# Applications of Derivatives

• Nov 24th 2006, 01:30 PM
Jacobpm64
Applications of Derivatives
A dose, $D$ of a drug causes a temperature change, $T$, in a patient. For $C$ a positive constant, $T$ is given by

$T = ( \frac{C}{2} - \frac{D}{3} ) D^3$.

(a) What is the rate of change of temperature change with respect to dose?

(b) For what doses does the temperature change increase as the dose increases?

I think I have this one correct as well, I'm just kind of fuzzy on this material, so I'd like some reassurance.

(a) This part is easy.
$\frac{dT}{dD} = \frac{3C}{2} D^2 - \frac{4}{3} D^3$

(b) This part I'm not too sure about.
$\frac{3C}{2} D^2 - \frac{4}{3} D^3 > 0$
$D^2 ( \frac{3C}{2} - \frac{4}{3} D) > 0$
$D < \frac{9C}{8}$

Is this correct? Thanks.
• Nov 25th 2006, 06:16 AM
earboth
Quote:

Originally Posted by Jacobpm64
A dose, $D$ of a drug causes a temperature change, $T$, in a patient. For $C$ a positive constant, $T$ is given by

$T = ( \frac{C}{2} - \frac{D}{3} ) D^3$.

(a) What is the rate of change of temperature change with respect to dose?

(b) For what doses does the temperature change increase as the dose increases?

I think I have this one correct as well, I'm just kind of fuzzy on this material, so I'd like some reassurance.

(a) This part is easy.
$\frac{dT}{dD} = \frac{3C}{2} D^2 - \frac{4}{3} D^3$

(b) This part I'm not too sure about.
$\frac{3C}{2} D^2 - \frac{4}{3} D^3 > 0$
$D^2 ( \frac{3C}{2} - \frac{4}{3} D) > 0$
$D < \frac{9C}{8}$

Is this correct? Thanks.

Hello,

your calculations are correct. To make sure that your final result isn't misread I would write: $0 < D < \frac{9C}{8}$

(I know it is quite impossible to get a negative dose, but only to be on the safe side I would write it this way)

EB