Math 122 - Sec 2: Quotient Rule Worksheet
Spring 2016
Professor Bennett
For this worksheet, we will examine two physiological formulae and the associated equations
and graphs: Mean Arterial Pressure and Muscle dynamics.
Mean arterial pressure: Mean arterial pressure MAP (in mg of mercury) is given by multiplying the cardiac output CO (L/min) by the systemic vascular resistance SV R (mmHg · min/L)
and then adding the central venous pressure, and in equation form
MAP = (CO × SVR) + CVP.
Since the central venous pressure is usually much smaller than the other term, in practice, we
use
MAP = CO × SVR.
While measure the MAP can be done easily using diastolic and systolic blood pressure, in terms
of understanding how the body works, we want to understand the physics and mathematics of
the system. Physics tells us that fluid moving through a vessel has resistance (our SVR) given
by the formula
kLη
SVR = R = 4
r
where k is a constant, L is the length of the blood vessels, η is the blood thickness and r is the
average radius of the blood vessel.
The cardiac output is given by the heart rate H (beats/min) multiplied by the stroke volume
V (L/beat), i.e.,
CO = H × V.
1. Using the above, write a formula for MAP involving k, L, η, r, H, and V .
MAP =
HV kLη
.
r4
2. When the body is trying to regulate MAP, which of the above factors is it likely to be able
to change in a short time span? Explain.
This is complicated as there are actually three different time scales that take place. The body
can very quickly change the heart rate (as I think everyone was happy with). The vascular radius
changes relatively quickly in people, as muscle activation creates such a change. It also changes
in response to environmental effects, such as when we are cold, we have vasoconstriction (they
narrow keeping more of the warmth in the core of the body) and we have vasodilation when our
core temperature is going up (to cool the body off ).
The stroke volume changes in the matter of a minute or so, although when we exercise it
goes up, so this would increase our MAP like the heart rate.
The blood ”thickness” changes more on the matter of hours. As the body becomes dehydrated,
the blood can become ”thicker”. Thus at the end of a marathon or other longer sporting event,
we get tired.
The length of blood vessels changes very slowly with growth, putting on (and losing) weight,
etc, so we are talking on the order of days (or longer) typically.
Comment: When I ask a question like this, part of my intent is for you to start thinking
about each of the terms and potentially ask questions about what’s possible. For many of you,
one of the important things to get out of this class is the ability to understand the difference
between constant parameters (the things that are unchanging in the context of the problem)
and the variables. Other important ideas are understanding of graphs and what they tell you,
particularly in terms of slope and area. In other words, it’s making the calculus knowledge a
useful aid in understanding and analysis.
While the technical skills will have some value (and for a few of you a great deal of value),
in general, it pales compared to the application and understanding of how to put these things
together. Thus, I think of questions 2, 3, and 5 as equally or even more important than questions
1 and 4.
3. Strokes are often caused when the Mean Arterial Pressure changes rapidly. Explain why this
means that we would care about the derivative of the MAP.
The derivative MAP measures the rate of change of the Mean Arterial Pressure. Thus the
derivative of the MAP is large and positive when the Mean Arterial Pressure is increasing rapidly.
We would hope that if we can figure out how to keep the derivative of the MAP low, then we
could decrease the risk of a stroke.
4. Calculate the (time) derivative of the MAP using the quotient rule, (assuming that items
the body cannot change are constants and those that it can are functions of time t).
HV kLη
From problem 1, MAP =
. Since H and r are both changing over the time scale we
r4
are looking at and are therefore functions of time, we will write them as H(t) and r(t). We note
that since on our time scale, V , k, L, and η are unchanging, we will treat them as constants.
The quotient rule tells us that
u 0 u0 v − uv 0
,
=
v
v2
so we want to look at what is the derivative of the numerator u = H(t)V kLη = (V kL)H(t) and
v = (r(t))4 . For u0 , we note that the constant rule for derivatives tells us that if a is a constant
and f (t) is differentiable, then (af )0 (t) = a · f 0 (t). Consequently, we just let a = V kLη in this
rule and we have
u0 = V kLηH 0 (t).
On the other hand, for v, we will need to use the chain rule, where the chain rule is given
differentiable functions f (x) and g(x), then (f ◦ g)0 (x) = f 0 (g(x)) · g 0 (x). In the case where
v = (r(t))4 , we have two functions f (x) = x4 and r(t) which plays the role of g(x) in the chain
rule above. Since f 0 (x) = 4x3 , substituting in for the chain rule we have
v 0 = 4(r(t))3 · r0 (t).
Thus, we have
MAP0 (t) =
=
=
=
=
u0 v − uv 0
v2
V kLηH 0 (t)(r(t))4 − V kLηH(t) · 4(r(t))3 r0 (t)
(r(t))8
V kLη · (H 0 (t)(r(t))4 − 4H(t) · (r(t))3 r0 (t))
(r(t))8
V kLη(r(t))3 · (H 0 (t)r(t) − 4H(t) · r0 (t))
factoring out (r(t))3
(r(t))8
V kLη · (H 0 (t)r(t) − 4H(t) · r0 (t))
canceling (r(t))3
(r(t))5
5. What happens to the heart rate when someone starts to exercise? Using the concepts of
calculus and the above derivative calculation, explain what the body must do to keep the mean
arterial pressure from increasing too much when someone starts to exercise.
When someone starts to exercise, the heart rate increases, which tells us that H 0 (t) is positive. But if the mean arterial pressure should not increase too much, then that means that the
derivative of MAP needs to be kept small (or even ideally 0). At a given time t0 ,
MAP0 (t0 ) =
V kLη (H 0 (t0 )r(t0 ) − 4H(t0 ) · r0 (t0 ))
,
(r(t0 ))5
Since V , k, L, η, H(t0 ), and r(t0 ) are all fixed at time t0 , this means that we need to have the
only terms that change cancel each other out (or at least keep the overall value small). That
means we are left with trying to keep H 0 (t0 )r(t0 ) − 4H(t0 )r0 (t0 ) as close to 0 as possible. Since
H 0 (t0 ) > 0, this means that we need
r0 (t0 ) =
H 0 (t0 )r(t0 )
<0
4H(t0 )
So r0 (t0 ) needs to be positive, which means that the mean radius of the blood vessels needs to
H 0 (t0 )r(t0 )
be increasing. Moreover, it should be increasing at roughly the rate of
to keep the
4H(t0 )
pressure from increasing too rapidly.
Comment: On a question like this, I really want you to be thinking and writing in terms of
the derivative and rates of change. The purpose is to practice using the language of calculus to
discuss and think about items.