Q. 8 - Calculus

Question

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

(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 (x 0, y 0)?

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

Step-by-Step Solution

Answer

 The slope of the slope will be. Its gradient and similar results are the same.

step:1 (part a) Graph of f plane

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

Its purpose is to explain why the curve function is a plane. Then let 

 $f(x, y)=z$ 

$z=2x+3y.$

 zThe formula contains three variable. Variable $z$ depends on the other two variable $x$ and $y.$

Each variable depends on the other two variables.

Therefore, the equation is in three dimensions, and therefore the graph is a plane.

step:2 (part b) direction increasing at points

 The aim is the direction in which the given function, most of the increases to the point $(-1,4)$quickly .

Since the slope of the function is the direction in which the function of most of the increases quickly, so you can find the slope of the function at the given point.

The slope of the function is

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

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

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

$=(2.1+3.0)i+(2.0+3.1)j$

$=2i+3j$

part:3 (part b) calculation

From (Step 2) in point$(-1,4)$ the gradient is as follows:                                

Therefore, the specified function is the most rapid point $(-1,4)$ in the direction of the

step:4 (part c) increase in direction

The aim is the direction in which the given function, most of the increases to the point quickly $\left(x_0,y_0\right)$

From $\left(2\right)$ at the point $\left(x_0,y_0\right)$the gradient is

Therefore, the specified function is the most rapid point $\left(x_0,y_0\right)$ in the direction of the

step:5 (part d) same answer for both parts

The objective is to explain why the result obtained in $\left(b\right)$ and $\left(c\right)$are same. From $\left(2\right)$you can notice that the gradient does not depend on $x$ and $y.$So, what might be the point of the course.? Hence, both the results are same