**CLOates**

**24-OCT-2006**

**A Few Solutions for Chapter 10 Problems**

**In
#5b on page 173**, we're asked to find the flow rate in gtt/min
necessary to infuse 150 mg of Tetracycline in 40 minutes.

First, when asked for a flow rate, start with a flow rate. So our
starting factor for this one is

150 mg / 40 min (that's a Drug
Infusion Rate in nursingspeak or a Weight Flow Rate
in engineerspeak).

Our answer units are gtt/min (according
to the statement in the gold box at the top of the page). Well, the
time part is easy. The starting factor is already "per
minute"

The 150 mg, however, is a bit more interesting. We need to convert
it to drops. That probably means we'll need to convert it first to mL and
then use the drop factor to convert mL to gtt.

Let's see, in Chapter 8 when they told us to give 150 mg of medication,
we looked on the medication label and found the medication strength in mg/mL
and then used that to convert mg to mL. We'll need to do that same kind
of thing here, but the strength we need is not the strength on the
label of the concentrated solution, but the strength of the medication in the
IV bag. That concentration is 150 mg / 100 mL, according to the problem
statement.

Now if we apply that concentration information to the drug infusion rate
above, we have

150 mg / 40 min * 100 mL /
150 mg (the surviving units here are mL / min).

That's close to the answer units, gtt/min.
If we apply the drop factor of 60 gtt / mL, we'll
have

150 mg / 40 min * 100 mL / 150
mg * 60 gtt / mL

= ( 150 x 100 x 60 ) / ( 40 x 150
) gtt / min

= 150 gtt / min

That rate passes the "ha-ha" test, in that it’s about 2.5
drops a second. That's moving right along (think of the beats of a
fast-tempo march like the circus march, "Barnum and Bailey's
Favorite"--you've heard it if you've ever been to a circus), but it's not
ridiculously fast. Checking the arithmetic again, seems to confirm the
numbers, and the answers in the back of the text agree.

----------------------------------

How about **p.
178 #6**? It's just a lot of extraneous information packed around 100 mL / 60 min = 100 mL / hr, right?

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With respect to **p****. 179, problem
9b**, consider the analogy I used in class. If it's
240 miles to

In dimensional analysis form, this is stated as

240 mi x 1 hr / 60
mi = 4 hr.

Now, in problem 9b, we're consuming what? Well, we could
view the infusion as a consumption of 50 mL of D5W. In that
view, we're consuming 50 mL of D5W at the rate of 300 mL / hr
(the rate calculated in 9a), for a total time of 1/6 hr or 10
min. That's the logic of "way 1."

Answer way 1 was 50mL X 1hr./300mL
X 60min/1hr = 10 min

On the other hand, we could view the infusion as a consumption of the
300 mg of Cleocin that's dissolved in the D5W.
In that view, we're consuming 300 mg of Cleocin at
the ordered rate of 30 mg / min. That's the logic of
"way2."

Answer way 2 was 300mg X 1min/30mg = 10 min

Either way works. Way 2 is a little safer, I suppose, in the sense
that it uses only quantities from the order itself in the calculation, instead
of depending on a previous calculation for the flow rate in mL / hr.

--------------------------------

You'd be surprised at how many people, sane or otherwise, describe the
med-math course as "fun." What you're picking up on, I suspect,
is the power of a relatively simple technique like dimensional analysis to
solve many kinds of problems--sometimes problems that are not in your area of expertise
and that you have no idea how to solve on first reading.

As an **example
from another field of endeavor**, every bear-in-the-air highway patrolman,
including my alter ego, Lt. Gona Pullham-Overnau, knows
that if (s)he times a car moving through a marked one-mile stretch of road, the
average speed of that vehicle is given by speed = 3600 /
#seconds. How'd they get that formula?

Well, let's state the problem the way we would see it in a cop-math
book. "Roger Ramjet rides his Harley through a marked mile
in 48 seconds. What is his average speed?"

Okay, we know that his speed is 1 mile / 48 seconds, but we usually
state speed in miles / hour, so with miles / hour as the desired answer unit,
we can write the conversion equation as

1 mile / 48 sec * 60 sec / 1 min
* 60 min / 1 hr (write it out by hand in
fraction form)

= (1 x 60 x 60 / 48) miles / hr

= 3600 / 48 miles / hr

= 75 miles / hr,

and, generally, speed in mph
= 3600 / <seconds to traverse one mile>.

We worked this problem without having to take Physics for Engineers and
Scientists I and II, or even OU's Physics for Poets
(Physics for Non-Majors) and Algebra II, for that matter. We didn't
say anything about distance = rate x time
and therefore rate = distance / time or anything
similar. We were able to figure it out just by looking at the units and
manipulating them. Pretty neat! Who says science and math are
boring?

Engineers and, particularly, experimental physicists do this all the
time when they investigate new, unknown, and puzzling phenomena, where the
applicability of any of the zillion formulas that describe known
phenomena is not obvious and may be non-existent. The joy of teaching the
med-math course is that I get to show a group of folks, most of whom would
probably never take a physics course, how to do this stuff. As spouse Sue
often reminds me, "Not all the compensation for this job [teaching] comes
in the pay envelope."

Let me know if the explanation of 9b isn't clear or if you have other
questions.**

________________________

**If the problems are still giving you a headache, take gr x aspirin