Q. 25 - Calculus

Question

In Exercises 21–26, find the discriminant of the given function.

$f\left(\theta ,\varphi \right)=\cos \theta \sin \varphi$.

Step-by-Step Solution

Answer

The answer is ${\sin }^2\varphi {\cos }^2\theta -{\cos }^2\varphi {\sin }^2\theta$.

Step 1. Given Information.

The function is $f\left(\theta ,\varphi \right)=\cos \theta \sin \varphi$.

Step 2. Explanation.

The discriminant is calculated by formula,

$det\left(H_f\right)=\frac{{\partial }^{2}f}{\partial {\theta }^2}·\frac{{\partial }^{2}f}{\partial {\varphi }^2}-{\left(\frac{{\partial }^{2}f}{\partial \theta \partial \varphi }\right)}^2$.

Calculate $\frac{\partial f}{\partial \theta }$, $\frac{\partial f}{\partial \varnothing }$, $\frac{{\partial }^{2}f}{\partial {\theta }^2}$, $\frac{{\partial }^{2}f}{\partial {\varphi }^2}$ and $\frac{{\partial }^{2}f}{\partial \theta \partial \varphi }$.

$\frac{\partial f}{\partial \theta }=-\sin \theta \sin \varphi$,  $\frac{{\partial }^{2}f}{\partial {\theta }^2}=-\cos \theta \sin \varphi$.

$\frac{\partial f}{\partial \varnothing }=\cos \varphi \cos \theta$,  $\frac{{\partial }^{2}f}{\partial {\varphi }^2}=-\sin \varphi \cos \theta$.

$\frac{{\partial }^{2}f}{\partial \theta \partial \varphi }=-\cos \varphi \sin \theta$.

Step 3. Calculation.

Calculate $det\left(H_f\right)=\frac{{\partial }^{2}f}{\partial {\theta }^2}·\frac{{\partial }^{2}f}{\partial {\varphi }^2}-{\left(\frac{{\partial }^{2}f}{\partial \theta \partial \varphi }\right)}^2$.

$$\begin{gathered} det\left(H_f\right)=\left(-\cos \theta \sin \varphi \right)\left(-\sin \varphi \cos \theta \right)-{\left(-\cos \varphi \sin \theta \right)}^2 \\ ={\sin }^2\varphi {\cos }^2\theta -{\cos }^2\varphi {\sin }^2\theta \end{gathered}$$