Q. 12 - Calculus

Question

Show that the function $h(x, y) = x^3 + y^3$ has a stationary point at the origin. Show that the discriminant det($H_h$(0, 0)) = 0. Show that there are points arbitrarily close to the origin such that h(x, y) > 0. Show that there are points arbitrarily close to the origin such that h(x, y) < 0. Explain why all this shows that h has a saddle at the origin .

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{All} \mathrm{the} \mathrm{points} \mathrm{near} \mathrm{the} \mathrm{origin} \mathrm{of} \mathrm{the} \mathrm{form} (-\mathrm{x},0) ,(-\mathrm{x}, -\mathrm{y}) \mathrm{and} (0, -\mathrm{y}), \mathrm{with} \mathrm{r}>0. \mathrm{y}>0, \\ \mathrm{h}(\mathrm{x},\mathrm{y})<0 \mathrm{and} \mathrm{for} \mathrm{the} \mathrm{points} \mathrm{of} \mathrm{the} \mathrm{form} (\mathrm{x},0), (\mathrm{x},\mathrm{y}) \mathrm{and} (0 , \mathrm{y} ) \mathrm{with} \mathrm{x}>0 ,\mathrm{y} >0, \mathrm{h}(\mathrm{x},\mathrm{y})>0. \\ \mathrm{Hence} \mathrm{near} \mathrm{the} \mathrm{origin} \mathrm{the} \mathrm{function} \mathrm{has} \mathrm{a} \mathrm{saddle}.. \end{gathered}$$

Step 1. Given

$Function h(x, y) = x^3 + y^3$

Step 2. Calculating critical point

$$\begin{gathered} \mathrm{Given} \mathrm{function} \mathrm{is} h(x, y) = x^3 + y^3 \mathrm{which} \mathrm{is} \mathrm{polynomial} \mathrm{and} \mathrm{hence} \mathrm{differentiable} . \mathrm{The} \mathrm{gradient} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{is} \\ \nabla h (x ,y, z ) =\frac{\partial h}{\partial x}i+\frac{\partial h}{\partial y}j. \\ =3x^{2}i+ 3y^{2}j \\ \mathrm{At} \mathrm{the} \mathrm{critical} \mathrm{points} \mathrm{the} \mathrm{gradient} \mathrm{of} \mathrm{a} \mathrm{function} \mathrm{vanishes} , \mathrm{that} \mathrm{is} \nabla h(x, y) =0, \\ That is 3x^2=0 .....\left(1\right) and 3y^2 =0.....\left(2\right) \\ \mathrm{The} \mathrm{solution} \mathrm{of} 1 \mathrm{and} 2 \mathrm{is} , x=0 and y=0 , \mathrm{So} \mathrm{the} \mathrm{critical} \mathrm{point} \mathrm{is} ( 0, 0\hspace{0.17em}). \end{gathered}$$

Step 3. Discriminate of function .

$$\begin{gathered} \mathrm{Now} \mathrm{the} \mathrm{second} \mathrm{order} \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{are} , \\ \frac{{\partial }^{2}h}{\partial x^2}=6x ,\frac{{\partial }^{2}h}{\partial y^2}= 6y , \frac{{\partial }^{2}h}{\partial x \partial y}=0 \\ \mathrm{Hence} \mathrm{the} \mathrm{discriminate} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{is}, \\ {H}_h( x , y ) = \frac{{\partial }^{2}h}{\partial x^2}\frac{{\partial }^{2}h}{\partial y^2}- {\left(\frac{{\partial }^{2}h}{\partial x \partial y}\right)}^2 = \left(6x\right)\left(6y\right) -0 \\ = 36xy \\ \mathrm{At} (0,0) {\mathrm{H}}_{\mathrm{h}}(0,0)=0. \mathrm{so} \mathrm{no} \mathrm{conclusion} \mathrm{can} \mathrm{be} \mathrm{drawn} \mathrm{using} \mathrm{the} \mathrm{discriminate} \\ \mathrm{Now} \mathrm{all} \mathrm{the} \mathrm{points} \mathrm{near} \mathrm{the} \mathrm{origin} \mathrm{of} \mathrm{the} \mathrm{form} (-\mathrm{x},0) ,(-\mathrm{x}, -\mathrm{y}) \mathrm{and} (0, -\mathrm{y}), \mathrm{with} \mathrm{r}>0. \mathrm{y}>0, \\ \mathrm{h}(\mathrm{x},\mathrm{y})<0 \mathrm{and} \mathrm{for} \mathrm{the} \mathrm{points} \mathrm{of} \mathrm{the} \mathrm{form} (\mathrm{x},0), (\mathrm{x},\mathrm{y}) \mathrm{and} (0 , \mathrm{y} ) \mathrm{with} \mathrm{x}>0 ,\mathrm{y} >0, \mathrm{h}(\mathrm{x},\mathrm{y})>0. \\ \mathrm{Hence} \mathrm{near} \mathrm{the} \mathrm{origin} \mathrm{the} \mathrm{function} \mathrm{has} \mathrm{a} \mathrm{saddle}. \end{gathered}$$