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7.3 Heavy Metal(#differentiation) The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,3, 2) is 100°C. (a) Derive a function T(x, y, z) that gives the temperature in the ball at the point (x, y, z). (b) Find the rate of change of T at (1, 3, 2) in the direction toward the point (0, 4, 2). (c) Show that at any point in the ball the direction of greatest increase in temper- ature is given by a vector that points toward the origin.

7.3 Heavy Metal(#differentiation) The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,3, 2) is 100°C. (a) Derive a function T(x, y, z) that gives the temperature in the ball at the point (x, y, z). (b) Find the rate of change of T at (1, 3, 2) in the direction toward the point (0, 4, 2). (c) Show that at any point in the ball the direction of greatest increase in temper- ature is given by a vector that points toward the origin.

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
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7.3 Heavy Metal(#differentiation) The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,3,2) is 100°C. (a) Derive a function T(x, y, z) that gives the temperature in the ball at the point (x, y, z). (b) Find the rate of change of T at (1,3,2) in the direction toward the point (0, 4, 2). (c) Show that at any point in the ball the direction of greatest increase in temper- ature is given by a vector that points toward the origin.
Transcribed Image Text:7.3 Heavy Metal(#differentiation) The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,3,2) is 100°C. (a) Derive a function T(x, y, z) that gives the temperature in the ball at the point (x, y, z). (b) Find the rate of change of T at (1,3,2) in the direction toward the point (0, 4, 2). (c) Show that at any point in the ball the direction of greatest increase in temper- ature is given by a vector that points toward the origin.
Question 1. The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1, 2, 2) is 120°. (a) Find the rate of change of T at (1, 2, 2) in the direction toward the point (2, 1, 3). (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin. Solution 1. (a) The distance from (x, y, z) to the center is √x² + y² + 22 so that where k is a constant. Since T(1, 2, 2) = 120 we T(x, y, z) = A vector from (1, 2, 2) to (2, 1, 3) is (2, 1, 3) Computing the gradient, we find so that T(x, y, z) = u= k √x² + y² + z2 have that k = 360, thus 360 == /x² + y² + 2² (1,2,2)= (1,-1, 1). Normalizing it we find (1,-1,1). 1 √3 VT(x, y, z) = -360(x² + y² + 2²)-(x, y, z), Du = VT u 1 = −360(x² + y² + 2²)−¹(x, y, z). (1, =(1,-1, 1) 360 √3 Plugging in (1, 2, 2) we find DuT(1,2,2)= 3√3 40 (b) The direction of greatest increase is parallel to the gradient of T. From (1) we see that the gradient points in the direction opposite to (x, y, z), hence toward the origin. (1) (x² + y² + ₂²)-³(x−y+z).
Transcribed Image Text:Question 1. The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1, 2, 2) is 120°. (a) Find the rate of change of T at (1, 2, 2) in the direction toward the point (2, 1, 3). (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin. Solution 1. (a) The distance from (x, y, z) to the center is √x² + y² + 22 so that where k is a constant. Since T(1, 2, 2) = 120 we T(x, y, z) = A vector from (1, 2, 2) to (2, 1, 3) is (2, 1, 3) Computing the gradient, we find so that T(x, y, z) = u= k √x² + y² + z2 have that k = 360, thus 360 == /x² + y² + 2² (1,2,2)= (1,-1, 1). Normalizing it we find (1,-1,1). 1 √3 VT(x, y, z) = -360(x² + y² + 2²)-(x, y, z), Du = VT u 1 = −360(x² + y² + 2²)−¹(x, y, z). (1, =(1,-1, 1) 360 √3 Plugging in (1, 2, 2) we find DuT(1,2,2)= 3√3 40 (b) The direction of greatest increase is parallel to the gradient of T. From (1) we see that the gradient points in the direction opposite to (x, y, z), hence toward the origin. (1) (x² + y² + ₂²)-³(x−y+z).
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