temperature gradient T(x,y,z) = 2x^2 - xyz

[cheffy]

The Celsius temperature in a region in space is given by:

\(\displaystyle \L T(x,y,z) = 2x^2 - xyz \\)

A particle is moving in this region and its position at time t is given by:

\(\displaystyle \L x = 2t^2 ,y = 3t,z = - t^2 \\)

where time is measured in seconds and distance in meters.

A) How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at the point P(8,6,-4)?
B) How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?

Am I supposed to find dT/ds if s: x=2t^2, y=3t, z=-t^2 for the distance? How would I do that? With a gradient? How would I find dT per second??

Thank you!

 

[Deleted member 4993]

Re: temperature gradient

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

A) How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at the point P(8,6,-4)?

B) How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?---> you need to find dT/dt

Am I supposed to find dT/ds if s: x=2t^2, y=3t, z=-t^2 for the distance? How would I do that? With a gradient? How would I find dT per second??

dT/dt = dT/dx * dx/dt + dT/dy * dy/dt + dT/dz * dz/dt

Thank you!

 

[cheffy]

Re: temperature gradient

dT/dt is the change per second, correct? How do I calculate the change per meter then?

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

A) How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at the point P(8,6,-4)?

B) How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?---> you need to find dT/dt

Am I supposed to find dT/ds if s: x=2t^2, y=3t, z=-t^2 for the distance? How would I do that? With a gradient? How would I find dT per second??

dT/dt = dT/dx * dx/dt + dT/dy * dy/dt + dT/dz * dz/dt

Thank you!

 

[cheffy]

I think I got part a.

But when I solved for dT/dt for part b, I get
(4x - xz)(4t)i - (3xz)j + (2txy)k

and when I plug in (8,6,-4), I get something ridiculous like 368. Help!

 

[Deleted member 4993]

I think I got part a.

But when I solved for dT/dt for part b, I get
(4x - xz)(4t)i - (3xz)j + (2txy)k

and when I plug in (8,6,-4), I get something ridiculous like 368. Help!

Please show detailed work so that we can catch your mistake.

By the way

dT/dx = 4x - yz .................different frm what you got

 

[cheffy]

I mistyped it. I got 4x-yz like I'm supposed to.

I substituted t=y/3 and then plugged in the point (8,6,-4)
(4x - yz)(4t)i - (3xz)j + (2txy)k

And then I get 224+96+48=368.

I think I got part a.

But when I solved for dT/dt for part b, I get


and when I plug in (8,6,-4), I get something ridiculous like 368. Help!

Please show detailed work so that we can catch your mistake.

By the way

dT/dx = 4x - yz .................different frm what you got

 

[Deleted member 4993]

I mistyped it. I got 4x-yz like I'm supposed to.

I substituted t=y/3 and then plugged in the point (8,6,-4)
(4x - yz)(4t)i - (3xz)j + (2txy)k.....Is dT/dt a vector?

If it is - then how do you justify your next step?

If not, then how did you get those i, j & k?

And then I get 224+96+48=368.