Question

Show that the curvilinear coordinate system defined by the following equations is
orthogonal:
x=uvcos(a)
y=uvsin(a)
z=1/2(u^2-v^2)
note here a stands for alpha

Answer 1

Answer on Question #40054 – Math - Other

Assignment

Show that the curvilinear coordinate system defined by the following equations is orthogonal:

$$\begin{array}{l} x = uv\cos(a) \\ y = uv\sin(a) \\ z = \frac{1}{2}(u^2 - v^2) \end{array}$$

note here a stands for alpha

Solution

$$\vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} uv\cos(\alpha) \\ uv\sin(\alpha) \\ \frac{1}{2}(u^2 - v^2) \end{pmatrix}$$

Derivatives of the radius vector:

$$\vec{r}_u^* = \begin{pmatrix} x_u \\ y_u \\ z_u \end{pmatrix} = \begin{pmatrix} v\cos(\alpha) \\ v\sin(\alpha) \\ u \end{pmatrix}; \vec{r}_v^* = \begin{pmatrix} x_v \\ y_v \\ z_v \end{pmatrix} = \begin{pmatrix} u\cos(\alpha) \\ u\sin(\alpha) \\ -v \end{pmatrix}; \vec{r}_\alpha^* = \begin{pmatrix} x_\alpha \\ y_\alpha \\ z_\alpha \end{pmatrix} = \begin{pmatrix} -uv\sin(\alpha) \\ uv\cos(\alpha) \\ 0 \end{pmatrix};$$

Scalar products:

$$\vec{r}_u^* \cdot \vec{r}_v^* = uv\cos^2(\alpha) + uv\sin^2(\alpha) - uv = 0$$ $$\vec{r}_u^* \cdot \vec{r}_\alpha^* = -uv^2\cos(\alpha)\sin(\alpha) + uv^2\cos(\alpha)\sin(\alpha) = 0$$ $$\vec{r}_v^* \cdot \vec{r}_\alpha^* = -u^2v\cos(\alpha)\sin(\alpha) + u^2v\cos(\alpha)\sin(\alpha) = 0$$

It means that $\vec{r}_u^*, \vec{r}_v^*$ and $\vec{r}_\alpha^*$ can be chosen as a basis and these vectors are orthogonal.