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Question 1088339: A drug is administered to a patient and the concentration of the drug in the bloodstream is monitored. At time t ≥ 0 (in hours since giving the drug), the concentration (in mg/L) is given by
c(t)= 5t / t^2 +1
a) Graph the function c(t).
b) What is the highest concentration of the drug that is reached in the patient’s bloodstream?
c) What happens to the drug concentration after a long period of time?
d) How long does it take for the concentration to drop below 0.3 mg/L?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
a) The graph is shown below. The graph was created by GeoGebra (free graphing software).
Other tools such as the Desmos Graphing Calculator (also free) are alternatives to get the same job done.
I'm using x instead of t because GeoGebra only accepts x as the variable when it comes to graphing.
To do this by hand, you can plug in various t values to get corresponding c(t) values. Each pair is essentially an (x,y) ordered pair. Plotting enough of these points will give you a good idea of what the graph looks like. The graph is the curve that goes through all of these points.
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b) The highest concentration that is reached in the patient's bloodstream is 2.5 mg/L
We get this answer using one of two methods
Method 1: Use a graphing calculator to look at the highest point on the function curve. Keep in mind that because  , this means  . The highest point is at (x,y) = (1,2.5) which is shown as point A below
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Method 2: If you are familiar with calculus, then you can find the derivative of  = \frac{5t}{t^2+1}) to get  = -\frac{5(t^2-1)}{(t^2+1)^2}) (I'm skipping a few steps). Then you solve  = 0) to get
-[5(t^2-1)]/[(t^2+1)^2] = 0
5(t^2-1)=0
t^2-1 = 0
t^2 = 1
t = 1 or t = -1
These t values are the critical values. We will ignore t = -1 because . Negative time values make no sense.
So the max happens at 
Plug this into the original function and we get
 = \frac{5*1}{1^2+1})
 = \frac{5}{2})
 = 2.5)
This confirms the answer found in method 1 above.
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c) As t gets larger, c(t) gets smaller and heads to 0. The graph in part a) shows this happening. The green curve slowly heads closer to the x axis but it will never touch the axis nor cross the axis.
With a calculus approach, we can use limits to get
=\lim_{t \to \infty}\left(\frac{\frac{1}{t^2}*5t}{\frac{1}{t^2}*(t^2+1)}\right))
=\lim_{t \to \infty}\left(\frac{\frac{5}{t}}{1+\frac{1}{t^2}}\right))
=\frac{\lim_{t \to \infty}\left(\frac{5}{t}\right)}{\lim_{t \to \infty}\left(1+\frac{1}{t^2}\right)})
=\frac{0}{1+0})
=0)
The result of 0 means that c(t) approaches 0 when t approaches infinity. In context of the problem, this means that the drug concentration will degrade/decay over time, which makes sense as the drug is leaving the system.
This also can be shown using a table. Make two columns for t and c(t). In the t column, write down values that get larger and larger such as 1, 5, 10, 100, 1000, 10000, etc. The c(t) column is the output of each t input
| t | c(t) |
|---|
| 1 | 2.5 | | 5 | 0.961538 | | 10 | 0.49505 | | 100 | 0.049995 | | 1000 | 0.005 | | 10000 | 0.0005 | | 100000 | 0.00005 |
As you can see, the c(t) column is getting closer and closer to 0
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d)
Plug in c(t) = 0.3 and solve for t
Use the quadratic formula (I'm going to skip a few steps here) to get the two approximate solutions
 or 
The first time value is when the concentration surpasses 0.3 mg/L while the second time value represents the point in time when the concentration drops below 0.3 mg/L
So t = 16.61 hours is the only answer for part d). This value is approximate.
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