1. ## Directional Derivatives/Gradient Vector

Here goes:

The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,2,2) is $120^{\circ}$.

(a) Find the rate of change of T at (1,2,2) in the direction toward the point (2,1,3).

(b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin.

2. Originally Posted by Aryth
Here goes:

The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,2,2) is $120^{\circ}$
so $T(x,y,z)=\frac{k}{\sqrt{x^2+y^2+z^2}},$ for some constant $k.$ since $120=T(1,2,2)=\frac{k}{\sqrt{1+4+4}}=\frac{k}{3},$ we get $k=360.$

(a) Find the rate of change of T at (1,2,2) in the direction toward the point (2,1,3).
first you need to find the unit vector in this direction: we have $\bold{v}=(2,1,3)-(1,2,2)=(1,-1,1).$ thus the unit vector is:

$\bold{u}=\frac{\bold{v}}{|\bold{v}|}=\frac{1}{\sqr t{3}}(1,-1,1).$ it's also clear that $\nabla T(x,y,z)= \frac{360}{(x^2+y^2+z^2)^{\frac{3}{2}}}(-x,-y,-z).$ now the rate of change of $T$ is:

$D_{\bold{u}}T(1,2,2)=\nabla T (1,2,2) \cdot \bold{u}=\frac{360}{27 \sqrt{3}}(-1,-2,-2) \cdot (1,-1,1)=\frac{-40}{3\sqrt{3}}.$

(b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin.
the direction of greatest increase at any point is the direction of $\nabla T$ at that point. but in part (a) we saw that $\nabla T(x,y,z)$

and $(-x,-y,-z)$ have the same direction and, clearly, the vector $(-x,-y,-z)$ points toward the origin. Q.E.D.