Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

Did I do this function rate of change question Calculus  ?

Suppose the function 

f(t)= 15t/1+0.25t^2

models a drug in milligrams entering the blood stream at time t in seconds from 0 to 25 seconds. 

What is the rate the drug is entering the body at 7 seconds?

I plugged in 7 for t and got  7.9245

How would I find the amount of drug in milligrams that is in the body at 7 seconds?

3 Answers

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  • 4 weeks ago

    You don't even need to use calculus. Just do a spreadsheet

    that automatically plugs in the values for each successive t

    and f(t) is the amount of the drug in the bloodstream in milligrams.

    f(t) = 15t / (1 + 0.25t²)

    Time (t) Drug f(t)

    seconds milligrams

    0 0.0000

    1 12.0000

    2 15.0000

    3 13.8462

    4 12.0000

    5 10.3448

    6 9.0000

    7 7.9245

    8 7.0588

    9 6.3529

    10 5.7692

    11 5.2800

    12 4.8649

    13 4.5087

    14 4.2000

    15 3.9301

    16 3.6923

    17 3.4812

    18 3.2927

    19 3.1233

    20 2.9703

    21 2.8315

    22 2.7049

    23 2.5891

    24 2.4828

    25 2.3847

  • 4 weeks ago

    I suggest before doing calculus, you focus on distinguishing between 15t/(1+0.25t^2) and 15t/1+0.25t^2.

    Nothing here makes sense. "The function models a drug"? What about the drug is it modeling?

    If the statement were

    The function models the rate, in mg/s, at which a drug is entering the system at time t in seconds, between 0 and 25 s.

    Then the answer to the first question is indeed about 7.92 mg/s.

    To find the amount of drug that has entered the system from 0 to 7 s, we would integrate f(t) between t=0 and t=7.

    77.52 mg

  • Ash
    Lv 7
    4 weeks ago

    f(t)= 15t/(1+0.25t²)

    Amount of drug at t seconds would be the area under the curve from t=0 to t=7

    ₀∫⁷f(t) dt =₀∫⁷ 15t/(1+0.25t²) dt

    Let u = (1+0.25t²)

    du = 0.5t dt

    30 du = 15t dt

    I will not change the limits to u as we will change back to t and then solve. However, you can alternatively change the limit to u and get the same answer.

    ∫ (30/u) du

    = 30∫ (1/u) du

    = 30 ln u + C

    Now plug back t

    = [30 ln(1+0.25t²)] ₀|⁷

    = 30 [ln(1 + 0.25(7²) - ln (1 + 0.25(0²))

    = 30[ln(13.25) - ln(1)]

    = 30[ln(13.25) - 0]

    = 30[ln(13.25)]

    = 77.5 mg

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