Q. 14. - Calculus

Question

Sketch level curves $z=1,4,9$ and 16 for the function

$z=x^2+y^2$. Include the graphs of three gradient vectors

on each level curve. What do you observe?

Step-by-Step Solution

Answer

The required graph of level curves $c=x^2+y^2$and $c=1,4,9$and $16$is  shown below.

 The gradient and tangent vectors are orthogonal

Step 1: The objective is to sketch the level curves z = 1 , 4 , 9 , 16   for the function.

The function is $z=x^2+y^2$

The graph of the function $z=x^2+y^2$is a paraboloid with a circle centered at the origin for each of its level curves.

The gradient is 

$$\begin{gathered} z=x^2+y^2 \\ \nabla z=2xi+2yj \end{gathered}$$

Hence, every gradient vector emanates from the origin in a direct line.

Step 2: Draw graph of level curves c = x 2 + y 2   for c = 1 , 4 , 9   and 16 and observe

For the point $\left(x_0,y_0\right)$

The gradient is $\nabla z\left(x_0,y_0\right)=2x_0\mathbf{i}+2y_0\mathbf{j}\mathbf{ }$and

a tangent vector to the level curve containing the point $\left(x_0,y_0\right)$is $y_0\mathbf{i}-x_0\mathbf{j}\mathbf{ }$

The level of curves $z=x^2+y^2$for $z=1,4,9 and 16$ is:

$$\begin{gathered} x^2+y^2=1 \\ {x}^2+y^2=4 \\ {x}^2+y^2=9 \\ {x}^2+y^2=16 \end{gathered}$$

The graph of level curves $c=x^2+y^2$for $z=1,4,9 and 16$is as shown below.

Therefore, the gradient and tangent vectors are orthogonal.