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.
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.
3 answers
Is the answer for part B -1C
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?
0 answers