1. ## Directional Derivatives and gradient vector killing me

The temperature at a point (x,y,z) is given by , where is measured in degrees Celsius and x,y, and z in meters. There are lots of places to make silly errors in this problem; just try to keep track of what needs to be a unit vector.
1) Find the rate of change of the temperature at the point (1, -1, 2) in the direction toward the point (-1, -5, 5).

2) In which direction (unit vector) does the temperature increase the fastest at (1, -1, 2)?

3)What is the maximum rate of increase of at (1, -1, 2)?
I got the second one right.
$<(-400e^{-49/36})/ \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2}),$ $100e^{-49/36}/ \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2}),$ $-(800/9)e^{-49/36}/ \sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2})>
$

but I do not get 1 and 3.

I tried to do both and i have
1) 0
3) $\sqrt(e^{-98/36}{(-400)^2+(100)^2+(-800/9)^2})$

But both are not right and I am totally out of idea how to solve it. Please someone help me

2. ## Directional Derivatives

To find the rate of change in f at point a in the direction v, we use the equation $D_vf(a)=\bigtriangledown f(a)\cdot v$

Here, $a=(1, -1, 2), v=(-1, -5, 5)-(1, -1, 2)=(-2,-4,3)$, whose unit vector is $\hat{v}=\frac{1}{\sqrt{29}}(-2,-4,3)$

Also, $\bigtriangledown T(a)=(T_x(a),T_y(a),T_z(a))$, so the first step in the problem is to find the three partial derivatives of T to give the gradient. Then, plug and chug: $D_vT(a)=\frac{1}{\sqrt{29}}(T_x(1, -1, 2)(-2)+T_y(1, -1, 2)(-4)+T_z(1, -1, 2)(3))$

The maximum/minimum rates of change will occur when $\bigtriangledown^2 T(a)=(T_{xx}(a),T_{yy}(a),T_{zz}(a))=\vec{0}$

See if this can get you started.

3. Originally Posted by Media_Man
The maximum/minimum rates of change will occur when $\bigtriangledown^2 T(a)=(T_{xx}(a),T_{yy}(a),T_{zz}(a))=\vec{0}$
I thought the magnitude of the gradient vector gives the max rate of change, doesnt it? Why square of gradient vector gives the max/min rate of change?

4. ## Good question

" $\bigtriangledown^2$" actually means "the gradient of the gradient" not "the square of the gradient." Consider an equation modeling a car traveling down a road. The position is the function. The velocity is the first derivative of the function. The acceleration is the second derivative. Hence, when $f''(x)=0$, that is a max or min rate of change. Therefore, $\bigtriangledown^2 T(a)=\vec{0}$ gives the point(s) of maximum or minimum velocity.

5. So how about taking magnitude of $
\bigtriangledown T(a)$
==> $
||\bigtriangledown T(a)||$
I thought solving for magnitude gives me the max rate of change.

How does it different from $
\bigtriangledown^{2} T(a)$
?

6. Ah, yes. Forgive me. I was doing it the hard way. There is indeed a theorem stating that the maximum rate of change occurs in the direction of the gradient vector. You are correct.