Question

The concentration C of certain drug in a patient's bloodstream t hours after injection is given by

C(t)= t/3t^2+5

Required:
a. Find the horizontal asymptote of C(t). (Answer must be in slope-intercept form.)
b. Determins what happens to the concentration of the drug as t increases. As t increases, what value will c(t) approach.
c. Determine the time at which the concentration is highest.

Answer 1

a) The horizontal asymptote of
`C(t)` is
`c = 0`.

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Explanation:

a) The horizontal asymptote of
`C(t)` is the horizontal line, to which the function converges when
`t` diverges to the infinity. That is:

`c = \lim _(t\to +\infty) (t)/(3\cdot t^(2)+5)` (1)

`c = \lim_(t\to +\infty)\left((t)/(3\cdot t^(2)+5) \right)\cdot \left((t^(2))/(t^(2)) \right)`

`c = \lim_(t\to +\infty)((t)/(t^(2)) )/((3\cdot t^(2)+5)/(t^(2)) )`

`c = \lim_(t\to +\infty) ((1)/(t) )/(3+(5)/(t^(2)) )`

`c = (\lim_(t\to +\infty)(1)/(t) )/(\lim_(t\to +\infty)3+\lim_(t\to +\infty)(5)/(t^(2)) )`

`c = (0)/(3+0)`

`c = 0`

The horizontal asymptote of
`C(t)` is
`c = 0`.

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

`C'(t) = ((3\cdot t^(2)+5)-t\cdot (6\cdot t))/((3\cdot t^(2)+5)^(2))`

`C'(t) = (1)/(3\cdot t^(2)+5)-(6\cdot t^(2))/((3\cdot t^(2)+5)^(2))`

`C'(t) = (1)/(3\cdot t^(2)+5)\cdot \left(1-(6\cdot t^(2))/(3\cdot t^(2)+5) \right)`

Now we equalize the expression to zero:

`(1)/(3\cdot t^(2)+5)\cdot \left(1-(6\cdot t^(2))/(3\cdot t^(2)+5) \right) = 0`

`1-(6\cdot t^(2))/(3\cdot t^(2)+5) = 0`

`(3\cdot t^(2)+5-6\cdot t^(2))/(3\cdot t^(2)+5) = 0`

`5-3\cdot t^(2) = 0`

`t = \sqrt{(5)/(3) }\,h`

`t \approx 1.291\,h`

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

`C''(t) = -(6\cdot t)/((3\cdot t^(2)+5)^(2))-((12\cdot t)\cdot (3\cdot t^(2)+5)^(2)-2\cdot (3\cdot t^(2)+5)\cdot (6\cdot t)\cdot (6\cdot t^(2)))/((3\cdot t^(2)+5)^(4))`

`C''(t) = -(6\cdot t)/((3\cdot t^(2)+5)^(2))- (12\cdot t)/((3\cdot t^(2)+5)^(2))+(72\cdot t^(3))/((3\cdot t^(2)+5)^(3))`

`C''(t) = -(18\cdot t)/((3\cdot t^(2)+5)^(2))+(72\cdot t^(3))/((3\cdot t^(2)+5)^(3))`

If we know that
`t \approx 1.291\,h`, then the value of the second derivative is:

`C''(1.291\,h) = -0.077`

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.