# Length of Curve Homework: Find Integral Solution

• WendysRules
In summary: I did not examine all your steps in detail; I just used Maple to do the computations, and arrived at the same final expression for ##ds## that you did.

## Homework Statement

Find the length of the curve:
##x=\frac{t}{1+t}##
##y=\ln(1+t)##
where ##0 \leq t \leq 2##
Length of curve integral ##\int^a_b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt##

## Homework Equations

##(\frac{dx}{dt})^2 = \frac{1}{(1+t)^4}## ##(\frac{dy}{dt})^2 = \frac{1}{(1+t)^2} ##

## The Attempt at a Solution

So, I think somewhere I missed a more efficient technique to use (or made an algebra mistake which is most likely), because otherwise I think this integral gets crazy too fast for a simple problem.

Start with ##\int^a_b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt## to which when we substitute our given problem, we get ##\int^2_0 \sqrt{\frac{1}{(1+t)^4}+\frac{1}{(1+t)^2}} dt## = ##\int^2_0 \sqrt{\frac{(1+t)^2+1}{(1+t)^4}} dt ## = ##\int^2_0 \frac{1}{(1+t)^2} \sqrt{(1+t)^2+1} dt##

To try to make this simpler, I said well, just let ##u = 1+t## therefore ##du=dt## so our integral looks like
##\int^2_0 \frac{1}{(u)^2} \sqrt{(u)^2+1} du## which then I believe calls for a trig substitution (let ##u=tan\theta##)... but that seems crazy! At this point why wouldn't I just de-parametrize the equation from the start if I wanted a crazy integral.

If anyone has a simpler way to go about this integral, I'd love to know. Otherwise if this seems to be the correct path, I'll just keep chugging along!

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WendysRules said:

## Homework Statement

Find the length of the curve:
##x=\frac{t}{1+t}## ##y=\ln(1+t)## where ##0 \leq t \leq 2##
Length of curve integral ##\int^a_b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt##

## Homework Equations

##(\frac{dx}{dt})^2 = \frac{1}{(1+t)^4}## ##(\frac{dy}{dt})^2 = \frac{1}{(1+t)^2} ##

## The Attempt at a Solution

So, I think somewhere I missed a more efficient technique to use (or made an algebra mistake which is most likely), because otherwise I think this integral gets crazy too fast for a simple problem.

Start with ##\int^a_b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt## to which when we substitute our given problem, we get ##\int^2_0 \sqrt{\frac{1}{(1+t)^4}+\frac{1}{(1+t)^2}} dt## = ##\int^2_0 \sqrt{\frac{(1+t)^2+1}{(1+t)^4}} dt ## = ##\int^2_0 \frac{1}{(1+t)^2} \sqrt{(1+t)^2+1} dt##

To try to make this simpler, I said well, just let ##u = 1+t## therefore ##du=dt## so our integral looks like
##\int^2_0 \frac{1}{(u)^2} \sqrt{(u)^2+1} du## which then I believe calls for a trig substitution (let ##u=tan\theta##)... but that seems crazy! At this point why wouldn't I just de-parametrize the equation from the start if I wanted a crazy integral.

If anyone has a simpler way to go about this integral, I'd love to know. Otherwise if this seems to be the correct path, I'll just keep chugging along!

What you have done so far looks correct. Yes, the integral is messy, but that's life: some problems just have messy solutions.

WendysRules
I don't agree with what you got for ## \frac{dy}{dt} ##. How did you get rid of the ## \ln(1+t) ## term that will show up in taking ## \frac{dy}{dt} ##?

I don't agree with what you got for ## \frac{dy}{dt} ##. How did you get rid of the ## \ln(1+t) ## term that will show up in taking ## \frac{dy}{dt} ##?

