Length of a Curve: Solve Homework Equation

In summary, the conversation involved finding the length of a curve using the arc length formula, but the student was stuck on understanding the relationship between the given equation and the variable t. Through the discussion, it was clarified that the equation was a function of the variable x, and using the fundamental theorem of calculus, the derivative of the function was found to be √(cos x). After some confusion, the correct integral was determined to be 2√(cos x + 1) * tan(x/2), and the length of the curve was calculated to be 4.
  • #1
Justabeginner
309
1

Homework Statement


Find the length of the curve given by the equation:

Homework Equations


[itex] y= \int_{-pi/2}^x √(cos t)\, dt [/itex] 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:

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

I don't know how to integrate that, and haven't learned it either.. Any assistance is much appreciated. Thank you!
 
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  • #2
You have not specified the equation of the curve.
 
  • #3
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
Is this ## y = \int_{\pi/2}^x \sqrt {\cos t} dt ## the equation of the curve?
 
  • #5
Yes that is sir.
 
  • #6
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
I think dy/dx is (sin x)/[-2 * sqrt(cos x)] .
 
  • #8
## - \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
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
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
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
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
Yes- and I think of that as similar to the method of improper integrals- can I make this connection?
 
  • #14
Are you seriously saying that ANY function can be taken to be a constant just because its domain is defined?
 
  • #15
No. I was just presuming that for this question, because of the lower and upper limits of integration.
 
  • #16
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
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
I am pretty sure you have studied the fundamental theorem of calculus.
 
  • #19
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
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
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
Justabeginner said:
∫sqrt(1 + cos^2(x)) dx

I do not understand how you got that.
 
  • #23
I thought I was supposed to use the arc length formula: ∫sqrt(1 + (dy/dx)^2)
 
  • #24
Justabeginner said:
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.
 
  • #25
Justabeginner said:
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##?
 
  • #26
Then the derivative of f(x) squared would just be cos(x) right?
 
  • #27
@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!
 
  • #28
Justabeginner said:
Then the derivative of f(x) squared would just be cos(x) right?

So do you see what voko is getting at?
 
  • #29
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?!
 
  • #30
Justabeginner said:
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) ##?
 
  • #31
cos(pi/2)= 0
cos(-pi/2)= 0
tan(pi/4)=1
tan(-pi/4)= -1 ?
 
  • #32
So (2) - (-2) = 4?
 
  • #33
Justabeginner said:
So (2) - (-2) = 4?

Yep.
 

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 represents the derivative of the curve function with respect to x.

2. How do I solve for the arc length of a curve?

To solve for the arc length of a curve, you will need to integrate the arc length formula with respect to x, using the limits of integration as the starting and ending points of the curve. This will give you the total length of the curve between those two points.

3. Can the arc length of a curve be negative?

No, the arc length of a curve cannot be negative. It represents a physical distance, and therefore, it is always a positive value.

4. What units are used to measure the arc length of a curve?

The units used to measure the arc length of a curve will depend on the units used for the curve function. For example, if the curve is measured in meters, then the arc length will also be measured in meters.

5. Is there a simpler method for finding the arc length of a curve?

Yes, there are some specific curves for which the arc length can be calculated using simpler methods, such as the Pythagorean Theorem or trigonometric functions. However, for most curves, the arc length formula is the most accurate method for finding the length.

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