Let f(x, y)=xex2−y and P=(9,81). (a) Calculate ∥∇fP∥. (b) Find the rate of change of f in the direction ∇fP. (c) Find the rate of change of f in the direction of a vector making an angle of 45∘ with ∇fP.

a)

b)

c)

Step-by-step explanation:

The given function is . Recall the following:

- (The gradient of f is defined as the vector whose components are the partial derivatives of the function with respect to each of its variables)

- Given a direction vector v, that is a vector that is unitary, the rate of change of the function f in the direction v is given by

- Recall that given two vectors a and b, the dot product between them is given by

- REcall that given a vector x, then

where theta is the angle between both vectors and ||a|| is the norm of the vector a

- Given a vector  of components (x,y) its norm is given by .

a)Let us calculate first the gradient of f and the calculate it at the given point. We will omit the inner steps of derivation, so you must check that the gradient of f is given by

. Since at P we have x=9, and y=81 the desired gradient is

and so the norm of the gradient at P is .

b) We want an unitary vector v for the gradient of f, so we take the gradient and divide it by its norm (i.e )

Hence, the rate of change is given by

c). We are given that . We consider a vector a that is unitary, hence, the rate of change of f in the direction of vector a is given by

38.6°

step-by-step explanation:

resultant is 10 pound

smaller force = 8 pound

angle formed = tan⁻¹(

                        =tan⁻¹(

                          =tan⁻¹(0.8)

                        =38.6°