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Hard Calculus

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A dose, D, of a drug causes a temperature change, T, in apatient. For C a positive constant, t is given by T=((C/2)-(D/3))D^3.

(a) What is the rate of change of temperature change with respect to dose.
(b) For what dose does the temperature change increase as the dose increases.

For part a I got CD^2-D^3=0
For part b I got c^2/4

Any help would be greatly appreciated.

  • Hard Calculus -

    dT/dD= (C/2-D/3)3D^2 + D^3(-1/3)
    = 3/2 CD^2 + D^3(2/3)
    check that.

    for the second part,
    set dT/dD to zero, and solve for D.

  • Hard Calculus -

    Is the answer for part B -1C

  • Hard Calculus -

    The temperature at 6:00 a.m. was 43.5° C and at 9:00 a.m it was 48° C. Assuming a constant rate of change, the temperature at 8:00 a.m was?

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