Integration (Deduce)

• Wireless-

  23 Aug 10, 16:30

  The answer for part (2ii) is enclosed in the pink box.

  If we let y=e^t, shouldnt the numerator be 1 instead of e^t in order to deduce?Correct me if im wrong. So is the question given wrong?

  

   

• Wireless-

  23 Aug 10, 17:11

   

  

  This Qn Dunno how to do

  a=6;

  v= Integ(6)dt;

  v=6t +K;

  s=Integ(6t +K)dt

  s=3t^2 + Kt +C

  Reaches bottom of chute in 4s:

  80 = 3(4)^2 + K(4) +C

  32 = 4K +C

  Initial Velocity, v= 6(0) +K

                            v=K

   

  Therefore at 4s, velocity , v= 24 +K

  Then im stuck

• Mad Hat

  23 Aug 10, 17:34

  >> If we let y=e^t, shouldnt the numerator be 1 instead of e^t in order to deduce? Correct me if im wrong. So is the question given wrong?

  The question is okay. If you are integrating with respect to y, you must have "dy" at the end of the integral and not "dt". The purpose of the e^t in the numerator is to combine with the "dt" to give you "dy".

  Since y = e^t
  ⇒ dy/dt = e^t
  ⇒ dy = e^t dt

• Wireless-

  23 Aug 10, 18:09

  Oh Yes! Ok i get it. How to do the other question?

• Mad Hat

  23 Aug 10, 18:46

  >> Oh Yes! Ok i get it. How to do the other question?

  Use the fact that when t = 0, s = 0. This information is basically given because we can assume that, at the beginning, the ice is at the top of the chute.

• Wireless-

  23 Aug 10, 19:14

  Rofl ok thanks got it