Q 61. - Calculus

Question

Ian is traveling along a glacier. The elevation of the glacier in his area is described by the function $f (x, y) = 1.2 - 0.2x 2 - 0.3y2 + 0.1xy - 0.25x$,

where x, y, and f are given in miles.

(a) What is the direction that Ian would need to go to descend most steeply if he is at the point $(0.5,-0.5)$?

(b) In what direction would Ian have to turn in order to contour (i.e., , neither ascend nor descend) across the glacier?

Step-by-Step Solution

Answer

Part (a) The person descend most steeply at the point $(0.5,-0.5)$ in the direction $⟨0.5,-0.35⟩$

Part (b) The required direction is

Part (a) Step 1: Given information

$f(x,y)=1.2-0.2x^2-0.3y^2+0.1xy-0.25x$

Part (a) Step 2: Calculation

The goal is to determine which way the individual declines the steepest at the point $(0.5,-0.5)$

The gradient of $f(x, y)$ is given by

$\nabla f(x,y)=f_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}+f_z(x,y)\mathrm{k}$

The given function's gradient is provided by

$$\begin{gathered} \nabla f(x,y)=f_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j} \\ =\frac{\partial }{\partial x}\left(1.2-0.2x^2-0.3y^2+0.1xy-0.25x\right)\mathrm{i}+\frac{\partial }{\partial y}\left(1.2-0.2x^2-0.3y^2\right. \\ +0.1xy-0.25x)\mathrm{j} \\ =(0-(0.2)2x-0+0.1y-0.25)\mathrm{i}+(0-0-(0.3)2y+0.1x-0)\mathrm{j} \\ =(-0.4x+0.1y-0.25)\mathrm{i}+(-0.6y+0.1x)\mathrm{j} \end{gathered}$$

At the point $(0.5,-0.5)$ the gradient is

$$\begin{gathered} \nabla f(0.5,-0.5)=\left\{\right(-0.4)0.5+0.1(-0.5)-0.25\}\mathrm{i}+\{-0.6(-0.5)+(0.1)0.5\}\mathrm{j} \\ =(-0.2-0.05-0.25)\mathrm{i}+(0.3+0.05)\mathrm{j} \\ =-0.5\mathrm{i}+0.35\mathrm{j} \\ =⟨-0.5,0.35⟩ \end{gathered}$$

In the direction $\nabla f(0.5,-0.5)=⟨-0.5,0.35⟩$ the function is increasing most rapidly at $(0.5,-0.5)$ In the opposite direction of the growth direction, the function declines.

Thus, the function decreases in the direction

$$\begin{gathered} -\nabla f(0.5,-0.5)=-⟨-0.5,0.35⟩ \\ =⟨0.5,-0.35⟩ \end{gathered}$$

As a result, the individual descends most sharply in the direction $⟨0.5,-0.35⟩$ of the location $(0.5,-0.5)$$\nabla f(0.5,-0.5)=⟨-0.5,0.35⟩$

Part (b) Step 1: Calculation

The gradient direction derivative at a point $\left(x_0,y_0\right)$ is given by

$D_{\mathbf{u}}f\left(x_0,y_0\right)=\nabla f\left(x_0,y_0\right)·\mathbf{u}$

Since the person neither ascends nor descends, the directional derivative at $(0.5,-0.5)$ is zero.

Thus, 

$$\begin{gathered} D_{\mathbf{u}}f\left(x_0,y_0\right)=\nabla f\left(x_0,y_0\right)·\mathbf{u} \\ =0 \end{gathered}$$

The function's gradient at $(0.5,-0.5)$ is

$\nabla f(0.5,-0.5)=⟨-0.5,0.35⟩$

Let  be the unit vector. Then,

Since  is a unit vector, so

$$\begin{gathered} \sqrt{u_1^2+u_2^2}=1 \\ {u}_1^2+u_2^2=1\dots \dots \left(2\right) \end{gathered}$$

Substitute $u_1=0.7u_2$ from ( 1 ) in ( 2)

$$\begin{gathered} {\left(0.7u_2\right)}^2+u_2^2=1 \\ 0.49u_2^2+u_2{}^2=1 \\ 1.49u_2^2=1 \\ {u}_2{}^2=\frac{1}{1.49} \end{gathered}$$

$u_2=\pm \frac{1}{\sqrt{1.49}}$

Put $u_2=\pm \frac{1}{\sqrt{1.49}}$ in $u_1=0.7u_2$ so

$u_1=\pm \frac{0.7}{\sqrt{1.49}}$

Therefore, the required direction is