Subsections

• Purpose
• Background
  • Partial derivatives
  
  
• The tangent plane for explicitly defined surfaces
• The tangent plane for implicitly defined surfaces
• Exercises

Partial Derivatives and the Tangent Plane

Purpose

The purpose of this lab is to acquaint you with using Maple to compute partial derivatives.

Background

When differentiating a function $$F(x,y)$$ of two (or more) variables, you need to specify which independent variable is being derived.

Partial derivatives

A differentiable function, $$F(x,y)$$, of two variables has two first order partial derivatives: $$\partial F /\partial x$$ and $$\partial F /\partial y$$. 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 $$\partial F /\partial x$$, you must treat the variable $y$ as if it was a constant and vice-versa when computing $$\partial F /\partial y$$.

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 commands below show examples of 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,y);
> D[1](f)(x,y);
> D[2,2](f)(x,y);

Note in the above D command that the 1 in the square brackets means x and the 2 means y. The next example shows how to evaluate the mixed partial derivative of the function given above at the point $$(-1,1)$$.

> eval(diff(f(x,y),x,y),{x=-1,y=1});
> D[1,2](f)(-1,1);

The tangent plane for explicitly defined surfaces

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 $$f(x)$$ at the point where $$x=a$$. Recall, the general equation of a line at the point $$(a,f(a)$$ having slope $$m$$ is $$y-f(a) = m(x-a)$$. This can be rewritten knowing that the derivative is the slope of a tangent line as $$y = f'(a)(x-a) + f(a)$$. Similarly for a funcion of two variables, the equation of the plane tangent to $z=f(x,y)$ at the point $(a,b)$ has the equation $$z = f_x(a,b)(x-a)+f_y(a,b)(y-b)+f(a,b)$$. The following examples will show you how to find the tangent plane to the function $$f(x,y)=\frac{1}{1+x^2+y^2}$$ at $$(\frac{1}{8},frac{1}{4})$$. 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 eval or subs command would be used.

> f:=(x,y)->1/(1+x^2+y^2);
> tp:=D[1](f)(1/8,1/4)*(x-1/8)+D[2](f)(1/8,1/4)*(y-1/4)+f(1/8,1/4);
> plot3d({f(x,y),tp},x=-1..1,y=-1..1);

To find a point, $$(x_0,y_0)$$, where the tangent plane is horizontal, you would need to solve where both first order partials are equal to zero simultaneously.

> solve({diff(f(x,y),x)=0,diff(f(x,y),y)=0},{x,y});

The horizontal plane at that point would simply be $$z=f(x_0,y_0)$$. Below is how to plot the surface and the horizontal tangent plane.

> tp:=f(0,0);
> plot3d({f(x,y),f(0,0)},x=-1..1,y=-1..1);

The tangent plane for implicitly defined surfaces

If $z$ is not explicitly solved for but assumed to be a function of $$x$$ and $$y$$, then the equation can be defined implicitly. Note the difference in the syntax for defining the surface for the implicitplot command. An equal sign must be included in the equation and defined as an expression, not a function. This gives the flexibility of being able to graph equations without having to solve for $$z$$ first.

>with(plots):
>surf:=x^2+y^2+z^2=1;
>implicitplot3d(surf,x=-2..2,y=-2..2,z=-2..2);

The tangent plane to an implicitly defined surface $$F(x,y,z)=0$$ is given below:

$$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0$$

To find the tangent plane to the sphere $$x^2+y^2+z^2=1$$ at the point $$(\frac{1}{2},-\frac{1}{2})$$ and $$z$$ is positive, you would first need to find the coordinating $$z$$ value for the $$(x,y)$$ ordered pair. Below is how you would do this in Maple as well as find and plot the tangent plane implicitly.

>with(plots):
>F:=x^2+y^2+z^2-1;
>solve(eval(F,{x=1/2,y=-1/2}),z);
>a:=eval(diff(F,x),{x=1/2,y=-1/2,z=sqrt(2)/2});
>b:=eval(diff(F,y),{x=1/2,y=-1/2,z=sqrt(2)/2});
>c:=eval(diff(F,z),{x=1/2,y=-1/2,z=sqrt(2)/2});
>tp:=a*(x-1/2)+b*(y+1/2)+c*(z-sqrt(2)/2)=0;
>implicitplot3d({F=0,tp},x=-2..2,y=-2..2,z=-2..2,numpoints=2000);

Exercises

1. Compute the three distinct second order partial derivatives of
   
   $$f(x,y)=\frac{\sin(x+y)}{(1+y^2)}$$
   
   
   at the point $(-1,1)$ using the diff command and then again using the D command.

2. Given:
   
   $$f(x,y)=\frac{3x^2}{5}+\frac{2y^2}{3}+1$$
   
   

   a)

   Find the plane tangent to the given surface at $(0.2,-0.3)$.

   b)

   Plot the surface $f(x,y)$ and the tangent plane on the same graph and rotate the 3-D plot to show the point of tangency. Use plotting ranges $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$.

   c)

   Find the point where the tangent plane to the given surface would be horizontal.

   d)

   Plot the surface $f(x,y)$ and the horizontal tangent plane on the same graph and rotate the 3-D plot to show the point of tangency. Use the same plotting ranges as above.

3. Use implicit methods to find and plot the plane tangent to the ellipsoid
   
   $$\frac{(x-1)^2}{3}+\frac{(y+1)^2}{2}+z^2=1$$
   
   
   at the point $(2,-1)$. Use plotting ranges $-1 \leq x \leq 3$, $-3 \leq y \leq 2$, and $-2 \leq z \leq 2$