The temperature at a point (x, y, z) is given by T(x, y, z) = 200e−x2 − 3y2 − 9z2 where T is measured in °C and x, y, z in meters. Required:
a. Find the rate of change of temperature at the point P(4, −1, 5) in the direction towards the point (5, −4, 6).
b. In which direction does the temperature increase fastest at P?
c. Find the maximum rate of increase at P.

a) Upq*gradT(P)  = -21,69 ⁰C/m

b) In direction of j  ( or  y )

c) gradT(P)  = - 84 ⁰C/m  is decreasing but has a maximum rate of increase in the direction of j (y)

Step-by-step explanation:

a) T (x,y,z) = 200*e∧-x²*i   -  3*y²*j   -  9*z²*k

We find a unit vector in the direction PQ

PQ = ( 5 , -4 , 6 ) - ( 4 , -1 , 5 )

PQ = ( 1 , -3 , 1 )   ||PQ|| = √ (1)² + (-3)² + (1)²    ||PQ|| = √ 11

||PQ|| = 3,32

Unit vector in the direction of PQ is

Upq = ( 1/3,32 , -3/3,32  , 1/3,32 )

grad T  = (  T´(x)  + T´(y) + T´(z) )

T´(x)  =  -400*x*e∧-x²

T´(y) = -6*y

T´(z) = -18*z

The rate of change of temperature at the point P in the direction of PQ is:

Upq*gradT = ( 1/3,32 , -3/3,32  , 1/3,32 )* ( -400*x*e∧-x² , -6*y , - 18*z )

Upq*gradT = ( -400*x*e∧-x²/3,32 ,  - (18/3,32)*y  - (18/3,32) * z )

At the point P  ( 4 , - 1 , 5 )

Upq*gradT(P)  = ( - 1600*e⁻16 /3,32  + 18*1/3,32  - 90/3,32 )

Upq*gradT(P)  = ( - 0,00018  + 5,4216 - 27,11 )

Upq*gradT(P)  = -21,69 ⁰C/m

b) In direction of j  ( or  y )

c) grad T  = (  T´(x)  + T´(y) + T´(z) )

grad T = ( -400*x*e∧-x² ,  -6*y  ,  -18*z )

At P ( 4 , -1 , 5 )

gradT(P)  = ( -400*4* e∧-4²  + 6  - 90 )

gradT(P)  = - 84 ⁰C/m

In the direction of y we find the maximum rate of increase at P