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Q. 33

Expert-verifiedFound in: Page 989

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.

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

$p=\left(3,\frac{\pi}{4}\right)\text{is}\mathrm{\nabla}f\left(3,\frac{\pi}{4}\right)=\u27e81,0\u27e9$

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

Gradient is given by,

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

Hence the gradient is, $\mathrm{\nabla}f(x,y)=\u27e8\mathrm{sin}y,x\mathrm{cos}y\u27e9$

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

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

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

So the direction in which f increases most rapidly at $p=\left(3,\frac{\pi}{4}\right)\text{is}\mathrm{\nabla}f\left(3,\frac{\pi}{4}\right)=\u27e81,0\u27e9$.

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