Posted by **George** on Sunday, November 9, 2008 at 6:27pm.

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 -
**bobpursley**, Sunday, November 9, 2008 at 6:52pm
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 -
**LOU**, Thursday, March 24, 2016 at 12:46pm
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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