The temperature at a point (x, y, z) is given by T(x, y, z) = 200e^-x^2 - 3y^2 -
ID: 2878770 • Letter: T
Question
The temperature at a point (x, y, z) is given by T(x, y, z) = 200e^-x^2 - 3y^2 - 7x^2 where T is measured in e^C and x, y, z in meters. Find the rate of change of temperature at the point P(2, -1, 5) in the direction towards the point (6, -3, 6). degree C/m In which direction does the temperature increase fastest at P? Find the maximum rate of increase at P. Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 4x^2 - 4xy + xyz Find the rate of change of the potential at P(4, 2, 6) in the direction of the vector v = i + j = k. In which direction does V change most rapidly at P? What is the maximum rate of change at P?Explanation / Answer
1)given T(x,y,z)=200e-x^2 -3y^2 -7z^2
gradient T=<-400xe-x^2 -3y^2 -7z^2,-1200ye-x^2 -3y^2 -7z^2,-2800ze-x^2 -3y^2 -7z^2>
at P(2,-1,5)
T=<-400*2e-2^2 -3(-1)^2 -7*5^2,-1200(-1)e-2^2 -3(-1)^2 -7*5^2,-2800(5)e-2^2 -3(-1)^2 -7*5^2>
T=<-800e-182,1200e-182,-14000e-182>
T=400e-182<-2,3,-35>
vector joining (2,-1,5),(6,-3,6) is u =<6-2,-3+1,6-5> =<4,-2,1>
|u|=[42+(-2)2+12] =21
rate of change of temperature =T.u/|u|
rate of change of temperature =400e-182<-2,3,-35>.<4,-2,1>/21
rate of change of temperature =400e-182[(-2*4)+(3*-2)+(-35*1)]/21
rate of change of temperature =400e-182(-49)/21
rate of change of temperature =-(19600/21)e-182
b)T=400e-182<-2,3,-35>
|T|=400e-182[(-2)2+32+(-35)2]
|T|=400e-1821238
direction of temperature increase fastest at P is T/|T|
direction of temperature increase fastest at P is [400e-182<-2,3,-35>]/[400e-1821238]
direction of temperature increase fastest at P is <-2/1238,3/1238,-35/1238>
c) maximum rate of increase at P =|T|=4001238 e-182
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