Consider the function f(x, y,z)equalse Superscript xyz minus 3, the point P(0 comma 3 comma negative 2 ), and the unit vector Bold uequalsleft angle seven ninths comma four ninths comma four ninths right angle .

a. Compute the gradient of f and evaluate it at P.
b. Find the unit vector in the direction of maximum increase of f at P.
c. Find the rate of change of the function in the direction of maximum increase at P.
d. Find the directional derivative at P in the direction of the given vector.

See the step by step solution.

Step-by-step explanation:

The main topics of this question are gradient and directional derivative of a multivariable function. Given a function f(x,y,z), its' gradient is defined by

. (i.e each of the components correspond to the partial derivative of the function f with respect each the its' variables).  The gradient is a vector function, which means it could be evaluated at a given point by replacing x,y,z in the formulas with respectively.

a. The given function is and the given point is P=(0,3,-2).

The partial derivatives of f are: .

Then, the gradient function of f is . So, in this case, we want to know which is

b. The gradient function evaluated at a given point gives the vector in which there is maximum increase. On the other hand, an unit vector is a vector, whose norm is equal to 1. Then, we just need to divide the previous answer by its' norm. So, the unit vector is the correct answer.

For questions c and d: The directional derivative (or rate of change) of a function f in the direction of vector v is given by the dot product of v and the gradient of f.

c. From b) we got that the vector of maximum increase at P is . So, the rate of change of f in that direction is (-6)*(-1)+0*0+0*0=6

d. Using the vector the rate of change of is given by (-6)*7/9+0*4/9+0*4/9 = -42/9.

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