Solving Arc Length Integral with Trigonometric Substitution

[everestwitman]

∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4

Could someone help me solve this? I can't seem to find a substitution that works, or find the square root of (x^4/4 + 1/(x^4). Any help would be very appreciated.

Thanks in advance!

 

[Mark44]

∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4

Could someone help me solve this? I can't seem to find a substitution that works, or find the square root of (x^4/4 + 1/(x^4). Any help would be very appreciated.

Thanks in advance!

What's inside the radical is almost a perfect square, so it's possible that you have made an error in the work leading up to the integral. Please show the function whose arc length you're trying to find, and what you did to get your integral.

 

[everestwitman]

What's inside the radical is almost a perfect square, so it's possible that you have made an error in the work leading up to the integral. Please show the function whose arc length you're trying to find, and what you did to get your integral.

Here is the work that i have done so far:

1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4

L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4
L = ∫sqrt(1 + (x^2/2-1/(2 x^2))^2) dx from x = 1 to 4
L = ∫sqrt(1 + x^4/4 + 1/(x^4) - 1/2) dx from x = 1 to 4
L = ∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4

 

[Mark44]

Here is the work that i have done so far:

1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4

L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4

Your mistake is in the next line. Notice that you have a sign error and an error in the coefficient.
If y = (1/6)x³ + 1/(2x), y' = (1/2)x² - 1/(2x²)

L = ∫sqrt(1 + (x^2/2-1/(2 x^2))^2) dx from x = 1 to 4
L = ∫sqrt(1 + x^4/4 + 1/(x^4) - 1/2) dx from x = 1 to 4
L = ∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4

Also, I hope this isn't how you're doing your work. Instead of starting each line with L = ... and dragging the integral along at each step, it's much easier, and easier to read, if you write down what you need for the step at hand. In other words, find the derivative, square it, add 1, and take the square root. That will be your integrand. Each step from then on until you evaluate the integral will show the integral symbol, but you don't need it in the steps leading up to getting the integrand.

I see that you didn't use the homework template, which has a section for the problem statement. Without knowing what the original problem was, I was not able to tell where you went wrong. This is one reason why we ask that you include this information. Please include the problem statement in any future posts.

 

[everestwitman]

Your mistake is in the next line. Notice that you have a sign error and an error in the coefficient.
If y = (1/6)x³ + 1/(2x), y' = (1/2)x² - 1/(2x²)


Also, I hope this isn't how you're doing your work. Instead of starting each line with L = ... and dragging the integral along at each step, it's much easier, and easier to read, if you write down what you need for the step at hand. In other words, find the derivative, square it, add 1, and take the square root. That will be your integrand. Each step from then on until you evaluate the integral will show the integral symbol, but you don't need it in the steps leading up to getting the integrand.

I see that you didn't use the homework template, which has a section for the problem statement. Without knowing what the original problem was, I was not able to tell where you went wrong. This is one reason why we ask that you include this information. Please include the problem statement in any future posts.

Thanks a lot for your help, Mark44! I will be sure to in the future.

 

[everestwitman]

Now how would I go about factoring the expression that I get when I correct my mistakes? What I have now is

∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4

 

[Mark44]

Now how would I go about factoring the expression that I get when I correct my mistakes? What I have now is

∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4

You still have mistakes. It's possible you're skipping steps.

Starting from y = x³/6 + 1/(2x), show me what you get for the following:
1. y'
2. (y')²
3. 1 + (y')²

If you do the three steps above correctly, what you get is actually fairly easy to factor inside the radical. Only then should you worry about the radical and the integral.

 

[everestwitman]

Here is the work that i have done so far, corrected for the mistake that you pointed out earlier.


1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4

L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4

(d(x^3/6 + 1/(2x)/dx) = (x^3/6 + 1/(2x)
((x^3/6 + 1/(2x))^2 = x^6/36 + x^2/6 + 1/(4x^2)

Putting it together,

∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4 = ∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4

 

[Mark44]

Here is the work that i have done so far, corrected for the mistake that you pointed out earlier.1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4

L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4

(d(x^3/6 + 1/(2x)/dx) = (x^3/6 + 1/(2x)
((x^3/6 + 1/(2x))^2 = x^6/36 + x^2/6 + 1/(4x^2)

Putting it together,

∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4 = ∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4

No.

Please do as I suggested in my previous post.

Starting from y = x³/6 + 1/(2x), show me what you get for the following:
1. y'
2. (y')²
3. 1 + (y')²

If you do the three steps above correctly, what you get is actually fairly easy to factor inside the radical. Only then should you worry about the radical and the integral.

Your first line should NOT start with "L = ∫<something>".

 

[Mark44]

Your differentiation is incorrect here.

(d(x^3/6 + 1/(2x)/dx) = (x^3/6 + 1/(2x)

 

[Mark44]

Also, it's probably easier to read if you write the left side like so:
(d/dx)(x³/6 + 1/(2x))

 

[Alborz]

Hint: You can rewrite the stuff inside the square root as a three term polynomial over 4x^4.

 

[everestwitman]

How's this?

1. (d/dx)((x^3/6 + 1/(2x)) = x^2/2 - 1/(2x^2)
2. (dy/dx)^2 = (x^4 -1)^2/4x^4 = ((x - 1)^2 (x+1)^2 (x^2+1)^2) / 4x^2
3. (dy/dx)^2 + 1 = (x^4 + 1)^2 / (4x^4)
4. sqrt((dy/dx)^2 + 1) = (x^4+1) / 4x^2 = x^2 / 4 + 1 / (4x^2)
5.∫ x^2 / 4 + 1 / (4x^2) dx from x = 1 to 4 = x^3/12 - 1/ (4x) from x = 1 to 4 = (4^3/12 - 1/ (4(4))) - (1^3/12 - 1/ (4)) = 87 / 16

Thank you for being so patient with me. I'm 15 and have very little clue what I'm doing. I skipped Algebra II and Precalculus (testing out of them by the skin of my teeth.)

 

[Mark44]

How's this?

1. (d/dx)((x^3/6 + 1/(2x)) = x^2/2 - 1/(2x^2) OK[/color]

It's helpful to simplify the above by factoring out 1/2. This gives you (1/2)(x² - 1/x²), which will make the following step easier.

2. (dy/dx)^2 = (x^4 -1)^2/4x^4 = ((x - 1)^2 (x+1)^2 (x^2+1)^2) / 4x^2 Not OK[/color]

I can't tell what you did on the right side of the first =.
[(1/2)(x² - 1/x²)]² = (1/4)(x⁴ - 2 + 1/x⁴)
When you add 1 to the above (= 4/4), you get a perfect square trinomial, which is easy to factor.

3. (dy/dx)^2 + 1 = (x^4 + 1)^2 / (4x^4)
4. sqrt((dy/dx)^2 + 1) = (x^4+1) / 4x^2 = x^2 / 4 + 1 / (4x^2)
5.∫ x^2 / 4 + 1 / (4x^2) dx from x = 1 to 4 = x^3/12 - 1/ (4x) from x = 1 to 4 = (4^3/12 - 1/ (4(4))) - (1^3/12 - 1/ (4)) = 87 / 16

Thank you for being so patient with me. I'm 15 and have very little clue what I'm doing. I skipped Algebra II and Precalculus (testing out of them by the skin of my teeth.)

That was probably not a good idea. In any case, for the areas in which you're weak, it would be very helpful for you to review. You could pick up a precalculus textbook from Amazon or look at the algebra and precalc topic on khanacademy.org. Some time spent on a regular basis would be a very good investment.