Tutor.com The Definite Integral Session

Apr. 3, 2012

  • Poor
  • Fair
  • Average
  • Good
  • Excellent
  • Click to rate this Poor
  • Click to rate this Fair
  • Click to rate this Average
  • Click to rate this Good
  • Click to rate this Excellent


Session Transcript - Math - Calculus, 3/14/2012 4:48PM - Tutor.comSession Date: 3/14/2012 4:48PM
Length: 23.7 minute(s)
Subject: Math - Calculus


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] arc length/curvature [ File > http://lhh.tutor.com/SharedSessionFiles/bbf7f236-5b29-4125-8344-8f17ed4d17be_vector_calc.pdf ]

Kasia T (Tutor)
[00:00:05] Welcome to Tutor.com !

Guest (Customer)
[00:00:04] Hey!
[00:00:06] no need to open taht file

Kasia T (Tutor)
[00:00:08] My name is Kasia.

Guest (Customer)
[00:00:08] that file*
[00:00:14] not the right one

Kasia T (Tutor)
[00:00:20] Great, cause it's upside down for me

Guest (Customer)
[00:00:21] are you familiar with arc length and curvature?
[00:00:22] lol

Kasia T (Tutor)
[00:00:30] I am

Guest (Customer)
[00:00:33] great
[00:00:37] I was wondering if we could work through some together
[00:00:46] the first ... calculate the length of the curve over the given interval

Kasia T (Tutor)
[00:00:51] I'd be glad to
[00:00:53] Go ahead

Guest (Customer)
[00:00:58] r(t) = <2t, lnt, t^2> over [1.4]
[00:01:59] so it says .. the length is found by
[00:02:02] the magnitude of the derivatives
[00:02:06] from a to b

Kasia T (Tutor)
[00:02:16] Is this arc length from t=1 to t=4, or something else?

Guest (Customer)
[00:02:22] 1 to 4

Kasia T (Tutor)
[00:02:27] Got it

Guest (Customer)
[00:02:40] there's a hint that says "look for a perfect square
[00:02:48] let's just find the derivative first

Kasia T (Tutor)
[00:02:59] Okay

Guest (Customer)
[00:03:06] 2, 1/t, 2t

Kasia T (Tutor)
[00:03:08] Would you like to try working on the board?
[00:03:18] Good, that's the three derivatives
[00:03:20] Nice job

Guest (Customer)
[00:03:21] right
[00:03:21] so now
[00:03:25] perfect square...
[00:03:26] one sec
[00:03:30] so the mag is just
[00:03:31] sqrt of those
[00:03:32] squared

Kasia T (Tutor)
[00:04:54] Pretty much
[00:05:04] Have you plugged these derivatives into the arc length formula,
[00:05:13] or shall we do that together?

Guest (Customer)
[00:05:27] well all we have to do
[00:05:28] is ..
[00:05:31] take the integral of that ...
[00:05:33] from a to b
[00:05:37] doesn't look very nice right now though
[00:05:45] 2 + 1/t^2 + 2t^2
[00:05:47] sqrt

Kasia T (Tutor)
[00:05:53] That's very close.

Guest (Customer)
[00:05:53] 4 i mean

Kasia T (Tutor)
[00:05:59] Exactly

Guest (Customer)
[00:06:20] so it says look for a perfect square ...

Kasia T (Tutor)
[00:06:22] Both of the 2's get squared to 4s
[00:06:54] Yes, we can simplify this integral because the expression under the square root
[00:06:58] is a perfect square.

Guest (Customer)
[00:07:04] so everytihng to the one half power?
[00:07:06] and simplify?

Kasia T (Tutor)
[00:07:13] Not exactly.
[00:07:33] The square root of a sum is not equal to

Guest (Customer)
[00:07:36] kk

Kasia T (Tutor)
[00:07:39] the square root of each term.

Guest (Customer)
[00:07:45] right
[00:07:48] so .. hm

Kasia T (Tutor)
[00:07:50] We'll have to factor the sum under the square root

Guest (Customer)
[00:07:54] okay

Kasia T (Tutor)
[00:07:57] The fraction makes things complicated.
[00:08:07] How about we factor out a 1/t^2,

Guest (Customer)
[00:08:11] okay

Kasia T (Tutor)
[00:08:20] to make it easier to see what we're dealing with.

Guest (Customer)
[00:08:21] so we're left with 4t^2
[00:08:26] and

Kasia T (Tutor)
[00:08:31] Good, and then?

Guest (Customer)
[00:08:33] 4t^4

Kasia T (Tutor)
[00:08:48] That's right, and one more term.

Guest (Customer)
[00:08:55] + 1

Kasia T (Tutor)
[00:08:59] Perfect!

Guest (Customer)
[00:09:14] oh
[00:09:14] so now ...

Kasia T (Tutor)
[00:09:19] Yes ?

