Good approximation - multivariable calculus

In summary, the Taylor series expanded at 3 and set to 3.01 gives the desired result but is marked as wrong. There is also a question about what statement is correct: (1+\Delta x, 1+\Delta y) is nearly parallel or perpendicular to (4,5). I thought the vector (0.37887, -0.54038) was nearly perpendicular. Well...In summary, the Taylor series expanded at 3 and set to 3.01 gives the desired result but is marked as wrong. There is also a question about what statement is correct: (1+\Delta x, 1+\Delta y) is nearly parallel or perpendicular to (4,5). I thought the vector
  • #1
Poetria
267
42
Homework Statement
##f(x,y) = x^3+x*y^2+y^3##
Then
f(1,1)=3
##f_x (1,1) = 4##
##f_y (1,1) = 5##
##\nabla f(1,1) = (4,5)##

Find the closest point to (1,1) at f(x,y) = 3.01, i.e. ##(1+\Delta x, 1+\Delta y)##
Relevant Equations
Linear approximation
I tried to use a Taylor series expanded at 3 and set to 3.01:
https://www.wolframalpha.com/input/?i=27+++9+(-3+++x)^2+++(-3+++x)^3+++3+y^2+++y^3+++(-3+++x)+(27+++y^2)=3.01

I got the vector ## (\Delta x, \Delta y)= (0.37887, -0.54038)##
It does give a desired result but it is marked as wrong.

There is also a question about what statement is correct:

##(1+\Delta x, 1 + \Delta y)## is nearly parallel or perpendicular to ##(4,5)##
Or
##(\Delta x, \Delta y)## is nearly parallel or perpendicular to ##(4,5)##

I thought the vector (0.37887, -0.54038) was nearly perpendicular. Well...
 
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  • #2
Poetria said:
Homework Statement:: ##f(x,y) = x^3+x*y^2+y^3##
Then
f(1,1)=3
##f_x (1,1) = 4##
##f_y (1,1) = 5##
##\nabla f(1,1) = (4,5)##

Find the closest point to (1,1) at f(x,y) = 3.01, i.e. ##(1+\Delta x, 1+\Delta y)##
Relevant Equations:: Linear approximation

I tried to use a Taylor series expanded at 3 and set to 3.01:
https://www.wolframalpha.com/input/?i=27+++9+(-3+++x)^2+++(-3+++x)^3+++3+y^2+++y^3+++(-3+++x)+(27+++y^2)=3.01

I got the vector ## (\Delta x, \Delta y)= (0.37887, -0.54038)##
It does give a desired result but it is marked as wrong.

There is also a question about what statement is correct:

##(1+\Delta x, 1 + \Delta y)## is nearly parallel or perpendicular to ##(4,5)##
Or
##(\Delta x, \Delta y)## is nearly parallel or perpendicular to ##(4,5)##

I thought the vector (0.37887, -0.54038) was nearly perpendicular. Well...
Imaging we look at the surface ##(x,y,f(x,y))## from the top. Then it looks like

1629037224236.png

Hence we are looking for a tangent spaces to the curve ##\{(x,y)\,|\,f(x,y)=3.01\}##. And then the one at ##(x_p,y_p)## among them for which tangent space and ##(1-x_p,1-y_p,3.01)## are perpendicular.
 
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  • #3
I am not sure if I have understood you.
In that case, wouldn't the vector (1.44385, -1.44385) be a better solution?

I have tried to use
 
  • #4
I cannot guess the numbers and haven't made the calculation.

If you want to approach it numerically, then minimize the distance between ##(1,1,3)## and ##(x,y,3.01)## under the condition that ##x^3+xy^2+y^3=3.01##.
 
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  • #5
fresh_42 said:
I cannot guess the numbers and haven't made the calculation.

If you want to approach it numerically, then minimize the distance between ##(1,1,3)## and ##(x,y,3.01)## under the condition that ##x^3+xy^2+y^3=3.01##.
Thank you very much. I will try.
 
  • #6
This is the formula for the distance squared but I have no idea how to add the constraint:
##(1-x)^2+(1-y)^2+(3.01-3)^2##
 
  • #8
Poetria said:
This is the formula for the distance squared but I have no idea how to add the constraint:
##(1-x)^2+(1-y)^2+(3.01-3)^2##
The keyword for approximations under constraints is the Lagrange multiplier. It is a technique to solve such problems. Wikipedia explains it quite well.

If you want to follow your path, then you'll get an equation ##g(x,y)=x^2+y^2+ax+by+c## which represents a paraboloid (opened at top). The solution will be its minimum.
 
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  • #9
I don't think anything too complicated is required. See if these (non-mathematician's) thoughts helps.

Point 1

Think of f(x,y) as a formula for the height of a hill, the xy plane being sea-level (ignore earth’s curvature). And imagine you have a map with contour lines.

Do know that ∇f(x,y) is a vector which points in the direction of ‘steepest ascent’ (maximum rate of change of f)?

If you are at P, above (1,1), your height is 3. Now imagine a contour line on your map for height = 3.01. You want the shortest way to get from P to the 3.01 contour line.

The shortest route will be the steepest – so your height will change most rapidly. The direction of the steepest route is given by vector ∇f(1,1) which is <4,5>. You need to move such that when your x-position changes 4 units, your y-position changes by 5 units.

That means Δy = (5/4)Δx

(We’re assuming ∇f(x,y) doesn’t change significantly while you move; it remains <4,5>. This is valid for small enough movements).
__________

Point 2

Have you met the ‘total differential’ (or ‘total derivative’)? In this case it is given by:

##df = \frac {∂f}{∂x}dx + \frac {∂f}{∂y}dy##

It gives the change in f resulting from changes in x and y. We are using Δx and Δy, so providing these values are small enough we can write:
##Δf = \frac {∂f}{∂x}Δx + \frac {∂f}{∂y}Δy##

Note that we we have Δf = 3.01 – 3.00 = 0.01

Edit: typo' corrected.
 
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Related to Good approximation - multivariable calculus

1. What is a good approximation in multivariable calculus?

A good approximation in multivariable calculus is a value that is close to the exact value of a function or expression. It is used to estimate the behavior of a function or expression at a certain point or within a certain range.

2. How is a good approximation calculated in multivariable calculus?

A good approximation is calculated using various methods such as Taylor series, linearization, and numerical methods like Euler's method or Runge-Kutta method. These methods involve using derivatives to estimate the behavior of a function at a given point.

3. Why is a good approximation important in multivariable calculus?

A good approximation is important in multivariable calculus because it helps us understand the behavior of complex functions and make predictions about their values at certain points. It also allows us to solve problems that would otherwise be too difficult to solve using exact methods.

4. What are some applications of good approximation in multivariable calculus?

Good approximation is used in various fields such as physics, engineering, economics, and statistics. It is used to solve optimization problems, model physical systems, and make predictions about the behavior of complex systems.

5. How can I improve my skills in finding good approximations in multivariable calculus?

To improve your skills in finding good approximations in multivariable calculus, it is important to have a strong understanding of derivatives and their applications. Practice using different methods of approximation and solving problems involving multivariable functions. Additionally, studying real-world applications of good approximation can help improve your skills in this area.

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