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1. The air pressure in a particular room varies from point to point and is given by p(x, y, 2) = 3x + xỉ exp(-2 y), where (x, y, z) describes a general point (with reference to a Cartesian coordinate system). (a) At the point (1,0,2), calculate the rate of change of pressure per unit distance in the direction of the vector i- 2j+ 2k . (b) At (1,0,2) , find all directions along which the rate of change of pressure per unit distance is zero. (Give the directions using vectors in the form ai + bj+ ck .) (c) At (1,0,2), find the direction along which the rate of increase in the pressure per unit distance is the greatest. What is the greatest rate of increase in the pressure per unit distance at (1,0,2) ? Solution. We need Vpvdr002) in all the parts. Vp=Pi+ ốp j+ = [6x+ 2x2 exp(-2 y)]i – 2x z² exp(-2y)j+2x² zexp(-2y)k =141 – 8j+ 4k at (x, y, z) = (1,0, 2) (a) The vector i– 2j+ 2k has magnitude 1? + (-2)² + 2² = 3. The required rate is given by 2 2, 38 (141 – 8j+ 4k) • j+=k 3

1. The air pressure in a particular room varies from point to point and is given by p(x, y, 2) = 3x + xỉ exp(-2 y), where (x, y, z) describes a general point (with reference to a Cartesian coordinate system). (a) At the point (1,0,2), calculate the rate of change of pressure per unit distance in the direction of the vector i- 2j+ 2k . (b) At (1,0,2) , find all directions along which the rate of change of pressure per unit distance is zero. (Give the directions using vectors in the form ai + bj+ ck .) (c) At (1,0,2), find the direction along which the rate of increase in the pressure per unit distance is the greatest. What is the greatest rate of increase in the pressure per unit distance at (1,0,2) ? Solution. We need Vpvdr002) in all the parts. Vp=Pi+ ốp j+ = [6x+ 2x2 exp(-2 y)]i – 2x z² exp(-2y)j+2x² zexp(-2y)k =141 – 8j+ 4k at (x, y, z) = (1,0, 2) (a) The vector i– 2j+ 2k has magnitude 1? + (-2)² + 2² = 3. The required rate is given by 2 2, 38 (141 – 8j+ 4k) • j+=k 3

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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hello, i got the workings for this question already, however i am still unsure about it, is it possible to explain it in depth to me? thanks !

1. The air pressure in a particular room varies from point to point and is given by p(x, y, z) = 3x + xỶ exp(-2 y), where (x, y, z) describes a general point (with reference to a Cartesian coordinate system). %3D (a) At the point (1,0, 2) , calculate the rate of change of pressure per unit distance in the direction of the vector i- 2j+ 2k . (b) At (1,0, 2) , find all directions along which the rate of change of pressure per unit distance is zero. (Give the directions using vectors in the form ai+ bj+ ck .) (c) At (1,0,2), find the direction along which the rate of increase in the pressure per unit distance is the greatest. What is the greatest rate of increase in the pressure per unit distance at (1,0,2) ? Solution. We need Vpy.=(1.0.2) in all the parts. (x,y.z)=(1,0,2) др Vp=Pi+Pj+Pk %3 [6х+2x7 еxp(-2 у)! - 2x 2 еxp(-2 у) j+2x zexp(-2 y)k = 141– 8j+ 4k at (x, y, z) = (1,0, 2) (a) The vector i- 2j+ 2k has magnitude 1? + (-2)² + 2² = 3. The required rate is given by %3D 2. j+=k 3 2 38 (141 – 8j+ 4k) • 3
Transcribed Image Text:1. The air pressure in a particular room varies from point to point and is given by p(x, y, z) = 3x + xỶ exp(-2 y), where (x, y, z) describes a general point (with reference to a Cartesian coordinate system). %3D (a) At the point (1,0, 2) , calculate the rate of change of pressure per unit distance in the direction of the vector i- 2j+ 2k . (b) At (1,0, 2) , find all directions along which the rate of change of pressure per unit distance is zero. (Give the directions using vectors in the form ai+ bj+ ck .) (c) At (1,0,2), find the direction along which the rate of increase in the pressure per unit distance is the greatest. What is the greatest rate of increase in the pressure per unit distance at (1,0,2) ? Solution. We need Vpy.=(1.0.2) in all the parts. (x,y.z)=(1,0,2) др Vp=Pi+Pj+Pk %3 [6х+2x7 еxp(-2 у)! - 2x 2 еxp(-2 у) j+2x zexp(-2 y)k = 141– 8j+ 4k at (x, y, z) = (1,0, 2) (a) The vector i- 2j+ 2k has magnitude 1? + (-2)² + 2² = 3. The required rate is given by %3D 2. j+=k 3 2 38 (141 – 8j+ 4k) • 3
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