Question

1. the concentration c, in milligrams per liter, of an antibiotic in a patients bloodstream t hours after injection is given by ( c(t)=\frac{60t}{t^{2}+25} ).

a. what happens to the concentration of the drug as t increases? think about end behavior.
b. sketch the function from ( t = 0 ) to ( t = 20 ) in the grid provided at right.
c. use your calculator to determine the time at which the concentration is highest.

Explanation:

Part A

Step1: Analyze the end - behavior of the function

We want to find \(\lim_{t
ightarrow\infty}C(t)=\lim_{t
ightarrow\infty}\frac{60t}{t^{2}+25}\).
Divide both the numerator and denominator by \(t^{2}\):

$$
LATEXBLOCK0
$$

Step2: Evaluate the limit

As \(t
ightarrow\infty\), \(\frac{60}{t}
ightarrow0\) and \(\frac{25}{t^{2}}
ightarrow0\). So \(\lim_{t
ightarrow\infty}\frac{\frac{60}{t}}{1+\frac{25}{t^{2}}}=0\).

Part C

Step1: Use the quotient rule

If \(C(t)=\frac{60t}{t^{2}+25}\), where \(u = 60t\) and \(v=t^{2}+25\). The quotient rule is \(C^\prime(t)=\frac{u^\prime v - uv^\prime}{v^{2}}\).
We know that \(u^\prime=60\) and \(v^\prime = 2t\).

$$
LATEXBLOCK1
$$

Step2: Set the derivative equal to zero

Set \(C^\prime(t)=0\), then \(1500 - 60t^{2}=0\).

$$
LATEXBLOCK2
$$

Since \(t\geq0\) (time), \(t = 5\)

Answer:

Part A

As \(t\) increases, the concentration of the drug approaches \(0\) milligrams per liter.

Part C

The time at which the concentration is highest is \(t = 5\) hours.