Let f(x, y) = (x² + y²)e=(x²+y´ -5). Find the rate of change of f at X = (1, –2) in thedirection v pointing toward the origin, that is, compute 4 f(X + tv) t-0 for an appropriateunit length vector v.

Answered: : (x² + y²)e-(x*+y -5), Find the rate…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Let f(x, y) = (x² + y²)e=(x²+y´ -5). Find the rate of change of f at X = (1, –2) in the
direction v pointing toward the origin, that is, compute 4 f(X + tv) t-0 for an appropriate
unit length vector v.

Expert Solution

Step 1

Given that $f (x, y) = (x^{2} + y^{2}) e^{- (x^{2} + y^{2} - 5)}$

The objective is to find the rate of change of f at $X = (1, - 2)$ in the direction v pointing towards the origin.

Now, $v = (0, 0) - (1, - 2) = (- 1, 2)$ then $\hat{v} = \frac{1}{\sqrt{1 + 4}} (- 1, 2)$

Consider, $f (x, y) = (x^{2} + y^{2}) e^{- (x^{2} + y^{2} - 5)}$

Then

$\begin{array}{rcl} \frac{\partial f}{\partial x} & = & 2 x e^{- (x^{2} + y^{2} - 5)} - 2 x^{3} e^{- (x^{2} + y^{2} - 5)} - 2 x y^{2} e^{- (x^{2} + y^{2} - 5)} \\ = & 2 x e^{- (x^{2} + y^{2} - 5)} (1 - x^{2} - y^{2}) \\ \frac{\partial f}{\partial y} & = & - 2 y x^{2} e^{- (x^{2} + y^{2} - 5)} + 2 y e^{- (x^{2} + y^{2} - 5)} - 2 y^{3} e^{- (x^{2} + y^{2} - 5)} \\ = & 2 y e^{- (x^{2} + y^{2} - 5)} (- x^{2} + 1 - y^{2}) \\ = & 2 y e^{- (x^{2} + y^{2} - 5)} (1 - x^{2} - y^{2}) \end{array}$

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