Finding Length of a Curve: y2 = (x-1)3

In summary: OK, maybe it's been a while, since I've done some mathematics :DIn summary, at the point (1,0), y=1 and x=0.
  • #1
EEristavi
108
5

Homework Statement



I have to find length of the curve: y2 = (x-1)3 from (1,0) to (2,1)

Homework Equations



s = ∫ √(1 + (f '(x) )2 ) dx where we have integral from a to b

The Attempt at a Solution



I'm bit confused:
I'm thinking of writing function regarding x, f(x). However, I can also write for y, f(y).
Which is better? and what values should I take for definite integral ("limits")
 
Physics news on Phys.org
  • #2
EEristavi said:

Homework Statement



I have to find length of the curve: y2 = (x-1)3 from (1,0) to (2,1)

Homework Equations



s = ∫ √(1 + (f '(x) )2 ) dx where we have integral from a to b

The Attempt at a Solution



I'm bit confused:
I'm thinking of writing function regarding x, f(x). However, I can also write for y, f(y).
Which is better? and what values should I take for definite integral ("limits")

Just try it in the form ##y=f(x)##. And you are given a starting point and an ending point. What does that tell you about limits on ##x##?
 
  • #3
Dick said:
What does that tell you about limits on xxx?

well, from the problem statement x should change from 0 to 1, but its not correct (as I see from the solution), integral is taken from 1 to 2 (and I can't figure it out why)
 
  • #4
EEristavi said:
well, from the problem statement x should change from 0 to 1, but its not correct (as I see from the solution), integral is taken from 1 to 2 (and I can't figure it out why)

You do know that, e.g., at the point ##(1,0)## the value of ##x## is 1, right?
 
  • #5
Dick said:
You do know that, e.g., at the point (1,0)(1,0)(1,0) the value of xxx is 1, right?

OK, now I think, I need more clarification... I will write what is in my mind:

point (1, 0) means that y=1, x=0
but I see from function y = f (x), that if x = 0 => y = 0

So I guess, here is my problem of understanding...
 
  • #6
EEristavi said:
OK, now I think, I need more clarification... I will write what is in my mind:

point (1, 0) means that y=1, x=0
but I see from function y = f (x), that if x = 0 => y = 0

So I guess, here is my problem of understanding...

Where I come from the point ##(1,0)## means ##x=1## and ##y=0##. That looks like the definition the problem is using also. I don't think I've ever heard of a different convention.
 
  • Like
Likes CWatters and EEristavi
  • #7
Dick said:
Where I come from the point ##(1,0)## means ##x=1## and ##y=0##. That looks like the definition the problem is using also. I don't think I've ever heard of a different convention.
Ok, maybe it's been a while, since I've done some mathematics :D

Thank you :)
 

Related to Finding Length of a Curve: y2 = (x-1)3

1. What is the formula for finding the length of a curve?

The formula for finding the length of a curve is known as the arc length formula, which is given by:
L = ∫√(1 + (dy/dx)^2) dx, where dy/dx is the derivative of the curve's equation.

2. How do you find the derivative of a curve's equation?

To find the derivative of a curve's equation, you can use the power rule, chain rule, or product rule depending on the complexity of the equation. In this case, the derivative of y2 = (x-1)3 is 3(x-1)2.

3. Can you find the length of a curve without calculus?

No, the arc length formula involves integration, which is a fundamental concept in calculus. Therefore, it is not possible to find the length of a curve without using calculus.

4. What is the significance of the equation y2 = (x-1)3?

This equation represents a cubic curve that is shifted one unit to the right. It can be used to model various real-life situations, such as the growth of a population or the trajectory of a projectile.

5. Are there any limitations to using the arc length formula?

Yes, the arc length formula can only be used to find the length of a curve with a continuous and differentiable equation. It also requires the use of calculus, which may not be suitable for all situations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
389
  • Calculus and Beyond Homework Help
Replies
1
Views
758
  • Calculus and Beyond Homework Help
Replies
13
Views
368
  • Calculus and Beyond Homework Help
Replies
6
Views
589
  • Calculus and Beyond Homework Help
Replies
8
Views
528
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
531
  • Calculus and Beyond Homework Help
Replies
14
Views
361
  • Calculus and Beyond Homework Help
Replies
1
Views
860
  • Calculus and Beyond Homework Help
Replies
9
Views
598
Back
Top