MAT 17A - DISCUSSION #5
Problem 1. Wound healing
Suppose you are part of a research group tasked to determine the effectiveness of a
drug that accelerates wound healing.
In the study, a surgeon gives a (brave) volunteer a rectangular wound on day 0, and
the wound is photographed 10 times a day for the next three days. Since skin is
not a uniform material, it heals faster in some directions than others. Knowing this,
the surgeon orients the wound so that it heals faster along its height (h(t)) than its
width (w(t)).
Since the drug might not have the same effect on healing along the wounds width
and height, you decide to keep track of the wound’s area, A(t) = w(t)h(t).
Consider the following (idealized) pictures of the wound:
Figure 1: A picture of the wound at t = 1 day (left), and a picture of the wound at
t + ∆t = 1.1 day (right).
a) Fill out the following table, given that, when t = 1 day and t + ∆t = 1.1 day,
h(t) = 2 mm, h(t + ∆t) = 1.9 mm, w(t) = 3 mm, and w(t + ∆t) = 2.95 mm.
Quantity
Change in height (mm)
Change in width (mm)
Change in time (days)
Symbol
∆h
∆w
∆t
In terms of h(t), w(t), ∆t, and t
h(t + ∆t) − h(t)
value
−0.1 mm
Table 1: Fill out the remainder of this table to complete problem 1a.
b) The change in area, ∆A = A(t + ∆t) − A(t), is shown on the right of Fig. 1 as
the shaded region. It is made up of three rectangles. Give expressions for the areas
of these three rectangles using only the variables h(t + ∆t), w(t + ∆t), ∆h and ∆w,
in order to find an expression for ∆A.
c) Check your answer to part b), by calculating the change in area using the given
values of h(t) = 2 mm, h(t + ∆t) = 1.9 mm, w(t) = 3 mm, and w(t + ∆t) = 2.95
mm (recalling that t = 1 day and t + ∆t = 1.1 day, ). Should the change in area be
positive or negative?
d) Divide your expression for the change in area from part b) by ∆t, which gives
∆A/∆t, an approximation of the rate the area of the wound changes with time (why
is this just an approximation?). Your answer should contain three terms, each one
corresponding to the area of one of the shaded rectangles divided by ∆t. Using the
numbers you calculated in Table 2, evaluate each of the three terms. Is one much
smaller than the others? Why do you think that is? Suppose you were to decrease
∆t. What do you think would happen to each of the three terms?
e) Suppose we have a model for h(t) and w(t). That is, suppose, based on fits to the
data, or an underlying physical law, we find that h(t) = 3 − t and w(t) = 3.5e−0.155t .
i. Verify that these functions give the correct values for h(1), w(1), h(1.1) and
w(1.1) used throughout this problem.
ii. Recalling your expressions in Table 2, re-calculate ∆h and ∆w for ∆t = 0.01,
0.001 and 0.0001 days. Plug these values into your answer for part d) (remembering
to change ∆t when you divide) and plot the areas of the three terms as a function
of log10 (∆t) using R/R-studio.
iii. Does your plot support your answer to part d)? Discuss.
f) Now, take the limit as ∆t goes to zero in your answer to part d). Note that this
makes
dA
∆A
=
lim
∆t→0 ∆t
dt
the derivative of the wound’s area with respect to time. Each term in your answer
to part d should also contain the derivative of a function with respect to time.
g) Based on your work above, what is
dA
d
w(t)h(t) =
=
dt
dt
Discussion Questions (optional): Suppose that you are investigating two drugs that
facilitate wound healing. You find the following models describe the width and height
of the wounds:
Treatment
None
Drug 1
Drug 2
Equation for w(t)
w(t) = 3.5e−0.155t
w(t) = 3.5e−0.16t
w(t) = 3.5e−0.3t
Equation for h(t)
h(t) = 3 − t
h(t) = 3 − 2t
h(t) = 3 − 1.25t
Table 2: Models describing wound healing in the presence and absence of drug
treatment.
Since large wounds are dangerous, you are most interested in developing a drug that
maximizes the initial rate that the wound closes. Which drug (if any) works the
best? Notice that the initial wound had a width of w(0) = 3.5 mm and a height of
h(0) = 3 mm. Does the best drug depend on the shape of the wound? Explain.
d ax
Note: It may be useful to know that
(e ) = aeax .
dx