anonymous
  • anonymous
derive this! dB/dt= k 100A/rW e^-kt - V
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
im pretty sure its -kte^-kt
anonymous
  • anonymous
how many variables are there? k,A,r,W and V are constants?
anonymous
  • anonymous
yeah they're constant, so i thought i could just eliminate

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dumbcow
  • dumbcow
is e^-kt in denominator ?
anonymous
  • anonymous
dB/dt is B'(t) so do you want B"(t)?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=What%20would%20you%20like%20to%20know%20about%3Fderive%20this!%20%20dB%2Fdt%3D%20k%20100A%2FrW%20e^-kt%20-%20V
dumbcow
  • dumbcow
-k^2(100A/rW)e^-kt
anonymous
  • anonymous
if k,A,r,W and V are constants and if you are looking for B"(t) then, k100A/rW still stay where is, e^-kt would be -k e^-kt, (t is a variable), and V will just disappear. so your answer would be \[d ^{2}B/d t^{2}=-(k ^{2}100/rW) e ^{-kt}\]
anonymous
  • anonymous
I forgot the A after the 100 of course, haha... more calculus and DE problems here would be nice...
anonymous
  • anonymous
ill put some more up then, thanks guys
anonymous
  • anonymous
hm, maybe tomorrrowws

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