Q. 9 - Calculus

Question

The function $g(x,y)=x^2-y^2$ is graphed in the figure as follows. use the definition of the partial derivatives to show that $f_x(0,0)=f_y(0,0)=0$ what are the equations of the lines tangent to the surface on the x and y directions at (0,0,0). what is the equation of the plane containing these two lines. 

Step-by-Step Solution

Answer

The value of the partial derivative is $$\begin{gathered} f_x(0,0)=0 \\ {f}_y(0,0)=0 \end{gathered}$$

The equation of the lines tangent to the surface in the x and y direction at (0,0,0) is 0.

Equation of the plane containing these two line is z=0.

Step 1: Given information

We are given an equation $g(x,y)=x^2-y^2$.

Step 2: Use the definition of partial derivatives to compute the partial derivatives

Using the definition we have

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{g(x+h,y)-g(x,y)}{h} \\ \underset{h\to 0}{\lim }\frac{{(x+h)}^2-y^2-x^2+y^2}{h} \\ \underset{h\to 0}{\lim }\frac{x^2+2xh+h^2-y^2-x^2+y^2}{h} \\ \underset{h\to 0}{\lim }2x+h \\ \underset{h\to 0}{\lim }2x \\ At point (0,0) we have this value \\ equal to zero \end{gathered}$$

Similarly we have,

$$\begin{gathered} \underset{k\to 0}{\lim }\frac{f(x,y+k)-f(x,y)}{k} \\ \underset{k\to 0}{\lim }\frac{x^2-{(y+k)}^2-x^2+y^2}{k} \\ \underset{k\to 0}{\lim }-2y+k \\ =-2k \\ At point (0,0) we have this value \\ equal to zero \end{gathered}$$

Step 3: Compute equations of the tangent line

The equation of tangent line in x direction is given as

$$\begin{gathered} z=f(0,0)+f_x(0,0) \\ z=0 \end{gathered}$$

The equation of tangent line in y direction is given as

$$\begin{gathered} z=f(0,0)+f_y(0,0)t \\ z=0 \end{gathered}$$

Now we find the equation of the plane containing these lines to do this we find the cross product of

$$\begin{gathered} N=(1,0,f_y(0,0\left)\right)\times (0,1,fx(0,0\left)\right) \\ N=(0,0,-1) \end{gathered}$$

And now we take dot product of N and vector (x,y,z)

We get,

$$\begin{gathered} (0,0,-1)(x,y,z) \\ -z=0 \end{gathered}$$