- #1
Alex Bard
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[itex]\partial[/itex]This is my first post, so I apologize for all my mistakes. Thank you for the help, in advance.
These are test review questions for Multi Variable Calculus.
Let f(x,y) = tan-1(y2 / x)
a) Find fx([itex]\sqrt{5}[/itex], -2) and fy([itex]\sqrt{5}[/itex], -2).
b) Find the rate of change of f at the point ([itex]\sqrt{5}[/itex], -2) in the direction toward the point (0,0).
c) Find the directional derivative in the direction of fastest increase at the point ([itex]\sqrt{5}[/itex], -2).
So from my understanding, part a, I would just have to do partial derivatives to x then y and then plug in the coordinates afterwards that are given.
Part b is the gradient vector.
Part c is just the largest value when getting the derivative.
a. [itex]\frac{∂f}{∂x}[/itex] = [1 / 1 + (y^2/x)^2] * y^2 = (y^2 * x^2)/1+y^4
when subbing in the points given, I get 20/17
[itex]\frac{∂f}{∂y}[/itex] = 1 / [1 + (y^2/x)^2] * 2y/x = (2y * x^2) / x(1 + y^4)
when subbing in the points given, I get 20/17√5
b. So here I would first make a vector from (√5, -2) going to (0,0), which is <-√5, 2>
Then unit vector is 1/√[(√5)2 + 22] = 1 / √9
= 1/3
Then 1/3(-√5, 2) is the unit vector
To get the Gradient, I use ∂ in respect to x and y AND using the calculations from part a, I get:
∇f(x,y) = <(y^2 * x^2)/1+y^4, (2y * x^2) / x(1 + y^4)>
now I'm a bit stuck. Do I just do the DOT product of ∇f and u? as in:
1/3(-√5, 2) ° <(y^2 * x^2)/1+y^4, (2y * x^2) / x(1 + y^4)>
or should I sub the points given of (√5, -2) into the ∇f and then dot them?
I haven't really gotten to part c yes since I believe I need the answer from b to do c.
All your help is appreciated.
These are test review questions for Multi Variable Calculus.
Homework Statement
Let f(x,y) = tan-1(y2 / x)
a) Find fx([itex]\sqrt{5}[/itex], -2) and fy([itex]\sqrt{5}[/itex], -2).
b) Find the rate of change of f at the point ([itex]\sqrt{5}[/itex], -2) in the direction toward the point (0,0).
c) Find the directional derivative in the direction of fastest increase at the point ([itex]\sqrt{5}[/itex], -2).
Homework Equations
So from my understanding, part a, I would just have to do partial derivatives to x then y and then plug in the coordinates afterwards that are given.
Part b is the gradient vector.
Part c is just the largest value when getting the derivative.
The Attempt at a Solution
a. [itex]\frac{∂f}{∂x}[/itex] = [1 / 1 + (y^2/x)^2] * y^2 = (y^2 * x^2)/1+y^4
when subbing in the points given, I get 20/17
[itex]\frac{∂f}{∂y}[/itex] = 1 / [1 + (y^2/x)^2] * 2y/x = (2y * x^2) / x(1 + y^4)
when subbing in the points given, I get 20/17√5
b. So here I would first make a vector from (√5, -2) going to (0,0), which is <-√5, 2>
Then unit vector is 1/√[(√5)2 + 22] = 1 / √9
= 1/3
Then 1/3(-√5, 2) is the unit vector
To get the Gradient, I use ∂ in respect to x and y AND using the calculations from part a, I get:
∇f(x,y) = <(y^2 * x^2)/1+y^4, (2y * x^2) / x(1 + y^4)>
now I'm a bit stuck. Do I just do the DOT product of ∇f and u? as in:
1/3(-√5, 2) ° <(y^2 * x^2)/1+y^4, (2y * x^2) / x(1 + y^4)>
or should I sub the points given of (√5, -2) into the ∇f and then dot them?
I haven't really gotten to part c yes since I believe I need the answer from b to do c.
All your help is appreciated.