Maybe my logic doesn't apply here, but I say let ##1+t=u## therefore ##\ln(1+t)=\ln(u)## then ##\frac{dy}{dt}=\frac{\frac{du}{u}}{dt}## = ##\frac{du}{udt} ##but since ##dt=du## in this case, our problem just becomes ##\frac{1}{u}## which just equals ##\frac{1}{1+t}##

WendysRules said:
Maybe my logic doesn't apply here, but I say let ##1+t=u## therefore ##\ln(1+t)=\ln(u)## then ##\frac{dy}{dt}=\frac{\frac{du}{u}}{dt}## = ##\frac{du}{udt} ##but since ##dt=du## in this case, our problem just becomes ##\frac{1}{u}## which just equals ##\frac{1}{1+t}##
I see my problem: You have no space between ## x=\frac{t}{t+1} ## and ## \\ ## ## y=\ln(1+t) ##,## \\ ## and I took it to mean ## x=(\frac{t}{t+1})y ##

WendysRules
I see my problem: You have no space between ## x=\frac{t}{t+1} ## and ## \\ ## ## y=\ln(1+t) ##,## \\ ## and I took it to mean ## x=(\frac{t}{t+1})y ##

I'll edit my post to make it more clear

I think I have a solution: Try ## u=\sinh(x) ##. If my algbra/calculus is correct, you get ## \int \coth^2(x) \, dx ##. See https://en.wikipedia.org/wiki/List_of_integrals_of_hyperbolic_functions for the integral. ## \\ ## And in a more clear form, I found it in an old CRC handbook: ## \int \coth^2(x ) \, dx=x-\coth(x) ##. ## \\ ## And presuming this is correct, it gets a little messy working backwards to ## u ##, etc., but it is workable.

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WendysRules said:
Maybe my logic doesn't apply here, but I say let ##1+t=u## therefore ##\ln(1+t)=\ln(u)## then ##\frac{dy}{dt}=\frac{\frac{du}{u}}{dt}## = ##\frac{du}{udt} ##but since ##dt=du## in this case, our problem just becomes ##\frac{1}{u}## which just equals ##\frac{1}{1+t}##

I did not examine all your steps in detail; I just used Maple to do the computations, and arrived at the same final expression for ##ds## that you gave.

I think I have a solution: Try ## u=\sinh(x) ##. If my algbra/calculus is correct, you get ## \int \coth^2(x) \, dx ##. See https://en.wikipedia.org/wiki/List_of_integrals_of_hyperbolic_functions for the integral. ## \\ ## And in a more clear form, I found it in an old CRC handbook: ## \int \coth^2(x ) \, dx=x-\coth(x) ##. ## \\ ## And presuming this is correct, it gets a little messy working backwards to ## u ##, etc., but it is workable.

I think PF prefers that we leave the solution for the OP to find, and only jump in with hints as needed. It sounds as though she did not need more help, just more encouragement to carry on despite the complexities.

Ray Vickson said:
I think PF prefers that we leave the solution for the OP to find, and only jump in with hints as needed. It sounds as though she did not need more help, just more encouragement to carry on despite the complexities.
I tried the ## u=\tan(\theta) ## also, which seemed to be a good choice. It led to what appeared to be something that would be very difficult to integrate. ## \\ ## And my apologies for perhaps supplying too much to the solution. This one I found very challenging myself=I was just happy that I found something that might work...

## 1. What is the purpose of finding an integral solution for the length of a curve?

The purpose of finding an integral solution for the length of a curve is to accurately measure the distance along a curved line. This is important in various fields such as mathematics, physics, engineering, and architecture, where precise measurements are needed for calculations and designs.

## 2. How is the length of a curve typically calculated?

The length of a curve can be calculated using integration. This involves breaking the curve into small segments and finding the length of each segment using the Pythagorean theorem. The lengths of these segments are then added together to find the total length of the curve.

## 3. What is the difference between a smooth curve and a piecewise curve?

A smooth curve is a continuous curve with no sharp corners or breaks, while a piecewise curve is made up of multiple segments with different equations. The length of a smooth curve can be found using a single integral, while the length of a piecewise curve requires multiple integrals to find the length of each segment.

## 4. Why is it important to use an integral solution for finding the length of a curve?

An integral solution is important because it takes into account the small variations in the curve and provides a more accurate measurement. Using simpler methods, such as measuring with a ruler, can lead to significant errors in the calculated length of the curve.

## 5. Can the length of a curve be negative?

No, the length of a curve cannot be negative. It is a physical quantity that represents a distance and therefore must be a positive value. If a negative value is obtained, it is an indication that an error has been made in the calculation or integration process.

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