Possible Gradient Vector question

[CAF123]

Homework Statement

The temperature T of a plate lying in the (x,y) plane is given by [tex] T(x,y) = 50 - x^2 - 2y^2. [/tex] A bug on the plate is intially at the point (2,1). What is the equation of the curve the bug should follow so as to ensure that the temperature decreases as rapidly as possible?

The Attempt at a Solution

So, [tex] \vec{∇}T(2,1) = <-4,-4> [/tex] If we want the temperature to decrease as rapdily as possible we are looking for when the directional derivative is minimal. So the direction is [itex] -\vec{∇}T(2,1) = <4,4>. [/itex] My professor gave a hint to find dy/dx and then solve for y. Using the Implicit Function theorem this gives; [tex] \frac{dy}{dx} = -\frac{-2x}{-4y}. [/tex] I don't see how this method takes into account that we are considering the minimal temperature? (In fact, the gradient vector is not used at all)?

 

[Ray Vickson]

Homework Statement

The temperature T of a plate lying in the (x,y) plane is given by [tex] T(x,y) = 50 - x^2 - 2y^2. [/tex] A bug on the plate is intially at the point (2,1). What is the equation of the curve the bug should follow so as to ensure that the temperature decreases as rapidly as possible?

The Attempt at a Solution

So, [tex] \vec{∇}T(2,1) = <-4,-4> [/tex] If we want the temperature to decrease as rapdily as possible we are looking for when the directional derivative is minimal. So the direction is [itex] -\vec{∇}T(2,1) = <4,4>. [/itex] My professor gave a hint to find dy/dx and then solve for y. Using the Implicit Function theorem this gives; [tex] \frac{dy}{dx} = -\frac{-2x}{-4y}. [/tex] I don't see how this method takes into account that we are considering the minimal temperature? (In fact, the gradient vector is not used at all)?

At every point on the path the bug's velocity should be along the negative of the gradient, so if the bug moves at constant speed v, it should move by dx = -v*(-2x)/D, dy = -v*(-4y)/D, where D = norm of the gradient vector = √(4x^2 + 16y^2). If you take dy/dx that gets rid of v and D.

RGV

 

[CAF123]

At every point on the path the bug's velocity should be along the negative of the gradient, so if the bug moves at constant speed v, it should move by dx = -v*(-2x)/D, dy = -v*(-4y)/D, where D = norm of the gradient vector = √(4x^2 + 16y^2). If you take dy/dx that gets rid of v and D.

RGV

Could you show how you got those eqns for dx and dy? By the Implicit function theorem, I get [tex] \frac{dy}{dx} = -\frac{x}{2y}, [/tex] while from the method above using those eqns derived, it is [itex] \frac{dy}{dx} = \frac{x}{2y} [/itex] out by a minus sign?

 

[Ray Vickson]

Could you show how you got those eqns for dx and dy? By the Implicit function theorem, I get [tex] \frac{dy}{dx} = -\frac{x}{2y}, [/tex] while from the method above using those eqns derived, it is [itex] \frac{dy}{dx} = \frac{x}{2y} [/itex] out by a minus sign?

Your dy/dx would take the bug along a level curve (a curve of constant T). You want the bug to move perpendicular to that direction.

RGV

 

[CAF123]

Oh I see. As said in my previous post, the directional derivative is minimal when we take the negative of the gradient vector. (The gradient vector points in the direction of increasing value for a function). From what RGV said, dy/dx will be parallel to -∇T and so in the same direction as <2x,4y> which means dy/dx = 4y/2x.
Could you still explain how your formulae for dx and dy are derived?