max min question calc 3

[mcwang719]

consider the function f(x,y)= x^3+y^3+3x^2-3y^2-8.
a. find the critical points
the critical points i got were (0,0) and (-2,2) by setting the partial deriv to 0. can anyone confirm.
b. analyze each critical point and determine whether it is a local max min or saddle point.
what i did was got the second partial derv. fxx, fxy, and fyy then used the second deriv test. for (0,0) i got -4 which is < 0 so that would be a saddle point. for (-2,2) i got the same thing -4 ( which doesn't seem right).
c find a unit normal to the curve at the point P(1,-1,f(1,-1)). ( isn't that the gradient vector?)
can anyone help? thanks!!!

 

[galactus]

Your function, \(\displaystyle f(x,y)=x^{3}+y^{3}+3x^{2}-3y^{2}-8\), has

first derivatives:

\(\displaystyle f_{x}(x,y)=3x^{2}-6x=0\) and \(\displaystyle 3y^{2}-6y=0\)

Solving these, respectively, for x and y yields:

x=0 or 2 and y=0 or 2. So, you have choices for relative extrema

(2,2); (2,0); (0,2); (0,0)

Taking the second derivatives:

\(\displaystyle f_{xx}(x,y)=6x-6\) and \(\displaystyle f_{yy}(x,y)=6y-6\) and \(\displaystyle f_{xy}(x,y)=0\)

\(\displaystyle D=f_{xx}(x,y)f_{yy}(x,y)-f^{2}_{xy}(x,y)\)

\(\displaystyle D=(6x-6)(6y-6)\).

Check D with your coordinates to see which is a min or max.

If \(\displaystyle D>0\ and\ f_{xx}(x,y)>0\), then you have a relative minimum.

If \(\displaystyle D>0\ and\ f_{xx}(x,y)<0\), then you have a relative maximum.

If D<0, saddle point.


As for the last problem, do you need the unit normal line at (1,1,-8)?.

 

[mcwang719]

problem c just states find a unit normal to the curve at the point P(1,-1,f(1,-1)).

 

[pka]

For part c.
\(\displaystyle \L
\begin{array}{l}
N = \frac{{T'}}{{\left\| {T'} \right\|}} \\
T = \frac{{R'}}{{\left\| {R'} \right\|}}\quad \Rightarrow \quad T' = \frac{{\left\| {R'} \right\|^2 R'' - \left( {R' \cdot R''} \right)R'}}{{\left\| {R'} \right\|^3 }} \\
\end{array}\).

 

[mcwang719]

for problem c finding the unit normal to the curve at the point P(1,-1,f(1,-1))
i have

N = (-f_x(a,b), -f_y(a,b), 1)
fx=3x^2+6x fy= 3y^2-6y
(-9,-9,1)
is that the correct answer?

am i on the right track? can anyone help. thanks!!!

 

[pka]

I am sorry that I did not read the question completely the first time.
When you said “normal to the curve”, I did not see that it was a level curve in R<SUP>2</SUP>.
So actually the question wants a unit gradient.

\(\displaystyle \L
\begin{array}{l}
\nabla f = < 3x^2 + 6x,3y^2 - 6y > \\
P = (1, - 1, - 8) \\
\nabla f(P) = < 9,9 > \\
u = \left\langle {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right\rangle \\
\end{array}\).