Finding Points on a Curve with Tangent Line Slope -1

Click For Summary
SUMMARY

The discussion focuses on finding points on the curve defined by the equation x²y² + xy = 2, where the slope of the tangent line is -1. The derivative was calculated as dy/dx = (-2xy² - y) / (2x²y + x). Participants clarified that to find the points where the slope equals -1, one must solve dy/dx = -1, leading to a simplified derivative when the factor (2xy + 1) is not zero. The conclusion emphasizes that multiple solutions exist due to the presence of two variables.

PREREQUISITES
  • Understanding of implicit differentiation
  • Familiarity with solving equations involving two variables
  • Knowledge of calculus concepts such as derivatives and slopes
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Practice implicit differentiation with different equations
  • Explore the concept of tangent lines in calculus
  • Learn about solving systems of equations with two variables
  • Investigate the implications of multiple solutions in calculus problems
USEFUL FOR

Students studying calculus, particularly those focusing on implicit differentiation and tangent line analysis, as well as educators seeking to enhance their teaching methods in these topics.

temaire
Messages
275
Reaction score
0

Homework Statement


Find all the points on the curve [tex]x^{2}y^{2}+xy=2[/tex] where the slope of the tangent line is -1.



The Attempt at a Solution


I differentiated both sides of the equation and got:
[tex]\frac{dy}{dx}=\frac{-2xy^{2}-y}{x^{2}2y+x}[/tex]

I know that [tex]\frac{dy}{dx}=-1[/tex], but if I substitute -1 in, I won't be able to go any further since I have two unknown variables. I would appreciate any help.
 
Last edited:
Physics news on Phys.org
[tex]\frac{dy}{dx}=\frac{-2xy^{2}-y}{2x^{2}y+x}~=~\frac{-y(2xy + 1)}{x(2xy + 1)}[/tex]

As long as 2xy + 1 [itex]\neq[/itex] 0, you can cancel the factors of 2xy + 1, leaving a much simpler derivative.

Also, you want to solve the equation dy/dx = -1, not dy/dx = 1, as you had. Notice that you still have two variables, but all that means is that there are lots of solutions.
 
Thanks, I got it.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K