the temperature at a point (x y z) is given by t(x y z)=1200e^(-x^2-2y^2-z^2) wh
ID: 2880158 • Letter: T
Question
the temperature at a point (x y z) is given by t(x y z)=1200e^(-x^2-2y^2-z^2) where T is measured in C and x, y, and z in meters.
Find the rate of change of the temperature at the point P(2,1,2)P(2,1,2) in the direction toward the point Q(3,3,3).
In what direction does the temperature increase fastest at P?
(1 point The temperature at a point Cx, y, z is given by T(x, y, z) 200e where T is measured in C and x, y, and z in meters. 1. Find the rate of change of the temperature at the point P(2, -1,2) in the direction toward the point Q(3, -3,3). Answer: PO 2,-1,2) D 2. In what direction does the temperature increase fastest at P? Answer: 3. Find the maximum rate of increase at P. Answer:Explanation / Answer
solution-: a)
T = < 2400x e^(x^22y^2z^2), 4800y e^(x^22y^2z^2), 2400z e^(x^22y^2z^2) >
T(2, 1, 2) = < 4800/e^10, 4800/e^10, 4800/e^10 >
Vector from (2, 1, 2) to (3, 3, 3) = < 32, 3+1, 32 > = < 1, 2, 1 >
Unit vector in direction toward point (3, 3, 3)
u = < 1, 2, 1 > / || < 1, 2, 1 > ||
u = < 1, 2, 1 > / 6
u = < 1/6, 2/6, 1/6 >
Rate of change of temperature at point (2, 1, 2) in direction toward point (3, 3, 3)
= T(2, 1, 2) • u
= < 4800/e^10, 4800/e^10, 4800/e^10 >*< 1/6, 2/6, 1/6 >
=4800/(e^10 6) 4800/(e^10 6) 4800/(e^10 6)
=14400/(e^10 6)
2.67 * 10¹
(b)
Temperature increases fastest in direction of gradient at (2, 1, 2)
= < 4800/e^10, 4800/e^10, 4800/e^10 >
= 4800/e^10 < 1, 1, 1 >
Unit vector in direction of fastest increase
= < 1, 1, 1 > / || < 1, 1, 1 > ||
= < 2, 3, 18 > / 3
= < 1/3, 1/3, 1/3 > answer
(c)
Maximum rate of increase at P
= < < 4800/e^10, 4800/e^10, 4800/e^10 > • < 1/3, 1/3, 1/3 >
= 4800/(e^10 3) + 4800/(e^10 3) + 4800/(e^10 3)
= 14400/(e^10 3)
= 48003 / e^10
3.78*10^-1 answer
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