Applied directional derivative problem

1. Dec 21, 2013

EngnrMatt

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

The temperature at a point (x,y,z) is given by T(x,y,z)=200e^[âˆ’x^(2)âˆ’y^(2)/4âˆ’z^(2)/9], where T is measured in degrees celcius and x,y, and z in meters.

Find the rate of change of the temperature at the point (0, -1, -1) in the direction toward the point (-2, 1, -4).

2. Relevant equations

A directional derivative with direction u is equal to âˆ‡f dotted with the unit vector u.

3. The attempt at a solution

For the gradient vector, I got:

âˆ‡T = <-400xe^[-x^(2)-y^(2)/4-z^(2)/9], -100ye^[''''], (-400/9)e^['''']>

evaluated at (0,-1,-1), I got âˆ‡T(0,-1,-1) = <0, 69.69, 30.9734>

Now, according to the problem, u = <-2, 1, -4>. This means that |u| = âˆš21, so the answer should be <0, 69.69, 30.9734> dotted with <-2/âˆš21, 1/âˆš21, -4/âˆš21> , which gives me -11.828, but this is apparently not the right answer. I would definitely like some help finding my mistake.

2. Dec 21, 2013

Ray Vickson

You are using the wrong vector u. Go back and re-read the question.