Question

show that the curvilnear coordinate system defined by the following equations is orthogonal. x=uvcosØ,
y=uvsinØ, z=1/2(u^2-v^2)

Answer 1

Answer on Question #41684 – Math – Analytic Geometry

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

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.

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