Implicit differentiation of one-parameter family

Click For Summary
SUMMARY

The discussion focuses on using implicit differentiation to demonstrate that the one-parameter family defined by the equation f(x, y) = c satisfies the differential equation dy/dx = -f_{x}/f_{y}. The key steps involve applying the chain rule for total derivatives, leading to the equation ∂f/∂x + ∂f/∂y (dy/dx) = 0. By isolating dy/dx, the solution confirms that dy/dx equals -f_{x}/f_{y}, where f_{x} and f_{y} represent the partial derivatives of f with respect to x and y, respectively.

PREREQUISITES
  • Understanding of implicit differentiation
  • Familiarity with partial derivatives
  • Knowledge of the chain rule in calculus
  • Basic concepts of differential equations
NEXT STEPS
  • Study the application of the chain rule in multivariable calculus
  • Learn about partial derivatives and their significance in implicit functions
  • Explore examples of differential equations derived from implicit functions
  • Investigate the geometric interpretation of implicit differentiation
USEFUL FOR

Students and educators in calculus, particularly those studying implicit differentiation and differential equations, will benefit from this discussion.

bleucat
Messages
4
Reaction score
0

Homework Statement



Use implicit differentiation to show that the one parameter family f(x, y)=c satisfies the differential equation dy/dx = -f_{x}/f_{y}, where f_{x}=\frac{\partial f}{\partial x} and f_{y}=\frac{\partial f}{\partial y}.

Homework Equations

The Attempt at a Solution



Well, my teacher said I need to use the chain rule, but I'm confused about how to differentiate something that is in the general form f(x, y). And if f(x, y)=c, doesn't the derivative trivially equal 0?

Thanks in advance for the help.
 
Physics news on Phys.org
From the chain rule for the total derivative with respect to x:
\frac{\partial f}{\partial x}\frac{dx}{dx} +\frac{\partial f}{\partial y}\frac{dy}{dx} = 0

\frac{\partial f}{\partial x} +\frac{\partial f}{\partial y}\frac{dy}{dx} = 0

Solving for dy/dx gives -fx/fy.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K