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# Calculus/Integration applications.

gabe wrote at 2008-12-01 23:43:50

pi(r^4) = kt + c

when t = 0, r = 1, therefore c = pi this is true

but

when t = 15, r = 2, c = pi

16pi = 15k + pi

15k = 15pi

k = pi

k does not equal one

starfish wrote at 2010-02-23 23:26:22
just gotta say whoever wrote the first answer and said

15pi = 15k

and then

k = 1

totally throwin me off.

which changes the answer that the second person got (80pi) to just plain 80

gallagherg wrote at 2011-01-13 01:37:44
the above answer is partially wrong.

15pi = 15k  (this is from Above)

k does not equal one

k equals pi

supahgeek wrote at 2011-05-02 02:17:36
actually, FYI, k=pi because 15(pi) divided by 15 equals pi, and 15k divided by 15 equals k, so k=pi.

Richard Evans wrote at 2012-03-06 19:11:17
I believe there is major mistake made in the solution of this problem.

In the steps solving for k the following was shown:

16 pi = 15k + pi

15 pi = 15k

k = 1

The answer for k should be pi not 1.

whit wrote at 2012-04-17 01:56:56
For part (a)  when you were solving for k, you had:

15 pi=15 k

but you had k=1. k actually equals pi. (k=pi)**

so the answer would end up being (for a):

pi r^4=pi t + pi

r^4=(pi t+pi)/pi

r^4=t+1

r=(t+1)^1/4

paragon wrote at 2014-02-07 02:34:10
I think you may have made a mistake. C/pi = C. Therefore in r^4= (kt +c)/pi simplifies to r^4= (kt/pi) + c. Plug in values of t=0 and r=1, 1= (k(0))/pi + c, c=1.

alias wrote at 2016-06-02 00:39:19
For part A, everything is correct up until

15 pi = 15k

k = 1

The correct answer is k = pi

Therefore the answer to part A is r = (t + 1)^1/4 after some simplifying

This will also change the answer of part B simply to t = 80

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