A severely ill patient is given an intramuscular injection of a highly potent antibiotic agent. It is known that the blood concentration, C, measured in micrograms per cc of the agent, increases according to the formula C = 2t/(t^2 + 9), where t is measured in hours after the time of injection. Correct to the nearest tenth of an hour, when will the concentration be maximal?
2006-10-22 04:57:45 · 5 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
Take the derivative of C(t) and then locate it's zeros... those will be times of maximum or minimum values.
Aloha
2006-10-22 05:02:09 · answer #1 · answered by Anonymous · 1⤊ 1⤋
Differentiate the function C(t) with respect to t twice. You'll have to use the quotient rule:
( f/g )' = ( g*f' - f*g' )/( g )^2
You'll get:
C'(t) = ( (t^2 + 9)*2 - (2*t)(2*t) )/( t^2 + 9 )^2
= ( 2*t^2 + 18 - 4*t^2 )/( t^2 + 9 )^2
= ( -2*t^2 + 18 )/( t^2 + 9 )^2
= -2*( t^2 - 9 )/( t^2 + 9 )^2
and for the second derivative:
C''(t) = ( (t^2 + 9)^2*( -4*t ) - [ -2*(t^2-9)*(t^2 + 9)*2t ] )/(t^2+9)^4
The concentration C(t) will be maximal when it flattens off and its slope becomes horizontal at some point. Thus, you want to solve for the places where C'(t)=0. That will happen when t^2=9. Thus, C'(t)=0 where t=-3 or t=3. The solution at t=-3 is invalid, so all you need to consider is t=3.
However, it's not enough that C'(t=3)=0 guarantees a local maximum. You also need to check the second derivative at that point. In that case, C''(t=3) = -1/27, which is a negative number. If the first derivative is 0 and the second derivative is negative, then the point MUST be a LOCAL maximum.
Now, you really should also be sure that the point C(t=0) doesn't beat the point C(t=3) because that boundary might just happen to be higher. In your case, C(t=0)=0 and C(t=3)=1/3. Clearly, C(t=3) is better than C(t=0).
Thus, your maximal concentration will be at t=3.0 hours, and the concentration will be 1/3 micrograms per cc of the agent.
2006-10-22 12:14:42 · answer #2 · answered by Ted 4 · 2⤊ 0⤋
Maxima when dC/dt = 0
Using Quotient Rule:
dC/dt = (2(t^2 +9) - 2t(2t)) / (t^2 +9)^2 = 0
(Denominator cant be zero)
So 2(t^2 +9) - 2t(2t) = 0
2t^2 + 18 - 4t^2 = 0
2t^2 = 18
t^2 = 9
t = 3
2006-10-22 12:04:07 · answer #3 · answered by Benny B 2 · 1⤊ 0⤋
Take the derivative, set to zero, and solve for the time
2006-10-22 11:59:41 · answer #4 · answered by arbiter007 6 · 0⤊ 1⤋
don't u think that home work is given to the students to improve their knowledge ? pl refer to your books, study well & do your home work yourself. nothing is difficult if u try.good luck.
2006-10-22 12:01:38 · answer #5 · answered by Anonymous · 1⤊ 3⤋