Q. 10 - Calculus

Question

10. Consider the function $f(x, y)=ax+by$, where neither a nor $b$ is zero.

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

(b) In what direction is $f$ increasing most rapidly at the point $(2,-3)$ ?

(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 of the function is a plane as it is in three dimensions

b, The direction of function increasing most rapidly at the point $(2,-3)$ is determined as $<a,b>$

c, The direction of function increasing most rapidly at the point $(x_0,y_0)$ is determined as$<a,b>$

d, The answers to Parts (b) and (c) are same as gradient values are same 

Step 1

(a)

Let the function be

$f(x, y)=ax+by$

Here, $a$ and $b$ are non-zero.

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

Let $f(x, y)=z$ then

$z=ax+by$

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 fast at $(2,-3)$.

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.

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

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

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

$=(a·1+b·0)\mathrm{i}+(a·0+b·1)\mathrm{j}$

$=ai+bj$

Step 3

From $( 2)$ at the point $(2,-3)$ the gradient is

$$\begin{gathered} \nabla z=ai+bj \\ =a, b \end{gathered}$$

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

Step 4

(c)

The goal is to determine which way the provided function increases the most fast at the given position$\left(x_0,y_0\right)$.

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

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

$$\begin{gathered} \nabla z=ai+bj \\ =<a, b> \end{gathered}$$

As a result, the given function increases most rapidly in the direction $<a, b> at point (x0,y0)$.

Step 5

(d)

The goal is to explain why the results in (b) and (c) are identical. You can see from (2) 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.

Explanation

The given data is the $f(x,y)=ax+by$

The objective is to find why is the function a plane, in which direction the increasing function is and to explain why the answers are same for parts (b) and (c)