Q. 45 - Calculus

Question

In Exercises 43–48: (a) Find the direction in which the given function increases most rapidly at the specified point. (b) Find the rate of change of the function in the direction you found in part (a). (c) Find the direction in which the given function decreases most rapidly at the specified point. Note: These are the same functions as in Exercises 37–42. 

$f\left(x,y\right)=\sqrt{x^2+y^2} \mathrm{at} \left(2,-3\right)$

Step-by-Step Solution

Answer

Part (a): The direction in which the given function increases most rapidly is .

Part (b): The rate of change of the function is $1$.

Part (c): The direction in which the given function decreases most rapidly is .

Step 1. Given information.

The given function is:

$f\left(x,y\right)=\sqrt{x^2+y^2}$

Part (a) step 1. Calculation.

First we find the gradient of the given function.

$\nabla f=\frac{\partial f}{\partial x}\hat{\mathrm{i}}+\frac{\partial f}{\partial y}\hat{\mathrm{j}}=\left(\frac{x}{\sqrt{x^2+y^2}}\right)\hat{\mathrm{i}}+\left(\frac{y}{\sqrt{x^2+y^2}}\right)\hat{\mathrm{j}}$

Now we find the gradient of the function at the point $\left(2,-3\right)$by putting $x=2 \mathrm{and} y=-3$we will get,

${\overline{)\nabla f}}_{\left(2,-3\right)}=\frac{2}{\sqrt{13}}\hat{\mathrm{i}}-\frac{3}{\sqrt{13}}\hat{\mathrm{j}}$

So, the direction in which the given function increases most rapidly is .

Part (b) Step 1. Calculation.

The rate of change of the function in $\left(\frac{2}{\sqrt{13}}\hat{\mathrm{i}}-\frac{3}{\sqrt{13}}\hat{\mathrm{j}}\right)$direction is:

$||\nabla f\left(2,-3\right)||=\sqrt{{\left(\frac{2}{\sqrt{13}}\right)}^2+{\left(\frac{-3}{\sqrt{13}}\right)}^2}=1$

Part (c) Step 1. Calculation.

The direction in which the given function decreases most rapidly at $\left(2,-3\right)$ is the opposite of direction in which the function increases the most rapidly that is:

So, the direction in which the given function decreases most rapidly is .