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Q. 32

Expert-verified
Found in: Page 989

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P. $f\left(x,y\right)=\frac{x}{{y}^{\mathrm{\prime }}}P=\left(4,3\right)$

See the step by step solution

## Step 1. Given Information

$\text{The gradient of a function}f\left(x,y\right)\text{is the vector function defined by}\mathrm{\nabla }f\left(x,y\right)=\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\mathbf{i}+\frac{\mathrm{\partial }f}{\mathrm{\partial }y}\mathbf{j}\text{,}\phantom{\rule{0ex}{0ex}}\text{Here}f\left(x,y\right)=\frac{x}{y}.$

## Step 2. Solution

So its gradient is given by,

$\text{As the gradient of a function}f\text{at a point}p\text{, points in the direction in which}f\text{increases most rapidly.}\phantom{\rule{0ex}{0ex}}\text{At}p=\left(4,3\right),\phantom{\rule{1em}{0ex}}\mathrm{\nabla }f\left(4,3\right)=\frac{1}{3}\mathbf{i}-\frac{4}{9}\mathbf{j}\phantom{\rule{0ex}{0ex}}\mathbf{\text{Hence the direction in which}}\mathbf{f}\mathbf{\text{increases most rapidly at}}\mathbf{p}\mathbf{=}\mathbf{\left(}\mathbf{4}\mathbf{,}\mathbf{3}\mathbf{\right)}\mathbf{\text{is}}\mathbf{\nabla }\mathbf{f}\mathbf{\left(}\mathbf{4}\mathbf{,}\mathbf{3}\mathbf{\right)}\mathbf{=}<\frac{1}{3},-\frac{4}{9}>$