# Partial Derivative Problem

1. Jan 26, 2014

### tacopwn

1. The problem statement, all variables and given/known data
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

2. Relevant 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

3. 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.

2. Jan 26, 2014

### 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?

3. Jan 26, 2014

### SammyS

Staff Emeritus

4. Jan 26, 2014

### tacopwn

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

5. Jan 26, 2014

### CAF123

6. Jan 26, 2014

### tacopwn

dF(x,y)=(-4x,-2x) = (-40, -20)

7. Jan 26, 2014

### 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.

8. Jan 26, 2014

### tacopwn

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

9. Jan 26, 2014

### SammyS

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

10. Jan 27, 2014

### tacopwn

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

11. Jan 27, 2014

### pasmith

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 $x(0) = y(0) = 10$.