Q. 21 - Calculus

Question

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

$f\left(x,y\right)=e^{2x}\cos y$.

Step-by-Step Solution

Answer

The answer is $-4e^{4x}$.

Step 1. Given Information.

The function is $f\left(x,y\right)=e^{2x}\cos y$.

Step 2. Explanation.

The discriminant is calculated by formula,

$det\left(H_f\right)=\frac{{\partial }^{2}f}{\partial x^2}·\frac{{\partial }^{2}f}{\partial y^2}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^2$

Find $\frac{\partial f}{\partial x}$, $\frac{{\partial }^{2}f}{\partial x^2}$, $\frac{\partial f}{\partial y}$, $\frac{{\partial }^{2}f}{\partial y^2}$ and $\frac{{\partial }^{2}f}{\partial x\partial y}$

$\frac{\partial f}{\partial x}=2e^{2x}\cos y$,  $\frac{{\partial }^{2}f}{\partial x^2}=4e^{2x}\cos y$

$\frac{\partial f}{\partial y}=-e^{2x}\sin y$,  $\frac{{\partial }^{2}f}{\partial y^2}=-e^{2x}\cos y$

$\frac{{\partial }^{2}f}{\partial x\partial y}=-2e^{2x}\sin y$

Step 3. Calculate Discriminant.

Calculate $det\left(H_f\right)=\frac{{\partial }^{2}f}{\partial x^2}·\frac{{\partial }^{2}f}{\partial y^2}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^2$

$$\begin{gathered} det\left(H_f\right)=\left(4e^{2x}\cos y\right)\left(-e^{2x}\cos y\right)-{\left(-2e^{2x}\sin y\right)}^2 \\ =-4e^{4x}{\cos }^{2}y-4e^{4x}{\sin }^{2}x \\ =-4e^{4x} \end{gathered}$$