Homework Help: Length of A Curve

1. Aug 4, 2013

Justabeginner

1. The problem statement, all variables and given/known data
Find the length of the curve given by the equation:

2. Relevant equations
$y= \int_{-pi/2}^x √(cos t)\, dt$ for x between -∏/2 and ∏/2

3. The attempt at a solution
y= sqrt (cos x)
dy/dx= (sin x)/[-2 * sqrt(cos x)]

So now applying the arc length formula of sqrt (1 + (dy/dx)^(2)), I get:

$\int_{-pi/2}^{pi/2} sqrt(1 + {(sin^2(x))/(4 cos(x))}) \, dx$
$\int_{-pi/2}^{pi/2} sqrt(1 + ({sin x * tan x}/4))\, dx$

I don't know how to integrate that, and haven't learnt it either.. Any assistance is much appreciated. Thank you!

Last edited: Aug 4, 2013
2. Aug 4, 2013

voko

You have not specified the equation of the curve.

3. Aug 4, 2013

Justabeginner

This is all the information I have. Where I am stumped is why y= sqrt(cos t) dt with the limits of integration being 0 and 1/2.

4. Aug 4, 2013

voko

Is this $y = \int_{\pi/2}^x \sqrt {\cos t} dt$ the equation of the curve?

5. Aug 4, 2013

Justabeginner

Yes that is sir.

6. Aug 4, 2013

voko

So you have to compute $ds = \sqrt {1 + (\frac {dy}{dx})^2 } dx$, and for that you need to find $\frac {dy}{dx}$ first. What is it?

7. Aug 4, 2013

Justabeginner

I think dy/dx is (sin x)/[-2 * sqrt(cos x)] .

8. Aug 4, 2013

voko

$- \frac {\sin x} {2 \sqrt {\cos x} }$ is the derivative of $\sqrt {\cos x}$.

But you need to differentiate $y(x) = \int_{\pi/2}^x \sqrt {\cos t} dt$.

9. Aug 4, 2013

Justabeginner

That is what I do not understand unfortunately. Y is supposed to be in terms of x, yet it is given in terms of t, and I do not know how to figure out x in terms of t, if I want to substitute. Am I not getting something?

10. Aug 4, 2013

voko

y is not given in terms of t. The variable t appears only inside the integral. The integral depends on the variable x - and this is what y depends on.

11. Aug 4, 2013

Justabeginner

Yes, and this is where I am confused. Since the limits of integration are -pi/2 to x and x is between -pi/2 and pi/2, can you not say that the second limit is pi/2?

12. Aug 4, 2013

voko

Given a function f(x), where x is between a and b, can you not say that it is just f(b) - a constant?

13. Aug 4, 2013

Justabeginner

Yes- and I think of that as similar to the method of improper integrals- can I make this connection?

14. Aug 4, 2013

voko

Are you seriously saying that ANY function can be taken to be a constant just because its domain is defined?

15. Aug 4, 2013

Justabeginner

No. I was just presuming that for this question, because of the lower and upper limits of integration.

16. Aug 4, 2013

voko

The upper limit is a variable. What you have on the right is a function of this variable. You need to fin the derivative of that function with respect to this variable.

17. Aug 4, 2013

Justabeginner

That is exactly where I am stumped. How do I know the relationship between f(b) and sqrt(cos t) or just t for that matter?

18. Aug 4, 2013

voko

I am pretty sure you have studied the fundamental theorem of calculus.

19. Aug 4, 2013

Justabeginner

Yes, I have.

I think applying that logic, I come up with this:

∫sqrt(cos t) dt= -2/3 sin (t)^(3/2) with the limits of integration being -pi/2 and x.
-2/3 sin (0)^(3/2) - (-2/3)sin(x)^(3/2)
2/3 sin (x)^(3/2)

Then is y= 2/3 sin (x)^(3/2)?
And, the length of the curve is therefore,
dy/dx= sqrt(sin x)

and sqrt(1 + sin^2(x))= (1+ sin^2(x)^(1/2)

If I take the integral of that, I have: 2/3 * (1+ sin^2(x))^(3/2) * [0.5x - 0.5sinx*cosx]
which is equal to 2/3 * (1 + sin^2(x))^(3/2) * 1/2 [x- sinx*cosx]
= 1/3 (1+ sin^2(x))^(3/2) * [x- sinx*cosx]
Am I correct so far or have I already gone wrong?

20. Aug 4, 2013

voko

The fundamental theorem of calculus states that if $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. This is exactly what you have: you have $y = F(x) = \int_a^x \sqrt {\cos t} dt$, so $f(x) = \sqrt {\cos x}$, so $\frac {dy} {dx} = \sqrt {\cos x}$ follows automatically and effortlessly. It should have, anyway.

I am not sure how you got $\sqrt {\sin x}$, it is certainly not correct.

21. Aug 4, 2013

Justabeginner

I realize my mistake now. I did not think of the part of FTC that states F'(x)= f(x).
I think then that: ∫sqrt(1 + cos^2(x)) dx with -pi/2 and pi/2 being the lower and upper limits
∫(1 + [(1/2) + (cos(2x))/2])^(1/2)
2/3 (1 + {cos(2x)/2})^(3/2) * -sin(2x)/4 with limits being -pi/2 and pi/2.
Then plug in:
I get: 2/3 (0.5^1.5) * 0 - [2/3 (0.5^1.5) * 0] = 0
I know I've done something wrong..

22. Aug 4, 2013

voko

I do not understand how you got that.

23. Aug 4, 2013

Justabeginner

I thought I was supposed to use the arc length formula: ∫sqrt(1 + (dy/dx)^2)

24. Aug 4, 2013

voko

Correct, but what you wrote earlier does not follow from that formula.

25. Aug 4, 2013

LCKurtz

If $f'(x)=\sqrt {\cos x}$ what is $(f'(x))^2$?