Tutor.com Functions Session

Mar. 2, 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, 3/1/2012 8:58PM - Tutor.comSession Date: 3/1/2012 8:58PM
Length: 18.4 minute(s)
Subject: Math - Calculus

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

Guest (Customer)
[00:00:00] Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x= 1+2t^.5, y=t^3-t, z=t^3+t; (3,0,2)

Neal H (Tutor)
[00:00:13] Welcome to Tutor.com ! How can I help you today?

Guest (Customer)
[00:00:48] hi
[00:00:56] i'm not sure how to solve this problem

Neal H (Tutor)
[00:00:59] Hello
[00:01:14] That's ok. Have you started it yet, or are you stuck?

Guest (Customer)
[00:01:29] i thought about it, but don't really know how to start

Neal H (Tutor)
[00:01:46] Ok, let me take a look at the problem, just a moment.
[00:00:48] I'm working on this...

Guest (Customer)
[00:01:52] sure

Neal H (Tutor)
[00:02:18] Ok, I'll write the problem on the board.
[00:03:35] alright. These three equations define a curve in 3d space.
[00:03:56] Now, we need to find the tangent line to this curve at the point(3,0,2).
[00:04:12] and keep it in parametric form.
[00:04:53] Ok, now, when we see "tangent" we should immediately be thinking of taking a derivative.
[00:05:18] In this problem, we need to take the derivative of each x,y and z component.

Guest (Customer)
[00:05:34] ok

Neal H (Tutor)
[00:05:38] now, do you see what the derivative of the x component is?
[00:06:07] or are you unsure?
[00:06:41] That's right!
[00:07:29] Good so far.
[00:07:52] Good.
[00:08:09] so, these equations give us the slope in each direction as a function of t.
[00:08:41] so this helps us define the tangent line.
[00:09:05] but we need the "t" value in order to get the slope at the point (3,0,2)
[00:09:20] do you see how to find the t value at this point?

Guest (Customer)
[00:09:42] i'm not sure

Neal H (Tutor)
[00:09:48] that's ok.
[00:10:20] we need (x(t), y(t), z(t)) = (3,0,2)
[00:10:39] so we need to find the t value that makes that happen.

Guest (Customer)
[00:11:02] how is that done?

Neal H (Tutor)
[00:11:18] ok, let's take the first component.
[00:11:27] we need x(t) = 3
[00:11:41] so we need to solve...
[00:12:36] can you guess the t value?

Guest (Customer)
[00:12:57] 1?

Neal H (Tutor)
[00:13:02] Right!
[00:13:14] at t=1 the curve will pass through that point!
[00:13:51] ok, now we can use that t value to find the slope in all three directions using the derivative equations.
[00:14:28] right.
[00:14:47] that's right.
[00:15:02] ok, now we have everything we need to find the parametric equations.
[00:15:26] we have the point the tangent line goes through, (3,0,2) and the slopes.
[00:15:39] so first, let's find x(t) for the tangent line.

Guest (Customer)
[00:16:26] i got it!

Neal H (Tutor)
[00:16:32] Good!

Guest (Customer)
[00:17:10] can we do one more problme please?

Neal H (Tutor)
[00:17:28] Oh, I'm sorry, I have other students waiting now.
[00:17:37] But you can log back in for another problem.

Guest (Customer)
[00:17:42] that's ok, thanks so much for your help!

Neal H (Tutor)
[00:17:47] You
[00:17:52] 're welcome!
[00:17:58] It was nice working with you,
[00:18:06] please fill out the survey, and visit us again

Guest (Customer)
[00:18:04] thanks, same here

Neal H (Tutor)
[00:18:13] ok, bye.

Guest (Customer)
[00:18:14] bye