Q. 23 - Calculus

Question

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

$f(s,t)=s^3{e}^{t/2}$

Step-by-Step Solution

Answer

The answer is $-\frac{3}{4}{s}^4{e}^t$.

Step 1. Given Information.

The function is $f(s,t)=s^3{e}^{t/2}$.

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 $\frac{{\partial }^{2}g}{\partial s\partial t}$.

$\frac{\partial g}{\partial s}=3s^2{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$,      $\frac{{\partial }^{2}g}{\partial s^2}=6s{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

$\frac{\partial g}{\partial t}=\frac{1}{2}{s}^3{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$,    $\frac{{\partial }^{2}g}{\partial t^2}=\frac{1}{4}{s}^3{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

$\frac{{\partial }^{2}g}{\partial s\partial t}=\frac{3}{2}{s}^2{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Step 3. Calculation.

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

$$\begin{gathered} det\left(H_g\right)=\frac{{\partial }^{2}g}{\partial s^2}·\frac{{\partial }^{2}g}{\partial t^2}-{\left(\frac{{\partial }^{2}g}{\partial s\partial t}\right)}^2 \\ =\left(6s{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\left(\frac{1}{4}{s}^3{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-{\left(\frac{3}{2}{s}^2{e}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2 \\ =\frac{6}{4}{s}^4{e}^t-\frac{9}{4}{s}^4{e}^t \\ =-\frac{3}{4}{s}^4{e}^t \end{gathered}$$