Question

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

$g\left(\theta ,\varphi \right)=cos\theta \cos \varphi$

Short answer

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

Step-by-step solution

Step 1. Given Information.

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

Step 2. Explanation.

The discriminant is calculated by formula, $det\left(H_g\right)=\frac{{\partial }^{2}g}{\partial {\theta }^2}·\frac{{\partial }^{2}g}{\partial {\varphi }^2}-{\left(\frac{{\partial }^{2}g}{\partial \theta \partial \varphi }\right)}^2$.

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

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

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

$\frac{{\partial }^{2}g}{\partial \theta \partial \varphi }=\sin \theta \sin \varphi$.

Step 3. Calculate.

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

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