Q. 55.

Question

In Exercises 55–60, find a function of two variables with the

given gradient

$\nabla f (x, y) = <- \frac{3}{x}, \frac{1}{y}>$

Step-by-Step Solution

Verified

Answer

The required function is $f (x, y) = \ln \frac{y}{x^{3}} + C$

1Step 1: Given information

The gradient is $\nabla f (x, y) = <- \frac{3}{x}, \frac{1}{y}> \dots \dots (1)$

2Step 2: The objective is to find the function from the given gradient.

Rewrite (1) as

$\nabla f (x, y) = - \frac{3}{x} i + \frac{1}{y} j f_{x} (x, y) i + f_{y} (x, y) j = - \frac{3}{x} i + \frac{1}{y} j f_{x} (x, y) = - \frac{3}{x} \dots \dots (2) and f_{y} (x, y) = \frac{1}{y} \dots \dots . (3)$

3Step 3: First check whether the function exists or not.

The function exists if,

$f_{x y} (x, y) = f_{y x} (x, y)$

Since $f_{x y} (x, y) = 0 = f_{y x} (x, y)$ so function exists.

Integrate $(3)$ with respect to $y$

$f (x, y) = \int \frac{1}{y} d y + q (x) = \ln y + q (x) \dots \dots (4)$

where $q (x)$ is an arbitrary function.

4Step 4: The objective is to find q ( x )   differentiate ( 4 ) partially with respect to x

$\frac{\partial}{\partial x} f (x, y) = \frac{\partial}{\partial x} \ln y + \frac{\partial}{\partial x} q (x) f_{x} (x, y) = 0 + q^{'} (x) - \frac{3}{x} = q^{'} (x) [F r o m (2); f_{x} (x, y) = - \frac{3}{x}] q^{'} (x) = - \frac{3}{x} \dots \dots . (5)$