Respuesta :
Answer:
a) [tex]\sqrt[]{163^3+9^2}[/tex]
b) [tex]\sqrt[]{163^3+9^2}[/tex]
c) [tex]\sqrt[]{163^3+9^2}\cdot \frac{\sqrt[]{2}}{2}[/tex]
Step-by-step explanation:
The given function is [tex]f(x) = xe^{x^2-y}[/tex]. Recall the following:
-[tex]\nabla f = (\frac{df}{dx}, \frac{df}{dy}[/tex] (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
[tex]\nabla f \cdot v [/tex]
- Recall that given two vectors a and b, the dot product between them is given by
[tex] a\cdot b = ||a|| ||b|| \cos(\theta)[/tex]
- REcall that given a vector x, then [tex] x \cdot x = ||x||^2[/tex]
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 [tex]\sqrt[]{x^2+y^2}[/tex].
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
[tex]\nabla f = ((2x^2+1)e^{x^2-y},-xe^{x^2-y})[/tex]. Since at P we have x=9, and y=81 the desired gradient is
[tex] \nabla f = (163,-9)[/tex] and so the norm of the gradient at P is [tex]\sqrt[]{163^2+9^2}[/tex].
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 [tex]\frac{\nabla f}{||\nabla f||}[/tex])
Hence, the rate of change is given by
[tex] \nabla f \cdot \frac{\nabla f}{||\nabla f||} = \frac{||\nabla f||^2}{||\nabla f||}=||\nabla f||[/tex]
c). We are given that [tex]\theta = \pm45 ^\circ [/tex]. We consider a vector a that is unitary, hence, the rate of change of f in the direction of vector a is given by
[tex] \nabla f \cdot a = ||a||||\nabla f||\cos(\pm 45 ^\circ) = \frac{\sqrt[]{2}}{2}||\nabla f||[/tex]
If [tex]f(x, y) = xe^{x^2-y}[/tex] and P=(9,81), then:
a) ∥∇fP∥ = 163.25
b) Rate of change of f in the direction ∇fP = 163.25
c) The rate of change of f in the direction of a vector making an angle of 45∘ with ∇fP = 115.43
The given function is:
[tex]f(x, y) = xe^{x^2-y}[/tex]
Find [tex]\frac{\delta f}{\delta x}[/tex] and [tex]\frac{\delta f}{\delta y}[/tex]
[tex]\frac{\delta f}{\delta x} = (2x+1)e^{x^2-y}\\\\ \frac{\delta f}{\delta y} = -xe^{x^2-y}[/tex]
Given P(9, 81)
[tex]\frac{\delta f}{\delta x} |P = [(2(9)+1]e^{9^2-81)[/tex]
[tex]\frac{\delta f}{\delta x} |P = 163[/tex]
[tex]\frac{\delta f}{\delta y} |P = [-9e^{9^2-81)[/tex]
[tex]\frac{\delta f}{\delta y} |P = -9[/tex]
[tex]\triangledown f_P = (\frac{\delta f}{\delta x} |P, \frac{\delta f}{\delta y} |P)\\\\\triangledown f_P = (163, -9)\\\\||\triangledown f_P|| = \sqrt{163^2+(-9)^2} \\\\||\triangledown f_P|| =163.25[/tex]
b) The rate of change of f in the direction ∇fP
This can be given by the equation [tex]\delta f_P . \frac{\delta f_P}{||\delta f_P||}[/tex]
[tex]\triangledown f_P . \triangledown f_P = ||\triangledown f_P||^2[/tex]
The rate of change becomes:
[tex]\frac{||\triangledown f_P||^2}{||\triangledown f_P||}[/tex]
Rate of change = [tex]||\triangledown f_P||[/tex]
Rate of change of f in the direction ∇fP = 163.25
C) The rate of change of f in the direction of a vector making an angle of 45∘ with ∇fP = [tex]||\triangledown f_P||cos 45[/tex]
The rate of change of f in the direction of a vector making an angle of 45∘ with ∇fP = 163.25(0.7071)
The rate of change of f in the direction of a vector making an angle of 45∘ with ∇fP = 115.43
Learn more here: https://brainly.com/question/15584134
