Multivariable calculus yummy in my tummy

In summary, the conversation discusses a problem given by a teacher and the method for solving it. The problem involves finding the direction, expressed as a unit vector, in which a function is increasing most rapidly at a given point on a surface. The difficulty in solving the problem arises from a discrepancy between the given gradient slope vector and the calculated unit vector. The conversation also mentions a similar technique for finding where the gradient of the function is equal to zero and the discrepancy does not occur in this case.
  • #1
mathwiz123
10
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?
 
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  • #2
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?
 
  • #3



Hi there! It looks like you have the right idea in solving this problem. Your method for finding the unit vector in the direction of maximum increase is correct, as you correctly determined the unit vector to be <-.9994,.0336>.

As for the discrepancy in your values for Uy, it could be due to rounding errors or a slight mistake in your calculations. It's always a good idea to double check your work and make sure your calculations are accurate.

The reason your second method for finding where ∇f = 0 worked is because you are essentially finding the critical points of the function, which is a different concept from finding the direction of maximum increase. In this case, the critical point is (2,0), which is where the gradient slope is equal to 0.

Overall, it seems like you have a good understanding of multivariable calculus and are on the right track. Keep up the good work!
 

1. What is multivariable calculus?

Multivariable calculus is a branch of mathematics that deals with functions of several variables, typically in three-dimensional space. It extends the concepts of single-variable calculus, such as derivatives and integrals, to functions with multiple variables.

2. How is multivariable calculus used in real life?

Multivariable calculus is used in various fields such as physics, engineering, economics, and statistics. It is used to model and solve real-world problems that involve multiple variables, such as optimizing the shape of a bridge or predicting the movement of planets.

3. What are the main topics covered in multivariable calculus?

The main topics covered in multivariable calculus include partial derivatives, multiple integrals, vector calculus, and theorems such as the gradient theorem and Stokes' theorem. It also involves applications of these concepts to optimization, curve/surface fitting, and line/surface integrals.

4. What are some challenges of learning multivariable calculus?

Some challenges of learning multivariable calculus include visualizing and understanding functions in three-dimensional space, grasping the concept of partial derivatives and multiple integrals, and applying vector calculus to real-life problems. It also requires a strong foundation in single-variable calculus.

5. How can I improve my understanding of multivariable calculus?

To improve your understanding of multivariable calculus, practice solving problems and visualizing functions in three-dimensional space. Make sure to have a solid understanding of single-variable calculus concepts before diving into multivariable calculus. Utilize resources such as textbooks, online tutorials, and practice quizzes to reinforce your knowledge.

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