{\partial(u,v)}$.
  6. Pick your own vector field $\vec F(x,y,z) = (?, ?, ?)$, compute the derivative, the divergence, and the curl. Use Mathematica to check your work. 
    F = {x^2 y, x y z, 3 x + 4 z }
D[F, {{x, y, z}}] // MatrixForm
Div[F, {x, y, z}]
Curl[F, {x, y, z}]