Question

he temperature at a point (x, y, z) is given by T(x, y, z) = 100e−x2 − 3y2 − 7z2 where T is measured in °C and x, y, z in meters. (a) Find the rate of change of temperature at the point P(4, −1, 4) in the direction towards the point (6, −4, 5). °C/m (b) In which direction does the temperature increase fastest at P? (c) Find the maximum rate of increase at P.

Answer 1

Answer with Step-by-step explanation:

We are given that

$T(x,y,z)=100e^{-x^2-3y^2-7z^2}$

a.We have to find the rate of change of temperature at the point P(4,-1,4) in the direction towards the point (6,-4,5).

$\Delta T(x,y,z)=100e^{-x^2-3y^2-7z^2}(-2x-6y-14z)$

Substitute x=4,y=-1 and z=4

$\Delta T(4,-1,4)=,100e^{-16-3-112}=100e^{-131}$

$u=$

$v==$

$\mid v\mid=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}$

$\hat{v}=\frac{v}{\mid v\mid}=\frac{1}{\sqrt{14}}$

Rate of change of temperature at point P(4,-1,4) in the direction of the point (6,-4,5)=$100e^{-131}\cdot \frac{1}{\sqrt{14}}$

Rate of change of temperature at point P(4,-1,4) in the direction of the point (6,-4,5)=$\frac{100}{\sqrt{14}}e^{-131}(-16-18-56)=-\frac{9000}{\sqrt{14}}e^{-131}$

b.The temperature increases fastest in the direction of $\Delta T(4,-1,4)$

c.Maximum rate of increase at P=$\mid\Delta T(4,-1,4)\mid$

Maximum rate of increase at P=$\sqrt{(e^{-131})^2}=2\sqrt{809}e^{-131}$