Use the contour diagram for f(x, y) shown below to estimate the directional derivative of f in the direction T at the point P.(a) At the point P = (2, 2) in the direction v = 7, the directional derivative isapproximately(b) At the point P = (3, 2) in the direction i = -i, the directional derivative is approximately(c) At the point P = (4,1) in the direction = (7 +j)/v2, the directional derivative is approximately(d) At the point P = (4,0) in the direction v = -i, the directional derivative is approximately 4.83.22.48.01.60.82.0-0.80.81.62.43.244.8-0.818.016.014.012.010.06,04.02.0002.00.0

Let the point P=(2,2)=2i→+2j→ In this we have to find the directional derivative o f in the…

Answered: Use the contour diagram for f(x, y)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Use the contour diagram for f(x, y) shown below to estimate the directional derivative of f in the direction T at the point P.
(a) At the point P = (2, 2) in the direction v = 7, the directional derivative is
approximately
(b) At the point P = (3, 2) in the direction i = -i, the directional derivative is approximately
(c) At the point P = (4,1) in the direction = (7 +j)/v2, the directional derivative is approximately
(d) At the point P = (4,0) in the direction v = -i, the directional derivative is approximately

4.8
3.2
2.4
8.0
1.6
0.8
2.0
-0.8
0.8
1.6
2.4
3.2
4
4.8
-0.8
18.0
16.0
14.0
12.0
10.0
6,0
4.0
2.0
00
2.0
0.0

Expert Solution

part(a) Answer

Let the point P=(2,2)= $2 \vec{i} + 2 \vec{j}$

In this we have to find the directional derivative o f in the direction of

$\vec{v} = \frac{\vec{P Q}}{\vec{||P Q||}} = \vec{i} \vec{P Q} = \vec{i} \vec{O Q} - \vec{O P} = \vec{i} \vec{O Q} = \vec{i} + \vec{O P} \vec{O Q} = \vec{i} + 2 \vec{i} + 2 \vec{j} s o, t h e v a l u e o f Q = (3, 2) T h e d i r e c t i o n a l d e r i v a t i v e s o f f i n t h e d i r e c t i o n o \vec{v} i s, f_{\vec{u}} (P) = \frac{f (Q) - f (P)}{\vec{||P Q||}} P o i n t o f P = (2, 2) a n d p o i n t o f Q = (3, 2) f_{\vec{u}} (2, 2) = \frac{f (3, 2) - f (2, 2)}{1} I n t h e a b o v e g r a p h t h e v a l u e o f f (3, 2) = 6 a n d f (2, 2) = 4 o n p u t t i n g g i v e n v a l u e s w e g e t, f_{\vec{u}} (2, 2) = \frac{6 - 4}{1} f_{\vec{u}} (2, 2) = 2$