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Answers without the blur. Sign up and see all textbooks for free! Q. 33

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. $f\left(x,y\right)=x\mathrm{sin}y,P=\left(3,\frac{\pi }{2}\right)$

$p=\left(3,\frac{\pi }{4}\right)\text{is}\mathrm{\nabla }f\left(3,\frac{\pi }{4}\right)=⟨1,0⟩$

See the step by step solution

## Step 1. Given Information

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

## Step 2. Solution

$\begin{array}{l}\mathrm{\nabla }f\left(x,y\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\mathrm{sin}y\right)\mathbf{i}+\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(x\mathrm{sin}y\right)\mathbf{j}\\ =\mathrm{sin}y\mathbf{i}+x\mathrm{cos}y\mathbf{j}\end{array}$

Hence the gradient is, $\mathrm{\nabla }f\left(x,y\right)=⟨\mathrm{sin}y,x\mathrm{cos}y⟩$

$\text{As the gradient of a function}f\text{at a point}p\text{points in the direction in which}f\text{increases most rapidly and}$

at $p=\left(3,\frac{\pi }{2}\right)$

$\begin{array}{l}\mathrm{\nabla }f\left(3,\frac{\pi }{2}\right)=\mathrm{sin}\frac{\pi }{2}\mathbf{i}+3\mathrm{cos}\frac{\pi }{2}\mathbf{j}\\ =\left(\mathbf{i}+0\right)\end{array}$

So the direction in which f increases most rapidly at $p=\left(3,\frac{\pi }{4}\right)\text{is}\mathrm{\nabla }f\left(3,\frac{\pi }{4}\right)=⟨1,0⟩$. ### Want to see more solutions like these? 