Question

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.

Answer 1

Answer:

a) $\sqrt[]{163^3+9^2}$

b) $\sqrt[]{163^3+9^2}$

c) $\sqrt[]{163^3+9^2}\cdot \frac{\sqrt[]{2}}{2}$

Step-by-step explanation:

The given function is $f(x) = xe^{x^2-y}$. Recall the following:

-$\nabla f = (\frac{df}{dx}, \frac{df}{dy}$ (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

$\nabla f \cdot v$

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

$a\cdot b = ||a|| ||b|| \cos(\theta)$

- REcall that given a vector x, then $x \cdot x = ||x||^2$

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 $\sqrt[]{x^2+y^2}$.

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

$\nabla f = ((2x^2+1)e^{x^2-y},-xe^{x^2-y})$. Since at P we have x=9, and y=81 the desired gradient is

$\nabla f = (163,-9)$ and so the norm of the gradient at P is $\sqrt[]{163^2+9^2}$.

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 $\frac{\nabla f}{||\nabla f||}$)

Hence, the rate of change is given by

$\nabla f \cdot \frac{\nabla f}{||\nabla f||} = \frac{||\nabla f||^2}{||\nabla f||}=||\nabla f||$

c). We are given that $\theta = \pm45 ^\circ$. We consider a vector a that is unitary, hence, the rate of change of f in the direction of vector a is given by

$\nabla f \cdot a = ||a||||\nabla f||\cos(\pm 45 ^\circ) = \frac{\sqrt[]{2}}{2}||\nabla f||$