Answer with Step-by-step explanation:
We are given that
[tex]T(x,y,z)=100e^{-x^2-3y^2-7z^2}[/tex]
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).
[tex]\Delta T(x,y,z)=100e^{-x^2-3y^2-7z^2}(-2x-6y-14z)[/tex]
Substitute x=4,y=-1 and z=4
[tex]\Delta T(4,-1,4)=,100e^{-16-3-112}<-8,6,-56>=100e^{-131}<-8,6,-56>[/tex]
[tex]u=<4,-1,4>[/tex]
[tex]v=<6-4,-4+1,5-4>=<2,-3,1>[/tex]
[tex]\mid v\mid=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}[/tex]
[tex]\hat{v}=\frac{v}{\mid v\mid}=\frac{1}{\sqrt{14}}<2,-3,1>[/tex]
Rate of change of temperature at point P(4,-1,4) in the direction of the point (6,-4,5)=[tex]100e^{-131}<-8,6,-56>\cdot \frac{1}{\sqrt{14}}<2,-3,1>[/tex]
Rate of change of temperature at point P(4,-1,4) in the direction of the point (6,-4,5)=[tex]\frac{100}{\sqrt{14}}e^{-131}(-16-18-56)=-\frac{9000}{\sqrt{14}}e^{-131}[/tex]
b.The temperature increases fastest in the direction of [tex]\Delta T(4,-1,4)[/tex]
c.Maximum rate of increase at P=[tex]\mid\Delta T(4,-1,4)\mid[/tex]
Maximum rate of increase at P=[tex]\sqrt{(e^{-131}<-8,6,-56>)^2}=2\sqrt{809}e^{-131}[/tex]
