Length of a Curve: Solve Homework Equation

[Justabeginner]

Homework Statement

Find the length of the curve given by the equation:

Homework Equations

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

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 learned it either.. Any assistance is much appreciated. Thank you!

 

[voko]

You have not specified the equation of the curve.

 

[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.

 

[voko]

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

 

[Justabeginner]

Yes that is sir.

 

[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?

 

[Justabeginner]

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

 

[voko]

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

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

 

[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?

 

[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.

 

[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?

 

[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?

 

[Justabeginner]

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

 

[voko]

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

 

[Justabeginner]

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

 

[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.

 

[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?

 

[voko]

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

 

[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?

 

[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.

 

[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..

 

[voko]

∫sqrt(1 + cos^2(x)) dx

I do not understand how you got that.

 

[Justabeginner]

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

 

[voko]

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

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

 

[LCKurtz]

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

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

 

[Justabeginner]

Then the derivative of f(x) squared would just be cos(x) right?

 

[voko]

@LCKurtz

$f(x)$ was used above to denote something particular. I would like to avoid making things more confusing than they already are here :) Thanks!

 

[LCKurtz]

Then the derivative of f(x) squared would just be cos(x) right?

So do you see what voko is getting at?

 

[Justabeginner]

Oh gosh I see what I messed up on.

sqrt(1 + cos(x)) with -pi/2 and pi/2 being the limits.
Then 2*sqrt (cos x + 1) * tan(x/2) with -pi/2 and pi/2 as the limits.
(2*-1) - (2.613)
-4.613
But that is negative. And a curve length cannot be negative?!

 

[voko]

Oh gosh I see what I messed up on.

sqrt(1 + cos(x)) with -pi/2 and pi/2 being the limits.
Then 2*sqrt (cos x + 1) * tan(x/2) with -pi/2 and pi/2 as the limits.

Good.

(2*-1) - (2.613)

Nope. What are the values of $\cos (\pi/2), \cos (-\pi/2), \tan (\pi/4), \tan (-\pi/4)$?

 

[Justabeginner]

cos(pi/2)= 0
cos(-pi/2)= 0
tan(pi/4)=1
tan(-pi/4)= -1 ?

 

[Justabeginner]

So (2) - (-2) = 4?

 

[voko]

So (2) - (-2) = 4?

Yep.