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 itís 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?





With respect to p. 179,  problem 9b, consider the analogy I used in class.  If it's 240 miles to Amarillo and my bucket-of-bolts jalopy will go only 60 mi / hr, then I'll "consume" the 240 miles at the rate of 60 miles each hour until 4 hours have passed and all the miles have been consumed.


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 PO q 6 h and e-mail me in the morning. :^)     --CLO