Math 2280-1
Wed January 14
HW due Friday ... problem session tomorrow Thursday 8:35-9:25 AM WBB 617
Quiz Friday ... see 2250 page from last spring to see level of questions
1.2 Differential equations of the form y# x = f x
(Also between today's and Monday's notes/discussion we will have introduced 1.3: slope fields, and 1.4:
separable differential equations. We will have much more to say about these two sections on Friday and
next week.)
,
Review from Monday. What were the main ideas we talked about?
,
Finish Newton's law of cooling, and the dead body mystery. Here's how far we'd gotten:
Exercise 4 Monday) Newton's law of cooling is a model for how objects are heated or cooled by the
temperature of an ambient medium surrounding them. In this model, the body temperature
T = T t changes at a rate proportional to to the difference between it and the ambient temperature A t . In
the simplest models A is constant.
a) Use this model to derive the differential equation
dT
=Kk T K A .
dt
solution: we converted the English sentence "the body temperature T = T t changes at a rate
proportional to the difference between it and the ambient temperature A t into a math equation phrase by
phrase:
"T= T t changes at a rate" / T # t
"proportional to" → =k ,
"the difference between it and the ambient temperature A t " / T K A .
This yields the differential equation
T# t = k T K A
which looks like the one we were trying for, except for the sign minus sign in front of the k. But we realize
that the "k" in our DE is negative, since when T O A, T # t must be negative, but T K A O 0 (and
when T ! A, T # t must be positive, but T K A ! 0 ). Thus it makes sense to write the constant of
proportionality as Kk, where k is a positive number, which yields the DE we wanted.
b) Would the model have been correct if we wrote
dT
= k T K A instead?
dt
yes, see above.
c) Use this model to partially solve a murder mystery: At 3:00 p.m. a deceased body is found. Its
temperature is 70 + F. An hour later the body temperature has decreased to 60 + . It's been a winter
inversion in SLC, with constant ambient temperature 30 + . Assuming the Newton's law model, estimate
the time of death.
If we set t = 0 to be 3:00 p.m. the IVP is
T # t =Kk T K A
T 0 = 70,
with A = 30. If we choose our time units to be hours, then we have the additional information that
T 1 = 60.
We cleverly used the chain rule backwards to "integrate" the DE and find a formula for T t . First divide
by T t K A to get everything involving the unknown function T t on the left:
T# t
=Kk.
T t KA
T# t
0
dt = Kk dt .
T t KA
0 ln T t K A =Kk t C C1 !!
Kk t C C
1
0 T t KA = e
Kk t
C
1
0 T t = A C C e , with C = e
0 T t = 30 C CeKk t .
.
T 0 = 70 0
70 = 30 C C 0 C = 40 .
0 T t = 30 C 40 eKk t .
T 1 = 60 0
60 = 30 C 40 eKk
3
0 30 = 40 eKk 0
= eKk .
4
So,
T t = 30 C 40 .75 t .
Finish!
Remark: Using the chain rule backwards didn't seem intuitive. The separation of variable method on the
next page uses differentials to get the same correct solutions to so-called "separable differential equations":
Definition: A separable first order DE for a function y = y x is one that can be written in the form:
dy
=f x f y .
dx
It's more convenient to rewrite this DE as
1 dy
=f x ,
(as long as f y s 0) .
f y dx
1
Writing g y =
the differential equation reads
f y
dy
g y
=f x .
dx
Solution (math justified): Suppose G y is an antiderivative with respect to y of g y and F x is an
antiderivative of f x with respect to x. By the chain rule, if y x is any solution to the DE on some
interval, then
d
d
G y x = G # y x y# x = g y x y# x = f x =
F x .
dx
dx
Thus
G y x = F x CC .
G y = F x CC
Expresses y implicitly as a function of x . You may be able to use algebra to solve this equation explicitly
for y = y x , and (working the computation backwards) y x will be a solution to the DE. (Even if you
can't algebraically solve for y x , this still yields implicitly defined solutions.)
Solution (differential magic): Treat
dy
as a quotient of differentials dy, dx , and multiply and divide the
dx
DE to "separate" the variables:
dy
f x
=
dx
g y
g y dy = f x dx .
Antidifferentiate each side with respect to its variable (?!)
g y dy = f x dx , i.e.
G y C C1 = F x C C2 0 G y = F x C C .
Agrees!
Exercise 1) Use separation of variables to re-find the solution functions T t to the Newton's Law of
cooling DE with constant ambient temperature:
T # t =Kk T K A .
It should seem more natural than when we had to figure out how to do the chain rule backwards.
Now return to Monday's notes, to discuss section 1.2. We will continue that discussion here:
Section 1.2 continued, after Monday's notes:
Exercise 2: A projectile with very low air resistance is fired almost straight up from the roof of a building
30 meters high, with initial velocity 50 m/s. Its initial horizontal velocity is near zero, but large enough so
that the object lands on the ground rather than the roof.
a) Neglecting friction, how high will the object get above ground?
b) When does the object land ?
Here's another fun example from section 1.2, which also reviews important ideas from Calculus - in
dy
particular we will see how the fact that the slope of a graph y = g x is the derivative
can lead to first
dx
order differential equations.
Exercise 3: (See "A swimmer's Problem" and Example 4 in section 1.2). A swimmer wishes to cross a
river of width w = 2 a , by swimming directly towards the opposite side, with constant transverse velocity
vS . The river velocity is fastest in the middle and is given by an even function of x , for Ka % x % a. The
velocity equal to zero at the river banks. For example, it could be that
x2
vR x = v0 1 K 2 .
a
See the configuration sketches below.
a) Writing the swimmer location at time t as x t , y t , translate the information above into expressions
for x# t and y# t .
b) The parametric curve describing the swimmer's location can also be expressed as the graph of a
function y = y x . Show that y x satisfies the differential equation
v0
dy
x2
=
1K 2 .
dx
vS
a
c) Compute an integral or solve a DE, to figure out how far downstream the swimmer will be when she
reaches the far side of the river.