The temperature at a point x y z is given by tx y z 400e x2...

The temperature at a point (x, y, z) is given by $$T(x, y, z) = 400e^{-x^2 - 3y^2 - 7z^2}$$ where T is measured in °C and x, y, z in meters. (a) Find the rate of change of temperature at the point P(2, -1, 5) in the direction towards the point (4, -5, 6). $$\frac{-3200e^{-4} - 24 - 70}{21}$$ °C/m (b) In which direction does the temperature increase fastest at P? $$<-1600e^{168}, 2400e^{168}, -28000e^{168}>$$ (c) Find the maximum rate of increase at P. $$\sqrt{2560000e^{-8} + 36 + 4900}$$

Transcript

00:01 In this problem, we are given the function t of xyz is equal to 400 e raised to negative x square minus 5y square minus 9 z square where t denotes the temperature at a point xyz and it is measured in degrees celsius while x, y and z are measured in meters.

00:25 The first question is to determine the rate of change of the temperature at the point p with coordinates 2 negative 1 -2 in the direction towards the point 5 -9 -45.

00:42 So here we are asked to determine the directional derivative of the function t at the point p in the direction of the unit vector u in the direction of the point 5.

00:55 Negative 4 5.

00:58 And this is evaluated as the gradient vector of the function t at the point b dot product with the vector u.

01:07 So here take v to be the position vector 5 negative 4 5.

01:18 Now find the unit vector in the direction of this vector for that find the magnitude of this vector that is square root of 25 plus negative 4 the whole square that is 16 plus 5 square 25 therefore we get the magnitude of this vector is square root of 66 so that the unit vector in the direction of vector v is 1 divided by square root of 66 times the vector 5 negative 4 5 divided by square root of 66 5 divided by square root of 66.

01:58 Now find the gradient vector del t at the point p.

02:05 It is evaluated as the vector do t by do x at the point p, do t by do y at the point p, do t by do y at the point p, where do t by do x, dot t by do y and do t by do y, so that are the partial derivatives of t with respect to x, y, and i said.

02:26 So find do t by do x, and that is negative 800 x, e raised to negative x square minus 5, y square minus 9 is that square.

02:39 So that dot t by do x at the point to negative 1 ,2 is negative 1 ,200, e raised to negative 45.

02:51 Similarly find do t by do y and that is negative 800 times 5 y e raised to negative x square minus 5 y square minus 9 is at square so that dot t by do y at the point p which is 2 negative 1 2 is negative positive 4 000 e raised to negative 45 finally find do t by do i said and that is negative 800 times 9 times z erased to negative x square minus 5 y square minus 9 is that square so that at the point to negative 1 2 it is evaluated to be 14 ,400 e raised to negative 45.

03:53 Negative 14 ,400, eras to negative 45 so that the directional derivative of t at the point to negative 1 2 ,2 in the direction of vector u is the vector negative, negative 600 erase to negative 45, 4 ,000 erased to negative 45, negative 14 ,400 erased to negative 45, negative 14 ,400 erased to negative 45, dot product with the vector, 5 divided by square root of 66, negative 4 divided by square root of 66, 5 divided by square root of 66.

04:33 Now the dot product of two vectors is the sum of product of corresponding components.

04:38 So this is evaluated as negative 800, e raised negative 45 divided by square root of 66 times 10 plus 20 plus 90...

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