Level Curves and Partial Derivatives

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SUMMARY

This discussion focuses on the relationship between level curves and partial derivatives of a function \( z = f(x,y) \). At point P, the partial derivative \( f_x(P) \) is negative, indicating a downward slope in the x-direction. The second partial derivative \( f_{xx}(P) \) is positive, suggesting a decrease in steepness as one moves left to right. Additionally, the partial derivative \( f_y \) is positive, indicating an upward slope in the y-direction, while the second partial derivative \( f_{yy} \) is also positive, reflecting an accelerating increase in the function's value as y increases.

PREREQUISITES
  • Understanding of partial derivatives and their geometric interpretation
  • Familiarity with level curves and contour plots
  • Basic knowledge of limits and slopes in calculus
  • Ability to analyze functions of two variables
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  • Study the implications of level curves on the behavior of multivariable functions
  • Learn about the Hessian matrix and its role in determining concavity
  • Explore the concept of directional derivatives and gradients
  • Investigate the relationship between level curves and optimization problems
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Students and professionals in mathematics, particularly those studying calculus and multivariable functions, as well as educators teaching these concepts.

Lancelot1
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Hello everyone,

I am trying to solve this wee problem regarding partial derivatives, and not sure how to do so.

The following image shows level curves of some function \[z=f(x,y)\] :

View attachment 7998

I need to determine whether the following partial derivatives are positive or negative at the point P:

\[f_{x} , f_{y} , f_{xx} , f_{yy} , f_{xy} , f_{yx}\]

I am not sure how to relate the partial derivatives to the level curves. I know that partial derivatives at a point are slopes of a curve created when we fix a plane such as x=a or y=b. Where and how do I see it in level curves ?

Cheers !
 

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Hi Lancelot,

Remember that a partial derivative is the limit of a slope.
More specifically:
$$
f_x(P)=\lim_{h\to 0}\frac{f(P+(h,0)) - f(P)}{h} \approx \frac{f(P+(h,0)) - f(P)}{h}
$$
Let's start at P and make a step with size $h>0$ to the right.
To the right of P we see that it has value 4.
And at P itself we have value 6.
So we have:
$$f_x(P)\approx \frac{4 - 6}{h} < 0$$
That is, the surface slopes down in the x-direction at point P.For the second partial derivative with respect to x we have:
$$f_{xx}(P) \approx \frac{f_x(P+(h,0)) - f_x(P)}{h}$$
We see that to the left the level curves are closer together than to the right.
That means that at left the slope is steeper than at the right.
So we take a small negative slope (at the right) minus a bigger negative slope (at the left), and end up with a positive number.
Or put otherwise, we begin with a steep downward slope, and have to add something positive to it to get a less steep slope.
So:
$$f_{xx}(P) > 0$$

How far do you get with the other partial derivatives?
 
When I try to look at the derivate by y, I go up and down the y-axis, the level curves are constant there, am I wrong ?
 
Lancelot said:
When I try to look at the derivate by y, I go up and down the y-axis, the level curves are constant there, am I wrong ?

If we go up the y-axis starting from some arbitrary point, the level curve constants go up, don't they?
It means that $f_y > 0$.
That leaves the question how fast they go up.
Since the level curves are closer together for higher y, it means that the rate they go up accelerates.
Consequently $f_{yy} > 0$.
 

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