18.02 MIT Multivariate calculus - Computing most rapid decrease using directional derivative

Question

I'm following the 18.02 course of MIT of multivariate calculus. Topic is directional derivatives.

Here is the question:

The function $T = x^2+2y^2+2z^2 $ gives the temperature at each point in space. At point $P=(1,1,1)$, in which direction should you go to get the most rapid decrease in $T$ What is the directional derivative in this direction?

Here is part of the answer:

We know that the fastest increase is in the direction of $\nabla T = \langle 2x,4y,4z \rangle $. At $P$, the fastest decrease is in the direction of $-\nabla T|_P= -\langle 1,2,2\rangle $. The unit vector in this direction is $\hat u = -\langle 1/3,2/3,2/3 \rangle$. The rate of change in this direction is $-|\nabla T|= -3$.

Here is what I don't understand:

It seems that there is a factor $1/2$ to obtain $-\nabla T|_P= -\langle1,2,2\rangle$ from $\nabla T = \langle 2x,4y,4z \rangle$. Where does it come from?

Answer 1

The answer you were given has a several mistakes. First, the gradient should be

$$ \nabla T(1,1,1) = \langle 2,4,4 \rangle $$

The direction vector is in the wrong direction $$ \hat{u} = -\frac{\nabla T}{|\nabla T|} = -\left\langle \frac13, \frac23, \frac23 \right\rangle $$

And the directional derivative is off by a factor of $2$

$$ \nabla T \cdot\hat{u} = -|\nabla T| = -6 $$

Answer 2

The gradient at $P(1,1,1)$ is $\nabla T_P= (2,4,4)$.

So, the magnitude of the rate of change is $$|\nabla T_P| = 2\cdot |(1,2,2)| = 6$$

Nevertheless, the direction of maximum decrease is in the direction of $-(2,4,4)$, which has the same direction as $-(1,2,2)$ which give the same unit vector $$u =-\frac{1}{3}(1,2,2)$$

Edit for completeness:

The directional derivative in the direction of $u$ is $$\nabla T_P\cdot u = -\frac{2}{3}|(1,2,2)|^2=-6$$