Vector Calc - Directional Derivative Question

At the point (8, 6), we have 36/2= 64/4+ c or c= 18. The equation is y^2= x^2+ 36. If you want an equation for a curve, it would be y= \pm\sqrt{x^2+ 36}. That's a hyperbola with center at the origin and asymptotes x= \pm\sqrt{36} or x= \pm 6. If you want to move toward the origin, take the negative branch. That is,
  • #1
at3rg0
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0

Homework Statement



Igor, the inchworm, is crawling along graph paper in a magnetic field. The intensity of the field at the point (x,y) is given by M(x,y) = 4x^2 + y^2 + 5000. If Igor is at the point (8,6), describe the curve along which he should travel if he wishes to reduce the field intensity as rapidly as possible.


Homework Equations


Gradient M = <8x, 2y>
To minimize, one must travel opposite the gradient

The Attempt at a Solution


I got the gradient as <8x, 2y> (just the partial derivatives as a vector). Also, at (8,6), the gradient is <64, 12>, which is in the direction of the unit vector 16i + 3j/sqrt(16^2 +3^2). So, the direction he wants to travel is the negative version of that vector. However, the problem wants me to arrive at the formula (y^3 = 27x towards the origin). How do I go from the vector to the curve?

This TEX is driving me crazy. If anyone can help me by editing my post, that would be great.

Thanks in advance!
 
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  • #2
You are correct that <64, 12> gives the direction he wants to go at that time. But as soon as he has moved slightly it will change. What you are saying is that he wants to move so that dx/dt= -8x and dy/dt= -2y. It's not difficult to solve both of those for x and y as functions of t but you don't need the time dependence- you only need the curve itself:
[tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{1}{4}\frac{y}{x}[/tex].
That's a simple separable equation. Separate the variables and integrate.
 

Related to Vector Calc - Directional Derivative Question

1. What is a directional derivative?

A directional derivative is a measure of the rate of change of a function in a particular direction. It represents how much a function changes as you move along a specific path or direction in its domain.

2. How is a directional derivative calculated?

The directional derivative is calculated using the gradient of the function and a unit vector representing the desired direction. The formula is: Duf = ∇f · u, where ∇f is the gradient of the function and u is the unit vector.

3. What is the relationship between directional derivatives and slopes?

Directional derivatives can be thought of as generalizations of slopes in multivariable calculus. Just like how the slope of a tangent line represents the rate of change of a function in one direction, the directional derivative represents the rate of change of a function in a specific direction.

4. Can the directional derivative be negative?

Yes, the directional derivative can be negative if the gradient of the function points in the opposite direction of the unit vector. This means that the function is decreasing in the direction of the unit vector.

5. How are directional derivatives used in real life?

Directional derivatives are used in physics, engineering, and other fields to model and analyze the behavior of systems. For example, in fluid dynamics, directional derivatives are used to determine the rate of change of fluid velocity in a specific direction. They are also used in optimization problems to find the direction of steepest ascent or descent for a function.

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