partial derivatives: classifying stationary points
- Thread starter judjud410
- Start date
Hello, judjud410!
Exactly where is your difficulty?
. . You don't know how to take partial derivatives?
. . You can't set them equal to zero and solve?
. . You don't know the Second Derivative Ttest?
\(\displaystyle z_x\:=\:-y\,+\,y^2\:=\:0\)
\(\displaystyle z_y\:=\:4y\,-\,x\,+\,2xy\:=\:0\)
The first equation gives us: \(\displaystyle \,y^2\,-\,y\:=\:0\;\;\Rightarrow\;\;y(y\,-\,1)\:=\:0\;\;\Rightarrow\;\;y\.=\,0,\;1\)
If \(\displaystyle y\,=\,0\), the second equation is: \(\displaystyle \,4\cdot0\,-\,x\,+\,2x\cdot0\:=\:0\;\;\Rightarrow\;\;x\,=\,0\)
\(\displaystyle \;\;\)We have a critical point: \(\displaystyle \,(0,\,0,\,0)\)
If \(\displaystyle y\,=\,1\), the second equation is: \(\displaystyle \,4\,-\,x\,+\,2x\:=\:0\;\;\Rightarrow\;\;x\,=\,-4\)
\(\displaystyle \;\;\)We have a critical point: \(\displaystyle \,(-4,\,1,\,2)\)
Second Derivative Test\(\displaystyle \;\;\)We will use: \(\displaystyle \,D\:=\
z_{xx})(z_{yy})\,-\,(z_{xy})^2\)
\(\displaystyle \;\;\;z_{xx}\:=\:0,\;\;\;z_{yy}\:=\:4\,+\,2x,\;\;z_{xy}\:=\:-1\,+\,2y\)
Hence: \(\displaystyle \,D\;=\;(0)(4\,+\,2x)\,-\,(-1\,+\,2y)^2\;=\;-(2y\,-\,1)^2\)
At \(\displaystyle (0,0,0):\;D\:=\:-(2\cdot0\,-\,1)^2\:=\:-1\) . . . saddle point
At \(\displaystyle (-4,1,2):\;D\:=\:-(2\cdot1\,-\,1)^2\:=\:-1\) . . . saddle point
Exactly where is your difficulty?
. . You don't know how to take partial derivatives?
. . You can't set them equal to zero and solve?
. . You don't know the Second Derivative Ttest?
\(\displaystyle z_x\:=\:-y\,+\,y^2\:=\:0\)
\(\displaystyle z_y\:=\:4y\,-\,x\,+\,2xy\:=\:0\)
The first equation gives us: \(\displaystyle \,y^2\,-\,y\:=\:0\;\;\Rightarrow\;\;y(y\,-\,1)\:=\:0\;\;\Rightarrow\;\;y\.=\,0,\;1\)
If \(\displaystyle y\,=\,0\), the second equation is: \(\displaystyle \,4\cdot0\,-\,x\,+\,2x\cdot0\:=\:0\;\;\Rightarrow\;\;x\,=\,0\)
\(\displaystyle \;\;\)We have a critical point: \(\displaystyle \,(0,\,0,\,0)\)
If \(\displaystyle y\,=\,1\), the second equation is: \(\displaystyle \,4\,-\,x\,+\,2x\:=\:0\;\;\Rightarrow\;\;x\,=\,-4\)
\(\displaystyle \;\;\)We have a critical point: \(\displaystyle \,(-4,\,1,\,2)\)
Second Derivative Test\(\displaystyle \;\;\)We will use: \(\displaystyle \,D\:=\
z_{xx})(z_{yy})\,-\,(z_{xy})^2\)\(\displaystyle \;\;\;z_{xx}\:=\:0,\;\;\;z_{yy}\:=\:4\,+\,2x,\;\;z_{xy}\:=\:-1\,+\,2y\)
Hence: \(\displaystyle \,D\;=\;(0)(4\,+\,2x)\,-\,(-1\,+\,2y)^2\;=\;-(2y\,-\,1)^2\)
At \(\displaystyle (0,0,0):\;D\:=\:-(2\cdot0\,-\,1)^2\:=\:-1\) . . . saddle point
At \(\displaystyle (-4,1,2):\;D\:=\:-(2\cdot1\,-\,1)^2\:=\:-1\) . . . saddle point