Question

Captain Ralph is in trouble near the sunny side of Mercury. The temperature of the ship's hull when he is at location (x, y, z) will be given by T (x, y, z) = e−x2 − 2y2 − 3z2, where x, y, and z are measured in meters. He is currently at (1, 1, 1). (a) In what direction should he proceed in order to decrease the temperature most rapidly?

Answer 1

The maximum rate of change occurs in the direction of the gradient vector at (1, 1, 1).

$T(x,y,z)=e^{-x^2-2y^2-3z^2}\implies\nabla T(x,y,z)=\langle-2x,-4y,-6z\rangle e^{-x^2-2y^2-3z^2}$

At (1, 1, 1), this has a value of

$\nabla T(1,1,1)=\langle-2,-4,-6\rangle e^{-6}$

so the captain should move in the direction of the vector $\langle-1, -2, -3\rangle$ (which is a vector pointing in the same direction but scaled down by a factor of $2e^{-6}$).