Brain Gains (Rapid Recall, Jivin' Generation)
- Consider the change-of-coordinates $x = u-3v$ and $y = 2u+4v$.
- Compute the differential $(dx,dy)$ and write it as a linear combination.
- Draw the region in the $xy$-plane given by $0\leq u\leq 1$ and $0\leq v\leq 1$.
- Compute the area of the parallelogram you just drew.
Solution
- 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 = {6 x + y^2, 2 x y - 4 y^3}
D[F, {{x, y}}] // MatrixForm
potential = DSolve[D[f[x, y], {{x, y}}] == F, f[x, y], {x, y}]
D[f[x, y] /. potential, {{x, y}}] // Flatten
- Let $\vec u = (1, 2, 3)$, $\vec v = (3, 0, -1)$ and $\vec w = (2, -1, 2)$. Compute the triple product $(\vec u\times\vec v)\cdot \vec w$. The absolute value of this quantity gives the volume of the parellelepiped formed by these 3 vectors.
Solution
The solution is -26. The volume of the parallelepiped formed by these three vectors is 26. We'll use this fact to perform 3D change-of-coordinates.
u = {1, 2, 3};
v = {3, 0, -1};
w = {2, -1, 2};
Cross[u, v]
Dot[Cross[u, v], w]
Graphics3D[Parallelepiped[{0, 0, 0}, {u, v, w}], Axes -> True, AxesOrigin -> {0, 0, 0}, AxesLabel -> Automatic]
Group Problems
- Compute the volume of the parallelepiped whose edges are given by the vectors $\vec u = (1, 0, 1)$, $\vec v = (0, -1, 2)$ and $\vec w = (-2, 1, 0)$. [Check: 4]
- Consider the change-of-coordinates $x=r\cos\theta$ and $y=r\sin\theta$.
- Compute the differential $(dx,dy)$ and write it as a linear combination.
- Compute the area of the parallelogram whose edges are given by the vectors $(\cos\theta, \sin\theta)$ and $(-r\sin\theta, r\cos\theta)$.
- 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.]
- $\vec F = (2y,3x)$
- $\vec F = (3y,3x)$
- $\vec F = (4x,5y,6z)$
- $\vec F = (4x,5z,6y)$
- $\vec F = (4x,5z,5y)$
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$