The problem was:
In the rocket example, use what you've learned to determine
the position function, x(t), of the rocket assuming
that x(0) = 0. First figure it out for
0 < t < 10,
then assume that after 10 seconds, the rocket's
velocity is constant, and use that to find the position function for
t > 10.
Step 1: Recall the function for the velocity.
v(t) = 100 ln
|
|
110
110 - 10t
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|
meters/second for 0 < t < 10 seconds
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and that
x(t) =
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t
0
|
v(u) du
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The lower limit of zero is established because the rocket starts out when
t = 0 at a position of zero (i.e.,
x(0) = 0). Observe the use of the dummy variable,
u inside the integral. This is to preserve our independent variable,
t, which is also the upper limit of the integral.
Step 2: Set up the integral.
This means using the v(t) from the first equation
above to form the integrand in the second equation above.
x(t) = 100
|
t
0
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ln
|
|
110
110 - 10u
|
|
du meters for 0 < t < 10 seconds
|
How you integrate this is not immediately clear, so
Step 3: Expand the log in the integrand into a difference of logs.
x(t) = 100
|
t
0
|
(ln(110) - ln(110 - 10u)) du meters for 0 < t < 10 seconds
|
Using the fact that the integral of the sum is the sum of the integrals:
x(t) = 100
|
t
0
|
ln(110) du - 100
|
t
0
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ln(110 - 10u) du for 0 < t < 10 seconds
|
Step 4: Integrate.
The first integral is a constant. You get for it
100
|
t
0
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ln(110) du = 100
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|
u ln(110)
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t
0
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= 100t ln(110)
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Now for the second integral:
Substitute the log argument in the second integral.
Let
s = 110 - 10u and
ds = -10du. The second integral
becomes
100
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|
(ln(110) - ln(110 - 10u)) du = -10
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|
ln(s) ds
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which you do by parts:
w = ln(s)
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|
dz = ds
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ds
dw =
s
|
|
z = s
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-10
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|
ln(s) ds = -10
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|
s ln(s) -
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|
s
ds
s
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= -10 (s ln(s) - s) + C
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Now back-substitute by replacing
s
with
110 - 10u and take it
over the limits
0 to
t.
-10
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t
0
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ln(110 - 10u) du = -10
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(110 - 10u) ln(110 - 10u) - (110 - 10u)
|
t
0
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Step 5: Apply the limits:
Applying the limits to the second integral and then eliminating the terms that cancel in the
difference:
-10
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t
0
|
ln(110 - 10u) du =
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-10
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|
(110 - 10t)ln(110 - 10t) + 10t - 110 ln(110)
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Step 6: Put it all together:
Finally taking the results of the first integral minus the results of the second
(and doing some algebra and applying the log-identity to simplify the expression):
x(t) =
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t
0
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ln
|
|
110
110 - 10u
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du = (100t - 1100)ln
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|
110
110 - 10t
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+ 100t
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which is the expression for the rocket's position,
x(t), in the first
10 seconds -- that is from its ignition to its burnout. At
t = 10 seconds you find that
x(10) = -100 ln(10) + 1000 = 769.74 meters
|
According to the original equation for
v(t) (first equation at the top
of this page), we find that
v(10) = 100 ln(10) = 230.26 meters/sec
For
t > 10 seconds the rocket maintains the speed
given above. Combining that with the constant velocity formula we have for
t > 10 seconds:
x(t) = 100 (t - 10) ln(10) - 100 ln(10) + 1000
= 1000 - 1100 ln(10) + 100 t ln(10)
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