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Math 116 Team Homework 2 Cover Sheet
Due Date: 21 Sept
Our first meeting was Tuesday after class in the first floor computer lab of the
Bob and Betty Beyster building. We were there from 9:30 untill0:30. In this time, we
mainly talked about problem one. We struggled with part 1 because we did not
understand how to compare the two integrals with you replaced t with -t. However,
after visiting the Math Lab, we were able to come to an understanding that the
function was
W all contributed to the answering of the problem in different
calculate the derivative of a function and I
ways. On one
would check that
results
the graph. On a different part, I would work out
integral on the board and get feedback from the group that I was doing it right.
took meticulous notes and copied down all of our findings along with
contributing when needed. We breezed through the rest of problem 1 after we figured
out part 1.
Our next meeting was brief on Thursday in the Bursley lobby just to relay the
information from Math lab to the rest of the team. After a general consensus, we
adjourned the meeting.
We then met again at the usual time of Thursday at 4:30 at the Coman
Computer Center at Baits. We went to the meeting having already tried problems 2, 3,
and 4. We discussed our results and realized that we got the same answers for 2 and 3.
We spent some time on problem 4, but eventually were able to solve the problem. We
looked over our final draft and decided that we were confident with all of our answers
and concluded the meeting.
This week's batch of problems was noticeably tougher than last week's. We
recognized that we may need to set aside more time to be able to fmish the team
homework for the weeks to come. Also, we determined that the Math Lab is a great
resource when struggling with a part of a problem.
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Problem 4: Section 7.1 #112
Assuming that wind resistance is proportional to velocity, the downward velocity v, of a vertically falling
kt
mass m is v =~9 (1 -
e -in) in which g =acceleration due to gravity, k =some constant. We are asked to
find h, a falling objects height above the earth's surface as a function of time, assuming the object sta rts
at ho.
We integrate the velocity function from ~0 to timet, and then subtract that from the original h~r, hO, to get the
height as a function oft.
'"-
kt) dt
h=ho-J0tmg
T ( 1-e-m
~ '. ·""-'
.,, [ \ <;(,
~(~
and
rtmg
Jo k
'.
1
-
'
'M..'
\>-~.
\
~
kt)
kt) dt= mg rt dt + mg rt e-ktfmdt = mg (t +_e--)=
-ktfm
rt (1- e-m
mg ((t
(1- e-m dt = mg
k Jo
k Jo
k Jo
k
k/m
k
+(~)e-ktfm) for [O,t])
mg
=(t
k
kt)
m
+ (m
-)e-iit
-k
k
Plug back into h:
h = hO- ~g t - ( 9:;
2
kt)
2
e -in + 9:; to get the height of a falling object as a function of time.