Need Help Integrating This Tricky Integrand

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In summary, the length of the given curve can be found by taking the integral of the modulus of the derivative of r(t) from 0 to 1. This can be simplified by completing the square and using trigonometric substitution to solve for the integral. After factoring out scalars from the vectors, the solution becomes more manageable.
  • #1
royblaze
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Find the length of the curve:

r(t) = <2t, t2, (1/3)t3>
r'(t) = <2, 2t, t2>

From bounds of t: 0 to 1.

So length = integral of the modulus of r'(t):

Integral of sqrt(t4+4t2+4)

I'm just dead stuck on how to attack it. I tried to make it integral of sqrt((t2+2)2), and then just getting rid of the square, but I'm feeling intrinsically unsure that that way will work.

Would setting a u = t4 help at all?

Any help is appreciated!
 
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  • #2
royblaze said:
Find the length of the curve:

r(t) = <2t, t2, (1/3)t3>
r'(t) = <2, 2t, t2>

From bounds of t: 0 to 1.

So length = integral of the modulus of r'(t):

Integral of sqrt(t4+4t2+4)

I'm just dead stuck on how to attack it. I tried to make it integral of sqrt((t2+2)2), and then just getting rid of the square, but I'm feeling intrinsically unsure that that way will work.

Yeah, sure, try it that way!
 
  • #3
I tried it and got a numerical value. So, I guess it's alright then.

Is it too much to ask for help on a different one? It's in the same form.

r(t) = <12t, 8t3/2, 3t2>
r'(t) = <12, 12t1/2, 9t>

Bounds of t = 0 to 1

Integral of sqrt(122 +(12t1/2)2 + (9t)2)dt

= integral sqrt(144 + 144t +81t2)dt.

I'm really just baffled; these are problems 1-10, which are technically supposed to be pretty simple. Am I just missing a glaringly easy way to integrate these things?
 
  • #4
royblaze said:
I tried it and got a numerical value. So, I guess it's alright then.

Is it too much to ask for help on a different one? It's in the same form.

r(t) = <12t, 8t3/2, 3t2>
r'(t) = <12, 12t1/2, 9t>

Bounds of t = 0 to 1

Integral of sqrt(122 +(12t1/2)2 + (9t)2)dt

= integral sqrt(144 + 144t +81t2)dt.

I'm really just baffled; these are problems 1-10, which are technically supposed to be pretty simple. Am I just missing a glaringly easy way to integrate these things?

Try to complete the square here. Prepare for a trigonometric substitution.
 
  • #5
I'm having trouble after completing the square. I don't remember too much about trigonometric substitution... :(

I factored out the 81, to move a constant out:

9 * (integral of sqrt(t2 + (16/9)t +16/9))

So I took the square of half of 16/9:

9 * (integral of sqrt([t2 + (16/9)t + (64/81)] + 80/81))

9 * (integral of sqrt((t+(8/9))2 + 80/81)

EDIT: And just for questioning's sake; the bounds are from 0 to 1, so I'm guessing the answer itself shouldn't be hard to compute?
 
  • #6
Factor out 80/81 and set

[tex]\tan^2{u}=\frac{81}{80}(t+\frac{8}{9})^2[/tex]
 
  • #7
It might be the energy drink crash I'm getting here, but I don't see how the 80/81 can be factored out...

Sorry :/
 
  • #8
[tex]\sqrt{x-a}=\sqrt{a}\sqrt{\frac{x}{a}-1}[/tex]

Do something like this.
 
  • #9
So I get

a = 80/81

sqrt(a) * (sqrt((t + 8/9)2)/a + 1)

I can't see how the sqrt goes away. Was I supposed to change the sign in b/t the binomal squared term and the 1?
 
  • #10
Now make an appropriate trigonometric substitution. Substitute tan in.
 
  • #11
Do you mean in the sense of the derivative of arctan? (1/(1+x2))?

If I substitute tan in,

u = tan(t+(8/9))2,

Is this the correct substitution? It seems different from yours.
 
  • #12
Ah ha! I found it. By factoring out scalars from the vectors, I made my life a ton easier.

I found it out after consulting my textbook.

Thanks though :)
 

FAQ: Need Help Integrating This Tricky Integrand

What is an integrand?

An integrand is a mathematical function that is being integrated with respect to one or more variables. It is the expression that is being evaluated in an integral.

Why is integrating tricky?

Integrating can be tricky because it requires a good understanding of mathematical concepts and techniques, as well as a lot of practice. It also involves a lot of steps and can be time-consuming.

How do I know which method to use to integrate a tricky integrand?

There are several methods for integrating tricky integrands, such as substitution, integration by parts, and trigonometric substitution. The best method to use depends on the form of the integrand and may require trial and error or a combination of methods.

What are some common mistakes to avoid when integrating tricky integrands?

Some common mistakes to avoid when integrating tricky integrands include incorrect use of integration rules, forgetting to include the constant of integration, and not simplifying the integrand before integrating.

How can I improve my integration skills for tricky integrands?

The best way to improve your integration skills for tricky integrands is to practice regularly and familiarize yourself with different integration techniques. You can also seek help from a tutor or consult a textbook or online resources for additional practice problems and explanations.

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