Topic 2.2 Extension 06
Definite integrals and differental eq'ns
To follow the worksheet "Area and the definite integral"
You may want to review your work on “Differential equations,” “Initial
value problems,” and “Area and the definite integral. ”
1)
Recall that the definite integral does not need the arbitrary
constant C. Why?
When you do the subtraction, it goes away.
2)
Consider the differential equation
a = dv/dt.
We separate the
a dt
variables to obtain dv = __________,
and then we place the integral
a dt
dv
symbol in front of each side to obtain ∫ __________
= ∫ __________.
(Do not actually integrate, yet).
3)
Since no initial values were given, integration would require
addition of the arbitrary constant C. Why?
When you take the derivative it goes away anyway. The arbitrary
constant gives the most general possible result.
4)
We can make the integral definite using a very simple procedure:
The left hand side contains the variable v. We arbitrarily assign the
initial value of this variable to be the symbol vi or vo, and the final
value to be just plain v. Thus the indefinite integral dv becomes the
definite integral
v
dv
v0
∫
<-----
place the final value of v here.
<-----
place the initial value of v here.
5)
What is the variable of integration on the right hand
What arbitrary initial value should we assign to it?
t
arbitrary final value should we assign to it? ____.
Now
indefinite integral (the right hand one) so that it is
integral:
t
a dt
t0
∫
6)
t
side? ____.
t0
____.
What
rewrite the
a definite
<-----
place the final value of t here.
<-----
place the initial value of t here.
Now integrate the left hand definite integral (number 4):
v
v
dv =
v
= v - v0
v0
v0
7)
Integrate the right hand integral (number 5) under the assumption
that the acceleration is constant:
t
t
a dt =
at
= at - at0
t0
t0
Why does a have to be constant in order to integrate?
Otherwise we need to know its time dependence.
∫
∫
[ ]
[ ]
8)
Since the two integrals were part of an equality, we can set the
results of 6) and 7) equal to each other. Solve this equation so that
v (without subscript) is isolated on the left hand side:
v - v0 = at - at0
v = v0 + at - at0
1
9)
In general, we choose the initial time to be
convenience. Rewrite your equation from 8) to reflect this:
v = v0 + at - a·0
v = v0 + at
zero
for
10)
Your results from 9) should look familiar.
Note that what we
have essentially done here is find the solution to a general or
arbitrary initial value problem. If given the actual initial values we
could substitute them into your general result to get a specific,
tailored equation. Suppose we then find out that the initial speed of
a particle is 12 m/s and its acceleration is a constant –10 m/s2. What
does your equation from 9) tailor to?
v = v0 + at
v = 12 + -10t
11)
Now we have an equation which will predict v for any time t.
What is the speed of the particle at t = 10 s?
v = 12 + -10(10)
v = -88 m/s
12)
Now follow the same procedure for the differential equation v =
dx/dt. Be sure to substitute the general result you got for v from 9)
and show all of your work. Again, assume a is a constant.
dx = v dt
∫dx = ∫v dt
∫dx = ∫(v0 + at) dt
x
dx
x0
∫
x
=
[ ]x
x
= x - x0
0
t
(v0 + at) dt
0
∫
[
=
t
]0
v0t + at2/2
= v0t + 12 at2
x - x0 = v0t + 12 at2
13)
Solve your equation for x (without the subscript) and assume the
initial time is zero, as before:
x - x0 = v0t + 12 at2
1
x = x0 + v0t + 2 at2
14)
Your results in 13) should look familiar, too.
Now suppose the
initial position is 35 m, the initial speed is 12 m/s and the
acceleration is –10 m/s2.
What does your general equation from 13)
tailor to?
x = x0 + v0t + 12 at2
x = 35 + 12t - 5t2
15)
What does your tailored equation predict
particle at t = 10 s will be?
x = 35 + 12(10) - 5·102
x = -345 m
2
the
position
of
the