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Subsections
The purpose of this lab is to acquaint you with using Maple to compute partial derivatives.
For a function
of a single real variable, the derivative
gives information on whether the graph of
is increasing or decreasing. Finding where the derivative is zero was important in finding extreme values. For a function
of two (or more) variables, the situation is more complicated.
A differentiable function,
, of two variables has two partial
derivatives:
and
. As you have learned in class, computing partial derivatives is
very much like computing regular derivatives. The main difference is
that when you are computing
, you must treat
the variable
as if it was a constant and vice-versa when computing
.
The Maple commands for computing partial derivatives are D
and diff. The diff command can be used on both expressions and functions whereas the D command can be used only on functions. The examples below show all first order and second order partials in Maple.
> f := (x,y) -> x^2*y^2-x*y;
> diff(f(x,y),x);
> diff(f(x,y),y);
> diff(f(x,y),x,x);
> diff(f(x,y),y,y);
> diff(f(x,y),x,y);
> D[1](f)(x,y);
> D[2](f)(x,y);
> D[1,1](f)(x,y);
> D[2,2](f)(x,y);
> D[1,2](f)(x,y);
The next example shows how to evaluate the mixed partial derivative of the function given above at the point
.
> der := diff(f(x,y),x,y);
> subs({x=-1,y=1},der);
> D[1,2](f)(-1,1);
The tangent plane like the tangent line to a single variable function is based on derivatives, however the partial derivatives are used for the tangent plane. Let's start with the equation of the tangent line to the function
at the point where
. Recall, the general equation of a line at the point
having slope
is
. This can be rewritten knowing that the derivative is the slope of a tangent line as
. Similarly for a funcion of two variables, the equation of the plane tangent to
at the point
has the equation
. The following example will show you how this can easily be translated to Maple syntax.
> g := x-> sin(x)-x^3/7+x^2;
> tanline := D(g)(5)*(x-5)+g(5);
> plot({g(x),tanline},x=-2..8);
The next example shows how to find the tangent plane to the function
at
. You could write the partials with diff or D. This example uses D as it is easier to plug in the the point with this syntax; with diff the subs command would be used.
> j:=(x,y)->1/(1+x^2+y^2);
> tp:=D[1](j)(1/8,1/4)*(x-1/8)+D[2](j)(1/8,1/4)*(y-1/4)+j(1/8,1/4);
> plot3d({j(x,y),tp},x=-1..1,y=-1..1,style=patchnogrid);
To find a point where the tangent plane to
is horizontal, you would need to solve where both first order partials are equal to zero simultaneously.
> solve({diff(j(x,y),x)=0,diff(j(x,y),y)=0},{x,y});
- Compute the three distinct second order partial derivatives of
at the point
using the diff command and then again using the D command.
- Given the single variable function
- a)
- Plot
over the interval
.
- b)
- Find the slope of
at
.
- c)
- Find the two
values that have the same slope as in part b.
- d)
- For each of the points from part c, find the equation of the line that is tangent to the graph of
at these points. Plot the function and the tangent lines on the same graph over the interval given in part a to show the lines have the same slope.
- Given:
- a)
- Find the tangent plane at
.
- b)
- Plot the function
and the tangent plane on the same graph and rotate to see the point of tangency.
- c)
- Find the point on the graph of
where the tangent plane is horizontal. Plot the function and this tangent plane on the same graph and rotate to see the point of tangency.
Next: About this document ...
Up: lab_template
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Dina J. Solitro-Rassias
2011-09-10