Is there a direction A in which the rate of change of the temperature function [tex]T(x,y,x)=2xy - yz[/tex] at P(1,-1,1) is -3ºC/ft? Give reasons for your answer.(adsbygoogle = window.adsbygoogle || []).push({});

For this problem I found the gradient of at the point P. So

[tex]\nabla f = 2y|_p \mathbf{i} + 2x-z|_p \mathbf{j} - y|_p \mathbf{k}[/tex]

which I then found was

[tex] \nabla f = -2 \mathbf{i} + 1 \mathbf{j} + 1 \mathbf{k} [/tex]

Then we want to know if the gradient, in the direction of A will be 3ºC, so

[tex] \nabla f \cdot \frac{A}{|A|} = \frac{3º}{ft}[/tex]

which means that we want an x, y, z, such that

[tex]\frac{-2x +y +z}{\sqrt{x^2 + y^2 + z^2}} = \frac{3º}{ft}[/tex]

So I belive this is right so far, but I don't think this eqn can be solved for.

Here is the other one:

The Celsius temperature in a region in space is given by [tex]T(x,y,z) = 2x^2 -xyz[/tex]. A particle is moving in this region and its position at time t is given by [tex]x=2t^2[/tex], [tex]y=3t[/tex], [tex]z=-t^2[/tex], where time is measured in seconds and distanace in meters.

a) How fast is the temp experienced by the particle changing in ºC/m when the particle is at the point P(8,6,-4)?

b) How fast is the temp experienced by the particle changing in ºC/sec at P?

So, using the chain rule

[tex] \frac{dT}{dt} = (4x-yz)4t \mathbf{i} - (xz)3 \mathbf{j} + (xy)2t \mathbf{k} [/tex]

For the change per meter I will just solve for one of the t eqns, and then substitute back in to the equation to find the gradient?

Then for the change per second, it will be the opposite?

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# Homework Help: Temperature Gradient Questions

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