# Math 2060 - Section 14.3

## 3013 days ago by jimlb

Section 14.3

Worksheet by Jim Brown

In this worksheet we give some graphs illustrating the partial derivatives of a function of two variables $z = f(x,y)$.

Example 1:  Consider the function $z = f(x,y) = -x^2 - 2 y^2$.  We have the point $(1,-1,-3)$ on the graph and consider the partial derivatives of $f(x,y)$ with respect to $x$ and $y$ at this point.  The point is in black on the graph.

var('x,y');P=plot3d(-x^2 - 2*y^2, (x,-2,2), (y,-2,2));P+=point((1,-1,-3),size=15,color='black');P.show()
 (x, y) (x, y)

To calculate the partial derivative of $f(x,y)$ with respect to $x$ at $(1,-1,-3)$, we fix $y = -1$ and consider the slope of the tangent line at the point $(1,-1,-3)$ to the curve $z = f(x,-1)$.  The curve is given in yellow and the tangent line is given in red.  Thus, we have $\frac{\partial f}{\partial x}(1,-1,-3) = -2$ is the slope of the red line.

P+=parametric_plot3d([x,-1,-x^2 -2], (x,-2,2), color='yellow',thickness=5);P+=parametric_plot3d([x,-1,-2*(x-1)-3], (x,-2,2), color='red', thickness=5);P.show()

We now add the curve $z = f(1,y)$ and the tangent line to this curve at $(1,-1,-3)$.  The curve is given in green and the tangent line in purple.  We have the quantity $\frac{\partial f}{\partial y}(1,-1,3)$ is the slope of the purple line.

P+=parametric_plot3d([1,y,-1-2*y^2],(y,-2,2), color='green', thickness=5);P+=parametric_plot3d([1,y,4*(y+1)-3],(y,-2,2), color='purple', thickness=5);P.show()