# Homework Help: Directionnal Derivative

1. Oct 11, 2006

### Lancen

This is a problem thats been bugging me. Suppose you are on the surface M(x,y)=3*x^2+y^2+5000 that describes a magnetic field, you are at the point (8,6), what is the curve along which you should travel so as to minimize field intensity as rapidly as possible? I am told I need to solve a differential equation, I am thinking I need to find the minimal directional derivative and somehow from that find the equation of the line that represents, and integrate it to get a curve. But I am at a loss as to how to start. Someone help?

2. Oct 11, 2006

### quasar987

You know that $-\nabla M(x,y)$ is the direction you need to go into in order for your field to decrease the most rapidly. And given a parametrized curve $\gamma (t) = (x(t),y(t))$, the speed $\dot{\gamma}$, is the direction in which the curve is "going". So if you'd have a curve whose speed is $-\nabla M(x,y)$, you'd have a curve that follows the path of fastest decrease, would you not?

3. Oct 11, 2006

### Lancen

Sorry I don't quite follow, the negative of the directional derivative is the directional I need to go, but what is the parametrized curve? Is the equation of setting up -M(x,y)=(x(t),y(t))?

4. Oct 11, 2006

### quasar987

A curve and a "parametrized curve" are practically the same thing. A curve is a function (typically smooth) $\gamma : (t_1, t_2)\rightarrow \mathbb{R}^2$, $\gamma(t)=(x(t),y(t))$.

This is a what a curve is. The other way to define a curve is as a "level curve", i.e. by finding a function f(x,y) such that f(x,y)=constant gives a certain relation btw x and y.

5. Oct 11, 2006

### quasar987

And just as the derivative of a single valued function f(x) gives the tangant to the curve, the derivative of the curve, typically noted $\dot{\gamma}$*, gives the tangeant to the curve, i.e. the direction where the curve is "heading".

*$$\dot{\gamma} =\frac{d}{dt}\gamma(t) = \left(\frac{dx(t)}{dt},\frac{dy(t)}{dt} \right)$$

Last edited: Oct 11, 2006
6. Oct 11, 2006

### Lancen

I am still a bit confused but I have something like this - (xn,yn) = (8,6)+[(-3x,2y)/sqrt(36x^2+4y^2)]*[v,w] where (xn,yn) is the new point after moving some distance, [v,w] the unit vector and [(-3x,2y)/sqrt(36x^2+4y^2)] the gradient divided by its length to give you the unit gradient, but how do I solve this?

7. Oct 11, 2006

### quasar987

I don't understand what you've done here. What do you not understand about the method I suggested?