(3.2) The directional derivative of f(x,y) at P(1,2) in the direction of i+j is 2v2and in the direction of –2j is -3. What is the directional derivative of f(x, y) in thedirection of -i– 2j? Show all working.

Answered: (3.2) The directional derivative of…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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(3.2) The directional derivative of f(x,y) at P(1,2) in the direction of i+j is 2v2
and in the direction of –2j is -3. What is the directional derivative of f(x, y) in the
direction of -i– 2j? Show all working.

Expert Solution

Step 1

Step:-1

Directional Derivative of f at point (p, q) in the direction of unit vector say $\hat{u} = <{\hat{u}}_{1}, {\hat{u}}_{2}>$ is

$D_{u} f (p, q) = <f_{x} (p, q), f_{y} (p, q)> . <{\hat{u}}_{1}, {\hat{u}}_{2}>$

Now, given that directional derivative of f(x, y) at P (1, 2) in the direction of $\hat{v} = i + j$ is $D_{u} f (1, 2) = 2 \sqrt{2}$

unit vector $u = \frac{v}{||v||} = \frac{i + j}{\sqrt{1^{2} + 1^{2}}} = \frac{i + j}{\sqrt{2}} = \frac{i}{\sqrt{2}} + \frac{j}{\sqrt{2}} = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}>$

using above formula, we have Directional Derivative of f(x, y) at P is

$D_{u} f (p, q) = <f_{x} (p, q), f_{y} (p, q)> . <{\hat{u}}_{1}, {\hat{u}}_{2}> D_{u} f (1, 2) = <f_{x} (1, 2), f_{y} (1, 2)> . <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}> D_{u} f (1, 2) = f_{x} (1, 2) \times \frac{1}{\sqrt{2}} + f_{y} (1, 2) \times \frac{1}{\sqrt{2}} = \frac{f_{x} (1, 2)}{\sqrt{2}} + \frac{f_{y} (1, 2)}{\sqrt{2}} D_{u} f (1, 2) = \frac{f_{x} (1, 2)}{\sqrt{2}} + \frac{f_{y} (1, 2)}{\sqrt{2}} b u t g i v e n t h a t D_{u} f (1, 2) = 2 \sqrt{2}, S o 2 \sqrt{2} = \frac{f_{x} (1, 2)}{\sqrt{2}} + \frac{f_{y} (1, 2)}{\sqrt{2}} \Rightarrow f_{x} (1, 2) + f_{y} (1, 2) = \sqrt{2} \times 2 \sqrt{2} \Rightarrow f_{x} (1, 2) + f_{y} (1, 2) = 4 - - - - - (1)$

Step:-2

Also, given that directional derivative of f(x, y) at P (1, 2) in the direction of ${\hat{v}}_{1} = - 2 j$ is $D_{u_{1}} f (1, 2) = - 3$

Similarly, as we do above, we have

unit vector $u_{1} = \frac{{\hat{v}}_{1}}{||{\hat{v}}_{1}||} = \frac{0 i - 2 j}{\sqrt{0^{2} + {(- 2)}^{2}}} = \frac{- 2 j}{2} = 0 i - j = <0, - 1>$

using above formula, we have Directional Derivative of f(x, y) at P is

$D_{u} f (p, q) = <f_{x} (p, q), f_{y} (p, q)> . <{\hat{u}}_{1}, {\hat{u}}_{2}> D_{u_{1}} f (1, 2) = <f_{x} (1, 2), f_{y} (1, 2)> . <0, - 1> D_{u_{1}} f (1, 2) = f_{x} (1, 2) \times 0 + f_{y} (1, 2) \times (- 1) = 0 - f_{y} (1, 2) D_{u_{1}} f (1, 2) = - f_{y} (1, 2) b u t g i v e n t h a t D_{u_{1}} f (1, 2) = - 3, S o - 3 = - f_{y} (1, 2) \Rightarrow f_{y} (1, 2) = 3 - - - - - (2)$

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