Multivariable calculus, gradient, directional derivatives

[miglo]

Homework Statement

the Celsius temperature in a region in space is given by T(x,y,z)=2x^2-xyz. a particle is moving in this region and its position at time t is given by 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 P(8,6.-4)?
b) how fast is the temperature experienced by the particle changing in degress celsius per second at P?

Homework Equations

directional derivatives?

The Attempt at a Solution

ill admit i haven't actually attempted anything because I am not sure how to start
this problem comes from a practice exam under the section on tangent planes and differentials but i don't see how that would be helpful
to me it seems like more of a directional derivative problem but I am not really sure
can someone please show me how to start this problem?

 

[mighty2000]

Hello! Great job identifying the topic of this problem - it is indeed related to directional derivatives. To begin, we can use the formula for directional derivatives:

Df(x,y,z) = ∇f(x,y,z) · v

Where ∇f(x,y,z) is the gradient of the function T(x,y,z) and v is the direction vector represented by the particle's position at time t.

a) To find the rate of change of temperature in degrees Celsius per meter, we need to find the directional derivative in the direction of motion of the particle at point P(8,6,-4). This can be done by finding the gradient of T(x,y,z) and plugging in the values for x=8, y=6, and z=-4. Then, we can plug in the direction vector v = <4t, 3, -2t> at time t=2 seconds (since the particle's position is given by x=2t^2, y=3t, z=-t^2). The resulting directional derivative will give us the rate of change of temperature in degrees Celsius per meter at point P.

b) To find the rate of change of temperature in degrees Celsius per second, we need to find the directional derivative in the direction of time at point P. This can be done by finding the gradient of T(x,y,z) and plugging in the values for x=8, y=6, and z=-4. Then, we can plug in the direction vector v = <4, 3, -2> (since time is measured in seconds, we can use the constant values for x, y, and z). The resulting directional derivative will give us the rate of change of temperature in degrees Celsius per second at point P.

I hope this helps you get started on the problem! Let me know if you have any further questions. Good luck!