Question

The concentration C of a certain drug in the bloodstream t hours after injection into muscle tissue is given by
c (t) = (4t) / (2+t^3)

Answer 1

when is concentration the greatest?
thanks!
:)

Answer 2

Do you know derivatives?

Answer 3

yes, i was able to get the first derivative but i got stuck at the second one
f'(x)=(-8t^3 + 8)/(2+t^3)^2

Answer 4

Why are you taking the second derivative?

Answer 5

I thought that would be the best way to find the absolute max

Answer 6

You can use your first derivative to find the max. Find the root(s). Then plug in the value for c(t) and that is your max.

Answer 7

find when is at max concentration*

Answer 8

by roots do you mean the result of setting f'(x)=0?
I got the cube root of 16, so is that the only thing i need to plug in to the original equation?

Answer 9

Yes, when f'(x) = 0

Answer 10

How did you get \(\sqrt[3]{16}\)????

Answer 11

\(-8t^3+8=0\) gave you \(t= \sqrt[3]{16}\)??

Answer 12

oh, it should have been 1

Answer 13

when i plugged in 1 to the original equation i got 4/3 but it wasnt the correct answer. . .

Answer 14

You have this: f'(x)=(-8t^3 + 8)/(2+t^3)^2

Why one earth are you playing with that -8?
The denominator never is zero, so focus on the numerator.
-8 is never zero, so get rid of it.

t^3 - 1 = 0

And the only Real solution is apparent.

Simplify your life. Don't make it unnecessarily complicated.

Answer 15

c(1) = 4/3, but why do we care? The question is WHEN, not HOW MUCH. We are done when we say t = 1 holds the greatest concentration.

Answer 16

thank you for your help I appreciate it

Answer 17

Note: It is generally considered bad form to change variable names mid problem.

If the concentration is C, it is not convenient to define the function c(t). A better choice would be C(t).

Answer 18

You should also check to see if it really is a max or min or neither. The 2nd derivative reveals this information.

c"(t) = 24(t^2)(t^3 - 4)/(t^3 + 2)^3

Thus, t = 0, t = cuberoot(4) are critical points, making t = 1 a min or max.
c"(1) < 0, making c(1) a max.

NOW it's done.