Guest (Customer)
[00:09:35] this is a perfect square

Kasia T (Tutor)
[00:09:41] Indeed.
[00:09:47] What is its square root?

Guest (Customer)
[00:09:53] ehh

Kasia T (Tutor)
[00:10:16] For a product,
[00:10:26] we can take individual square roots of each factor
[00:10:27] LIke this:

Guest (Customer)
[00:10:37] right

Kasia T (Tutor)
[00:10:49] Does that help?

Guest (Customer)
[00:10:58] ehh i'm trying to remember that perfect square stuff

Kasia T (Tutor)
[00:11:11] Let's start with the first root.
[00:11:16] What's the square root of 1/t^2 ?

Guest (Customer)
[00:11:31] 1/t

Kasia T (Tutor)
[00:11:36] You got it!
[00:12:00] The second root is a little easier
[00:12:07] if we write it from biggest exponent to smallest
[00:12:09] Like this:
[00:12:31] Do you see a way to factor it now?

Guest (Customer)
[00:12:42] something in the form ( ) ()
[00:12:44] lol

Kasia T (Tutor)
[00:13:01] Yes, that's just what we're looking for
[00:13:20] One more thing that might help:
[00:13:32] Notice that t^4 is the same thing as (t^2) ^2
[00:14:10] This expression is a quadratic, just instead of x we have t^2

Guest (Customer)
[00:14:28] okay ...
[00:14:32] for some resaon i'm not seeing it

Kasia T (Tutor)
[00:14:47] I'll temporarily replace t^2 with x :
[00:15:11] How would we factor this polynomial?

Guest (Customer)
[00:15:44] wowww something is not clicking today lol
[00:16:02] (2x+1)(2x+1)

Kasia T (Tutor)
[00:16:09] You got it!

Guest (Customer)
[00:16:09] 2x+1 squared
[00:16:11] lol...

Kasia T (Tutor)
[00:16:14] Good job!

Guest (Customer)
[00:16:16] wonder why that l took so llong
[00:16:17] lol

Kasia T (Tutor)
[00:16:31] That's alright, you got it and that's what counts
[00:16:45] Now we'll put back the t^2 instead of x, so it will be :

Guest (Customer)
[00:16:49] so here we have (2t^2+1)
[00:16:52] ...?

Kasia T (Tutor)
[00:17:07] Yes, that's it

Guest (Customer)
[00:17:09] ookay

Kasia T (Tutor)
[00:17:26] Now we can simplify the square root

Guest (Customer)
[00:17:31] right
[00:17:33] we can just cancel it
[00:17:36] the square

Kasia T (Tutor)
[00:17:39] Indeed
[00:18:11] Now the length integral looks a lot more doable.

Guest (Customer)
[00:18:14] 2t + 1/t

Kasia T (Tutor)
[00:18:19] Very good!

Guest (Customer)
[00:18:30] i think that's good enough then lol

Kasia T (Tutor)
[00:18:43] Would you like to finish this on your own ?

Guest (Customer)
[00:19:28] holy cow my brain is not working. i can't just split up the integrla right
[00:19:31] and distriubte it out

Kasia T (Tutor)
[00:19:45] No worries, let's do it together.
[00:20:05] For an integral of a sum, we just integrate each term separately.
[00:20:11] What's the antiderivative of 2t ?

Guest (Customer)
[00:20:11] right
[00:20:13] that's what i thought
[00:20:18] i can distriubte it out
[00:20:19] kk

Kasia T (Tutor)
[00:20:34] Alright, anything else I can help you with on this problem ?

Guest (Customer)
[00:20:44] k so we have
[00:20:48] t^2 and lnt
[00:20:50] integralso f each

Kasia T (Tutor)
[00:21:04] Yes, that's the right antiderivatives

Guest (Customer)
[00:21:09] okay so then
[00:21:13] we just take he net change..

Kasia T (Tutor)
[00:21:18] Very good
[00:21:22] What do you get for that?

Guest (Customer)
[00:21:52] 15 + [ln4-ln1]

Kasia T (Tutor)
[00:22:12] Just a minute, I'm checking

Guest (Customer)
[00:22:19] 13.61.
[00:22:20] ish
[00:22:40] oh wiat
[00:22:41] no
[00:22:48] 16.39

Kasia T (Tutor)
[00:23:00] That's right, 16.39
[00:23:09] The curve is about 16.4 units long
[00:23:20] Good job sticking with this problem!

Guest (Customer)
[00:23:19] okay great. thanks

Kasia T (Tutor)
[00:23:27] Do you have any questions about anything we've done?

Guest (Customer)
[00:23:27] mostly algebra lol
[00:23:31] nope, tanks!
[00:23:32] thanks
[00:23:35] see ya later!

Kasia T (Tutor)
[00:23:40] Thanks for using Tutor.com !