Finding the Path of a Particle in a Temperature Gradient

[tacopwn]

Homework Statement

Find an equation for the path of a particle that starts at P(10,10) and always moves in the direction of maximum temperature increase if the temperature in the plane is T(x,y) = 400-2x^2 -y^2

Homework Equations

T(x,y) = 400-2x^2 -y^2
dT/dx = -4x
dT/dy = -2y
d^2T/dx^2 = -4
d^2T/dy^2 = -2
maximum is at (0,0)=400

The Attempt at a Solution

I'm not sure how to solve this because the maximum temperature increase (second derivative) is always -4 in the x and -2 in the y.

 

[CAF123]

Yes, the maximum temperature is at the coordinates (0,0), however you are looking for the path that takes you from (10,10) to there via the path of greatest temperature. Have you come across the gradient vector?

 

[SammyS]

Homework Statement

Find an equation for the path of a particle that starts at P(10,10) and always moves in the direction of maximum temperature increase if the temperature in the plane is T(x,y) = 400-2x^2 -y^2

Homework Equations

T(x,y) = 400-2x^2 -y^2
dT/dx = -4x
dT/dy = -2y
d^2T/dx^2 = -4
d^2T/dy^2 = -2
maximum is at (0,0)=400

The Attempt at a Solution

I'm not sure how to solve this because the maximum temperature increase (second derivative) is always -4 in the x and -2 in the y.

What is the gradient ?

 

[tacopwn]

Yes, the maximum temperature is at the coordinates (0,0), however you are looking for the path that takes you from (10,10) to there via the path of greatest temperature. Have you come across the gradient vector?

The gradient plane at (10,10) is -40x-20y+700=z

 

[CAF123]

The gradient plane at (10,10) is -40x-20y+700=z

What is the gradient vector?

 

[tacopwn]

What is the gradient vector?

The gradient vector at (10,10):
dF(x,y)=(-4x,-2x) = (-40, -20)

 

[HallsofIvy]

Okay, the gradient vector at any point (x, y) is <-4x, -2y> and at (10, 10) the gradient vector is <-40, -20>. But no one asked about the gradient vector at (10, 10)! What is important is that the gradient points in the direction of maximum increase so you want dy/dx to give a curve in the same direction as the gradient. A vector <a, b> has length a in the x-direction and length b in the y direction.

Of course, the derivative, dy/dx, is the slope of the tangent line so that dy/dx= b/a.

 

[tacopwn]

Okay, the gradient vector at any point (x, y) is <-4x, -2y> and at (10, 10) the gradient vector is <-40, -20>. But no one asked about the gradient vector at (10, 10)! What is important is that the gradient points in the direction of maximum increase so you want dy/dx to give a curve in the same direction as the gradient. A vector <a, b> has length a in the x-direction and length b in the y direction.

Of course, the derivative, dy/dx, is the slope of the tangent line so that dy/dx= b/a.

Oh okay, so once you obtain dy/dx which is 2x/y, how do you make sure it starts at (10,10)?

 

[SammyS]

Oh okay, so once you obtain dy/dx which is 2x/y, how do you make sure it starts at (10,10)?

How do you write the equation of a path in general?

 

[tacopwn]

How do you write the equation of a path in general?

(x,y)=(10,10) + t(2,1)

 

[pasmith]

(x,y)=(10,10) + t(2,1)

That must be wrong: it heads away from the global maximum at the origin as $t$ increases (and doesn't pass through the origin at all).

You need to solve
$$\dot x = \frac{\partial T}{\partial x} = -4x, \\ \dot y = \frac{\partial T}{\partial y} = -2y$$
subject to $x(0) = y(0) = 10$.