anonymous
  • anonymous
An item costs $500 at time t=0 and costs $P in year t. When inflammation is r% per year, the price is given by P = 500e^(rt/100) a) if r is a constant, at what rate is the price rising ( in dollars per year) 1) initially 2) after 2 years b)now suppose that r is increasing by 0.3 per year when r=4 and t=2. At what rate is the price increasing at that time? I can do part a) of this question but cant figure out the solution to part b).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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perl
  • perl
inflammation?
anonymous
  • anonymous
inflation not inflammation sorry
perl
  • perl
b) use product rule

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More answers

perl
  • perl
show me your work for part a) first
perl
  • perl
nice pic
anonymous
  • anonymous
dp/dt = 5re^(rt/100) so initially is 5r dollars/year and after 2 years 5re^(r(2)/100)
perl
  • perl
good
perl
  • perl
now suppose r is not constant
anonymous
  • anonymous
yea what do I do with 0.3
anonymous
  • anonymous
do I do the dP/dt again with r=0.3?
anonymous
  • anonymous
what?*
perl
  • perl
so I assumed that r is a function of time
perl
  • perl
|dw:1333530222146:dw|
perl
  • perl
where r ' (t) is the derivative of r(t)
anonymous
  • anonymous
so r'(t) = 0.3 right?
perl
  • perl
now it says r is increasing .3 per year, that means dr/dt = .3
perl
  • perl
yes
perl
  • perl
and plug in r = 4 and t = 2 , and you are done
anonymous
  • anonymous
i got dp/dt = 500e^(rt/100)*1/100(r+r'(t)t)
anonymous
  • anonymous
what am i doing wrong?
perl
  • perl
woops, i forgot 1/100
perl
  • perl
you did it right i think
perl
  • perl
dp/dt = 5e^(rt/100)(r+r'(t)t)
anonymous
  • anonymous
yeah thats what i have
anonymous
  • anonymous
thanks
anonymous
  • anonymous
so the answer is 24.92?
perl
  • perl
let me check
perl
  • perl
yes

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