Contents of this Calculus 2 episode:
Partial differentiation, Systems of equation, Stationary points, Critical points, First order derivatives, Second order derivatives, Hessian matrix, Local minimum, Local maximum, Saddle point.
Now, let’s see how we can find local minima and maxima using partial differentiation.
Differentiation:
Solving the system of equations:
The resulting number pairs are points in the x,y plane.
These points are called stationary points, and at these points
the function can have a minimum, a maximum or a saddle point.
The solutions of the linear system are the stationary points
And now we can get to the second order derivatives.
We arrange them neatly in a matrix that is called a Hessian matrix.
And then we substitute the stationary points.
We have to take these matrices and look at their ... ahem ... determinants.
If somebody have not heard about the determinants of matrices yet, well, that is understandable, it is a very simple thing.
Here is a 2x2 matrix,
and its determinant is a number.
This number can be positive, negative or zero.
Let’s say for this matrix here,
the determinant is -14.
We calculate the determinant of the Hessian matrix, which can be positive, negative or zero.
If the determinant is positive, that means there is a minimum or a maximum.
.
If it is negative, then there is a saddle point.
If it is zero, then further investigation is necessary, but it doesn’t happen very often.
We will try summarizing it in this tiny space here.
Let's see what happens at the two stationary points.
Well, it seems is a saddle point.
And is a local minimum.
Let's see another one like this.
Let’s find the local extrema and saddle points of the following function.
Here are the stationary points:
And now come the second derivatives.
Next, let's see what happens at the stationary points.
Differentiation:
Solving the system of equations:
, ,
, ,
Two stationary points: and
Here comes the Hessian matrix:
Now let's see the stationary points!
First let's check .
Substitute zero for x, y and z:
This is indefinite, so is a saddle point.
Next, let's see .
Substitute one for x and y, and zero for z:
This is positive definite, so it is a local minimum.
Calculus 2 episode