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September 21, 2014

September 21, 2014

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

(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 6:52pmdT/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 -
**George**, Sunday, November 9, 2008 at 7:06pmIs the answer for part B -1C

- Hard Calculus -

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