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Math Help - integral word problem

  1. #1
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    integral word problem

    A population of bacteria is changing at a rate of dp/dt = (3000/e^3+3t) , where t is the time in minutes. If the initial population is 13000, find the population when t=5 minutes.

    so i have to use substitution on this...i tried let u = 3t, but the problem didnt work out right on the quiz and he corrected it saying it should have been u = e^3 + 3t, which differentiates into du=3dt => (1/3)du=dt ? what happened to the e^3? doesnt e^3 = e^3 after you differentiate it?
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  2. #2
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    Quote Originally Posted by jeph View Post
    A population of bacteria is changing at a rate of dp/dt = (3000/e^3+3t) , where t is the time in minutes. If the initial population is 13000, find the population when t=5 minutes.

    so i have to use substitution on this...i tried let u = 3t, but the problem didnt work out right on the quiz and he corrected it saying it should have been u = e^3 + 3t, which differentiates into du=3dt => (1/3)du=dt ? what happened to the e^3? doesnt e^3 = e^3 after you differentiate it?
    That differencial equation makes no sense.
    I think you are missing something.
    Because you have,
    e^3 maybe you mean,
    e^{3t}
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  3. #3
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    thats what it says....i took a picture



    i just dont get how you make du/dt = e^3 + 3t into (1/3)du = dx. the e^3 just disappears...he even substituted the whole bottom line (e^3 + 3t) with u
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  4. #4
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    What the grader wrote in red is absolutely correct.
    Perhaps what you have misunderstood is that e^3 is a constant!
    The derivative of e^3 + 3t is just 3.
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  5. #5
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    oh...

    that constant thing keeps getting me. this wasnt the first time i made this mistake =T
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