Q. 15. - Calculus

Question

Sketch level curves $z=19,16,21$ and $24$ for the function

$z=25-x^2-y^2$. Include the graphs of three gradient

vectors on each level curve. What do you observe?

16. Sketch level curves$z=9,16,21$, and $16$ for the function.

Step-by-Step Solution

Answer

The graph of level curves $z=25-x^2-y^2$for $z=9,16,21 and 24$is as shown below.

 The gradient and tangent vectors are orthogonal. 

Step 1: The objective is to sketch the level curves z = 9 , 16 , 21   a n d   24 for the function

The function is $z=25-x^2-y^2$

The graph of the function $z=25-x^2-y^2$concentric circle, each of whose level curves is a circle centered at the origin.

So the gradient is 

$$\begin{gathered} z=25-x^2-y^2 \\ \nabla z=-2xi-2yj \end{gathered}$$

Hence, the magnitude of the gradient vectors grows.

Step 2: Draw a graph of level curves z = 25 - x 2 - y 2   for z = 9 , 16 , 21   a n d   24 and observe

The level of curves $z=25-x^2-y^2$for $z=9,16,21 and 24$ is

$$\begin{gathered} x^2+y^2=16\left(\sin ce \left.9=25-x^2-y^2\right) \right. \\ {x}^2+y^2=9 \left(\sin ce \left.16=25-x^2-y^2\right) \right. \\ {x}^2+y^2=4 \left(\sin ce \left.21=25-x^2-y^2\right) \right. \\ {x}^2+y^2=1 \left(\sin ce \left.24=25-x^2-y^2\right) \right. \end{gathered}$$

The graph of level curves $z=25-x^2-y^2$for $z=9,16,21 and 24$ is shown below:

Hence, the gradient and tangent vectors are orthogonal.