Q. 8 - Calculus

Question

8. Consider the function $f(x, y)=2 x+3 y$

(a) Why is the graph of $f$ a plane?

(b) In what direction is $f$ increasing most rapidly at the point $(-1,4)$ ?

(c) In what direction is $f$ increasing most rapidly at the point $\left(x_0,y_0\right)$ ?

(d) Why are your answers to parts (b) and (c) the same?

Step-by-Step Solution

Answer

a, The graph is a plane because the graph has three dimensions

b, The direction in which $f$increasing most rapidly at the point $(-1,4)$is determined as $(2,3)$

c, The direction in which $f$increasing most rapidly at the point $(x_0,y_0)$is determined as $(2,3)$

d, The answers to parts (b) and (c) the same because the gradient is same whatever maybe the point

Introduction

The given data is the function$f(x,y)=2x+3y$

The objective is to find why is the function a plane, in what direction is the function increasing most rapidly and why is the parts same

Step 1

a,

Let the function be

$f(x, y)=2 x+3 y$

The goal is to explain why the function's graph is a plane.

Let $f(x, y)=z$ then

$z=2 x+3 y$

There are three variables in the equation. The variable $z$ is affected by the variables $x$ and $y$. Each variable is influenced by the other two variables.

As a result, the equation is three-dimensional, and the graph is flat.

Step 2

(b)

The goal is to determine which direction the provided function increases the most rapidly at $(-1,4)$.

Find the gradient of the given function at the given position, since the gradient of the function is the direction in which the function increases the most rapidly.

The function's gradient is.

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

$=\left\{\frac{\partial }{\partial x}(2x+3y)\right\}\mathrm{i}+\left\{\frac{\partial }{\partial y}(2x+3y)\right\}\mathrm{j}$ $=\left(2\frac{\partial }{\partial x}x+3\frac{\partial }{\partial x}y\right)\mathrm{i}+\left(2\frac{\partial }{\partial y}x+3\frac{\partial }{\partial y}y\right)\mathrm{j}$

$=(2·1+3·0)\mathrm{i}+(2·0+3·1)\mathrm{j}$

$=2\mathrm{i}+3j$

Step 3

From the equation $\left(2\right)$ at the point $(-1,4)$the gradient can be determined as follows:

$$\begin{gathered} \nabla z=2i+3j \\ =<2,3> \end{gathered}$$

As a result, the given function increases most rapidly in the direction $<2,3>$ at point $(-1,4)$.

Step 4

c.

The goal is to determine which way the provided function rises the most rapidly at the position $(x_0,y_0)$.

The gradient is $(x_0,y_0)$ from $\left(2\right)$ to the point $(x_0,y_0)$

$$\begin{gathered} \nabla z =2 i+3 j \\ =<2,3> \end{gathered}$$

As a result, the given function increases most rapidly in the direction $<2,3>$ at point $(x_0,y_0)$.

Step 5

(d)

The goal is to explain why the results in (b) and (c) are identical. You can see from $\left(2\right)$ that the gradient is independent of $x$ and $y$. So, regardless of the position, the gradient is the same.

As a result, both outcomes are identical.