Gradients.html

Gradients and directional derivatives

Gradients are defined in the VectorCalculus package.

> restart; with(LinearAlgebra): with(VectorCalculus): with(plots):

Warning, the names &x, CrossProduct and DotProduct have been rebound

Warning, the assigned names <,> and <|> now have a global binding

Warning, these protected names have been redefined and unprotected: *, +, ., D, Vector, diff, int, limit, series

Warning, the name changecoords has been redefined

The Gradient function takes gradients of expressions.

> f:= x*y*ln(x);

> Gradient(f,[x,y]);

> Gradient( sin(x*z)*cos(y+z), [x,y,z] );

The gradient is actually what is known as a vector field, meaning a vector whose components are functions of the space variables. (That's why the unit vectors above have a bar over them.) You can evaluate a field at a point to get a vector.

> evalVF( Gradient(f,[x,y]), <2,1> );

With a gradient vector you can get the derivative at a point in any direction. Remember to take the dot product with a unit vector.

> % . <1,-1>/sqrt(2);

You can see the special relationship between gradients and level curves using gradplot as follows.

> p1:= contourplot(f, x=1..2,y=0..1, contours=16):
p2:= gradplot(f, x=1..2,y=0..1, color=blue ):

> display({p1,p2},axes=boxed,scaling=constrained);

Examples

Gradient as an unknown

Suppose that the directional derivative of a function f(x,y) at (2,-1) equals 2*sqrt(2) in the direction of i+j and -3 in the direction of -j. What is the maximum possible directional derivative of f at (2,-1)?

We set up the gradient as an unknown. Once this is found, the answer is simple.

> gradf:= <a,b>;

Now for the data given in the problem:

> eq1:= gradf . (<1,1>/sqrt(2)) = 2*sqrt(2);

> eq2:= gradf . (<0,-1>) = -3;

We use these two pieces of information to find the unknowns.

> s:= solve( {eq1,eq2}, {a,b} );

The answer to the problem is thus

> Norm( subs(s,gradf), 2 );

Tangent line

Find an equation for the line tangent to the ellipse x^2 - x*y + y^2 = 7 at the point (-2,1).

> ellipse:= x^2 - x*y + y^2 = 7;

The ellipse is a level curve of the function

> f:= lhs(ellipse);

Hence the gradient of f is normal to the ellipse, and hence normal to all tangent lines for the ellipse.

> n:= evalVF( Gradient(f,[x,y]), <-2,1> );

To get an equation for a line, we need a vector parallel to it, not normal to it. A simple trick (in the 2D case) is to reverse the components and flip the sign of one of them.

> v:= <4,5>;