Q. 22 - Calculus

Question

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

$g\left(x,y\right)=e^{-3x}\sin \left(2y\right)$.

Step-by-Step Solution

Answer

The answer is $-36e^{-6x}$.

Step 1. Given Information.

The function is $g\left(x,y\right)=e^{-3x}\sin \left(2y\right)$

Step 2. Explanation.

The discriminant is calculated by formula, 

$det\left(H_g\right)=\frac{{\partial }^{2}g}{\partial x^2}·\frac{{\partial }^{2}g}{\partial y^2}-{\left(\frac{{\partial }^{2}g}{\partial x\partial y}\right)}^2$.

Calculate $\frac{\partial g}{\partial x}$, $\frac{{\partial }^{2}g}{\partial x^2}$,  $\frac{\partial g}{\partial y}$, $\frac{{\partial }^{2}g}{\partial y^2}$, and $\frac{{\partial }^{2}g}{\partial x\partial y}$.

$\frac{\partial g}{\partial x}=-3e^{-3x}\sin 2y$, $\frac{{\partial }^{2}g}{\partial x^2}=9e^{-3x}\sin 2y$

$\frac{\partial g}{\partial y}=2e^{-3x}\cos 2y$, $\frac{{\partial }^{2}g}{\partial y^2}=-4e^{-3x}\sin 2y$.

$\frac{{\partial }^{2}g}{\partial x\partial y}=-6e^{-3x}\cos 2y$

Step 3. Calulation.

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

$$\begin{gathered} det\left(H_g\right)=\left(9e^{-3x}\sin 2y\right)\left(-4e^{-3x}\sin 2y\right)-{\left(-6e^{-3x}\cos 2y\right)}^2 \\ =-36e^{-6x}{\sin }^{2}2y-36e^{-6x}{\cos }^{2}2y \\ =-36e^{-6x}\left({\sin }^{2}2y+{\cos }^{2}2y\right) \\ =-36e^{-6x} \end{gathered}$$