Multivariable calculus yummy in my tummy

Click For Summary
SUMMARY

The discussion centers on determining the direction of maximum increase for a function f(x,y) given the gradient slope <-56, 1.886> at the point (2,0). The user initially calculated the unit vector incorrectly, obtaining Ux = -0.9977 and Uy = 0.0672, instead of the correct values derived from normalizing the gradient vector. The discrepancy arose from misapplying the gradient's magnitude in the calculations, leading to confusion regarding the correct Uy value, which was found to be approximately half of the user's result.

PREREQUISITES
  • Understanding of multivariable calculus concepts, specifically gradients.
  • Familiarity with vector normalization techniques.
  • Knowledge of dot product calculations.
  • Ability to solve systems of equations involving vectors.
NEXT STEPS
  • Study the process of gradient vector normalization in multivariable calculus.
  • Learn about the geometric interpretation of gradients and their significance in optimization.
  • Explore the application of the dot product in determining angles between vectors.
  • Investigate common pitfalls in solving systems of equations involving vectors.
USEFUL FOR

Students and educators in mathematics, particularly those studying multivariable calculus, as well as anyone involved in optimization problems in higher dimensions.

mathwiz123
Messages
10
Reaction score
0
[Question]
Hi, my teacher gave us this problem, and he couldn't figure out why
method was incorrect and why I got the answer I did.

Given we know the gradient slope = <-56,1.886> at the point (2,0) on a
surface f(x,y), in what direction, expressed as a unit vector, is f
increasing most rapidly?





[Difficulty]
I solved the problem like this:

Max slope = magnitude of gradient slope
(gradient slope) dot (unit vector) = 56.03
-56.03Ux + 1.866Uy = 56.03
sqt(Ux^2 + Uy^2) = 1

Solving the system Ux= -.9977 Uy=.0672

The only problem is, by definition, I should be able to get the unit
vector by taking the gradient slope vector and dividing by the
magnitude of the gradient vector. or,

<-56/56.03, 1.886/56.03> = <Ux,Uy> = <-.9994,.0336>

The weird thing is, the correct Uy value is almost exactly half of
mine...what's going on?

[Thoughts]

I tried a similar technique for finding where the ∇f = 0

<-56, 1.866> dot (unit vector) = 0

-56.03Ux + 1.886Uy = 0
sqt(Ux^2 + Uy^2) = 1

Solving the system this time I got the correct answer, why here and
not there?
 
Physics news on Phys.org
mathwiz123 said:
[Question]
Hi, my teacher gave us this problem, and he couldn't figure out why
method was incorrect and why I got the answer I did.

Given we know the gradient slope = <-56,1.886> at the point (2,0) on a
surface f(x,y), in what direction, expressed as a unit vector, is f
increasing most rapidly?





[Difficulty]
I solved the problem like this:

Max slope = magnitude of gradient slope
(gradient slope) dot (unit vector) = 56.03
-56.03Ux + 1.866Uy = 56.03
sqt(Ux^2 + Uy^2) = 1

Solving the system Ux= -.9977 Uy=.0672
I understand that the 56.03 is the magnitude of the given vector but the coefficient of Ux should be -56, not -56.03.

The only problem is, by definition, I should be able to get the unit
vector by taking the gradient slope vector and dividing by the
magnitude of the gradient vector. or,

<-56/56.03, 1.886/56.03> = <Ux,Uy> = <-.9994,.0336>

The weird thing is, the correct Uy value is almost exactly half of
mine...what's going on?

[Thoughts]

I tried a similar technique for finding where the ∇f = 0

<-56, 1.866> dot (unit vector) = 0

-56.03Ux + 1.886Uy = 0
sqt(Ux^2 + Uy^2) = 1

Solving the system this time I got the correct answer, why here and
not there?
 

Similar threads

Undergrad Directional derivative and gradient definition confusion
  • · Replies 6 ·
Replies
6
Views
2K