Q. 24 - Calculus

Question

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

$g\left(s,t\right)=s^{2}t{e}^{-s{t}^2}$.

Step-by-Step Solution

Answer

The answer is $6t^6{s}^5{e}^{-s{t}^2}-25t^4{s}^4{e}^{-s{t}^2}+16t^2{s}^3{e}^{-2s{t}^2}-4s^2{e}^{-2s{t}^2}$

Step 1. Given Information.

The function is $g\left(s,t\right)=s^{2}t{e}^{-s{t}^2}$.

Step 2. Explanation.

The discriminant is calculated by formula,

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

calculate $\frac{\partial g}{\partial s}$, $\frac{\partial g}{\partial t}$, $\frac{{\partial }^{2}g}{\partial s^2}$, $\frac{{\partial }^{2}g}{\partial t^2}$ and $\frac{{\partial }^{2}g}{\partial s\partial t}$

$\frac{\partial g}{\partial s}=2st{e}^{-s{t}^2}-s^2{t}^3{e}^{-s{t}^2}$,   $\frac{{\partial }^{2}g}{\partial s^2}=2t{e}^{-s{t}^2}-4s{t}^3{e}^{-s{t}^2}+s^2{t}^5{e}^{-s{t}^2}$

$\frac{\partial g}{\partial t}=s^2{e}^{-s{t}^2}-2s^3{t}^2{e}^{-s{t}^2}$

$\frac{{\partial }^{2}g}{\partial t^2}=-6s^{3}t{e}^{-s{t}^2}+4s^4{t}^3{e}^{-s{t}^2}$

$\frac{{\partial }^{2}g}{\partial s\partial t}=2s{e}^{-s{t}^2}-2s^2{t}^2{e}^{-s{t}^2}+2s^3{t}^4{e}^{-s{t}^2}$

Step 3. Calculation.

Calculate $det\left(H_g\right)=\frac{{\partial }^{2}g}{\partial s^2}·\frac{{\partial }^{2}g}{\partial t^2}-{\left(\frac{{\partial }^{2}g}{\partial s\partial t}\right)}^2$

$$\begin{gathered} det\left(H_g\right)=\left(2t{e}^{-s{t}^2}-4s{t}^3{e}^{-s{t}^2}+s^2{t}^5{e}^{-s{t}^2}\right)\left(-6s^{3}t{e}^{-s{t}^2}+4s^4{t}^3{e}^{-s{t}^2}\right)-{\left(2s{e}^{-s{t}^2}-2s^2{t}^2{e}^{-s{t}^2}+2s^3{t}^4{e}^{-s{t}^2}\right)}^2 \\ =6t^6{s}^5{e}^{-s{t}^2}-25t^4{s}^4{e}^{-s{t}^2}+16t^2{s}^3{e}^{-2s{t}^2}-4s^2{e}^{-2s{t}^2} \end{gathered}$$