Chain Rule

May 2010
25
0
I have this question and I dont know where to start.

The average commission c (in dollars) per trade by an investment corporation decreased as more of its customers traded online, according to the formula

\(\displaystyle c(u) = 100u^2 - 160u + 110\) [units are dollars per trade] [u = fraction of trades done online]

During that time, the fraction of online trades increased according to

\(\displaystyle u(t) = 0.42 + 0.02t\) [t = months since January 1, 1998]

Use direct substitution to express the average commission per trade c as a function of time t (do not just simplify the expression) and then use the chain rule to derive the formula for estimate the rate of change of average commission per trade. Use this formula to estimate the rate of change of average commission per trade at the beginning of October, 1998.
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
Find \(\displaystyle c'(u)\) and \(\displaystyle u'(t)\)


Note that,

\(\displaystyle
c(u) = 100u^2 - 160u + 110 \implies c(t) = 100(0.42 + 0.02t)^2 - 160(0.42 + 0.02t) + 110
\)

\(\displaystyle c'(t) =c'(u) \times u'(t)\)
 
May 2010
25
0
ok so

\(\displaystyle c'(t) = (.02)(200u-160)\)
then I would use
\(\displaystyle c(t) = 100(0.42 + 0.02t)^2 - 160(0.42 + 0.02t) + 110 \)
to find the rate of change of average comission per trade that would be at the beginning of October, 1998 which would be


\(\displaystyle c(10) = 100(0.42 + 0.02(10))^2 - 160(0.42 + 0.02(10)) + 110\)

does this sound right?
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
\(\displaystyle c(u) = 100u^2 - 160u + 110 \implies c(t) \) \(\displaystyle = 100(0.42 + 0.02t)^2 - 160(0.42 + 0.02t) + 110 \implies c'(t)\) \(\displaystyle = 100\times 2\times 0.02 \times (0.42 + 0.02t) - 160\times 0.02
\)
 
May 2010
25
0
Ok so I think I finally have it.

\(\displaystyle c(u) = 100u^2-160u+110\)

and

\(\displaystyle u(t) = 0.42+0.02t\)

this would give you

\(\displaystyle c(t) = 100(0.42 + 0.02t)^2 - 160(0.42 + 0.02t) + 110\)
then \(\displaystyle c'(t) = 4(.42 + .02t)-3.2\)

my question is to find the rate of change of average commission per trade at the beginning of october, 1998 would you use c(9) or c(10) if t=months since January,1 1998?
 
Jan 2010
354
173
my question is to find the rate of change of average commission per trade at the beginning of october, 1998 would you use c(9) or c(10) if t=months since January,1 1998?
October is the 10th month of the year, and January is the 1st month. So, October is 9 months after January.
 
May 2010
25
0
So the final answer is this?

\(\displaystyle c(9)=100(0.42+0.02(9)^2-160(0.42+0.02(9))+110 = -60\)

\(\displaystyle c'(-60)=4(0.42+0.02(-60)-3.2 = - 15.92\)

So the estimated rate of change of average commission per trade at the beginning of October, 1998 is = -15.92

This seems right but I have the funny feeling it isnt.
 
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