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

Part a) The equation is in three dimensions and hence the graph is a plane.

Part b) The given function increases most rapidly at a point $(-1,4)$in the direction $⟨2,3⟩$

Part c) The given function increases most rapidly at the point $\left(x_0,y_0\right)$ in the direction $⟨2,3⟩$

 Part d) Both the results are the same.

Part (a) Step 1: The objective is to explain why the graph of the function is a plane.

The function is $f(x,y)=2x+3y\dots \dots .\left(1\right)$

Let $f(x, y)=z$then $z=2 x+3 y$

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

The other two variables have an impact on each variable.

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

Part (b) Step 1: The objective is to find the direction in which the given function increases most rapidly at the point ( - 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 gradient of the function is

$$\begin{gathered} \nabla z=f_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j} \\ =\left\{\frac{\partial }{\partial \mathrm{x}}(2\mathrm{x}+3\mathrm{y})\right\}\mathrm{i}+\left\{\frac{\partial }{\partial \mathrm{y}}(2\mathrm{x}+3\mathrm{y})\right\}\mathrm{j} \\ =\left(2\frac{\partial }{\partial \mathrm{x}}\mathrm{x}+3\frac{\partial }{\partial \mathrm{x}}\mathrm{y}\right)\mathrm{i}+\left(2\frac{\partial }{\partial \mathrm{y}}\mathrm{x}+3\frac{\partial }{\partial \mathrm{y}}\mathrm{y}\right)\mathrm{j} \\ =(2·1+3·0)\mathrm{i}+(2·0+3·1)\mathrm{j} \\ =2\mathrm{i}+3\mathrm{j}\dots \dots .\left(2\right) \end{gathered}$$

From ( 2 ) at the point $(-1,4)$ the gradient is

$$\begin{gathered} \nabla z=2\mathrm{i}+3\mathrm{j} \\ =⟨2,3⟩ \end{gathered}$$

Part (c) Step 1: The objective is to find the direction in which the given function increases most rapidly at the point x 0 , y 0  

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

$$\begin{gathered} \nabla z=2\mathrm{i}+3\mathrm{j} \\ =⟨2,3⟩ \end{gathered}$$

Hence, the given function increases most rapidly at the point $\left(x_0,y_0\right)$ in the direction $⟨2,3⟩$

Part (d) Step 1: The objective is to explain why the result obtained in (b) and (c) are the same.

 From $\left(2\right)$ that, the gradient is independent of $x$ and $y$

So, regardless of the position, the gradient is the same.

Hence, both the results are the same.