# Homework Help: Curve Length Question

1. Jan 21, 2008

### apoptosis

Help in Curve Length Question :D

1. The problem statement, all variables and given/known data
Find the length of the curve where x=cos(t) and y=t+sin(t) where 0$$\leq$$t$$\leq$$$$\pi$$

2. Relevant equations
$$\sqrt{(dx/dt)^2+(dy/dt)^2}$$

x=cos(t)
dx/dt=-sin(t)

y=t+sin(t)
dy/dt=1+cos(t)

3. The attempt at a solution
$$\sqrt{(-sin(t))^2+(1+cost(t))^2}$$
$$\sqrt{(sint)^2+1+2cost+(cos(t))^2}$$
$$\sqrt{2+2cos(t)}$$dt

where I am stuck is how to do substitution to find the length of the curve.

I found the antiderivative:
(-1/3sint)(2+2cost)^(3/2) but when I substitute the limits pi and zero I get a length of zero
any pointers to get me in the right direction? thanks

Last edited: Jan 21, 2008
2. Jan 21, 2008

### rock.freak667

I do not not think you integrated correctly. Show how you got that please.

3. Jan 21, 2008

### apoptosis

so, I substituted my two derivatives into the length equation, multiplied out my factors, and added sin(t)^2 with cos(t)^2 to get 1.
Thus, sqrt of 1+1+2cos(t)

is there a certain part of the integration that you see an error in?

Last edited: Jan 21, 2008
4. Jan 21, 2008

### rock.freak667

$$\int \sqrt{2+2cos(t)} dt$$

$$=\sqrt{2} \int \sqrt{1+cost} dt$$

$$=\sqrt{2} \int \sqrt{\frac{1-cos^2t}{1-cost} dt$$

$$=\sqrt{2} \int \sqrt{\frac{sin^2t}{1-cost} dt$$

$$=\sqrt{2} \int \frac{sint}{\sqrt{1-cost}} dt$$

Did you get that?

Let u=1-cost => du=sint dt

$$=\sqrt{2} \int u^{\frac{-1}{2}} du$$

$$= 2\sqrt{2}\sqrt{1-cost}$$

5. Jan 21, 2008

### apoptosis

Oh, I see! It's matter of taking a further step and factoring out the $$\sqrt{2}$$ from my initial integral.
One question, how did you go from 1+cost to (1-cost^2)/(1-cost)?

Thanks rockfreak. I'll try the question using that.

Last edited: Jan 21, 2008
6. Jan 21, 2008

### Mystic998

(a - b)*(a + b) = (a^2 - b^2). Always a good one to be able to quote off the top of your head.

7. Jan 21, 2008

### rock.freak667

yeah it's basically that.

Multiply the numerator and denominator by 1-cost

8. Jan 21, 2008

### dynamicsolo

I might mention that this is one of those parametric curves where it is important to investigate its behavior before evaluating definite integrals to find arclengths. The integral from 0 to pi for this problem will give a non-zero value for the integral function rock.freak667 shows. However, you would be in trouble if you were asked to go from 0 to 2(pi) and naively used that function.

I believe this curve is a cycloid; you would have to use a symmetry argument to take the arclength you find in going from 0 to pi and then doubling that result. This is a frequent potential pitfall in evaluating arclengths and enclosed areas for periodic curves (polar curves such as the rosettes often present such difficulties).

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