#$&*
course MTH 173
7/12
Week 2 Qz 2
A man reading a newpaper walks at a constant speed away from the lamp which illuminates the paper he is reading. The illumination from the lamp
at clock time t is given by I = 190 / (t+3)2, where I is illumination in watts/m2 and t is clock time in seconds.
At what average rate is the illumination from the lamp changing between clock times t = 17.5 and t = 17.6 seconds?
At what average rate is the illumination from the lamp changing between clock times t = 17.5 and t = 17.51 seconds?
At what average rate is the illumination from the lamp changing between clock times t = 17.5 and t = 17.501 seconds?
What do you estimate is the rate at which illumination is changing at clock time t = 17.5 seconds?
The rate at which the illumination a newspaper changes is given by Rate = 1040 / (t+3)3, where Rate is rate of change in watts/m2 per second and
t is clock time in seconds. How much do you think illumination will change between t = 17.5 and t = 35 seconds?
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(I(17.6)-I(17.5))/(17.6-17.5) = -.004379/.1 =~ -.0438W/(m*s)
(I(17.51)-I(17.5))/(17.51-17.5) = -.0004408/.1 =~ -.004408W/(m*s)
(I(17.501)-I(17.5))/(17.501-17.5) = -.000044105/.1 =~ -.00044105 W/(m*s)
dI/dt = -2(190)(t+3)^-3(1)
dI/dt = -2(190)(17.5+3)^-3 = -.0441 W/(m*s)
so for Rate = 1040 / (t+3)3, ave rate:
(Rate(35) -Rate(17.5))/(35-17.5) = .018953/(35-17.5) =~ .00108 W/(m*s)
so since .00108W = dI/dt, then .00108W/(m*s)*17.5sec = .01895 W/m
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Wk 3 Qz 1
Determine the average rate of change of the function y(t) = .5 t 2 + 56 t + -55 between x and x + D<="""" p="""">x.
What are the growth rate, growth factor and principal function P(t) for an initial investment of $ 760 which is compounded annually at 10%
interest?
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(.5(x + Dx)^2 + 56(x + Dx) -55 - (.5x^2 + 56x -55))/(x+Dx-x)
(.5(x^2 +2x*Dx + Dx^2) + 56x + 56Dx -55 -.5x^2 -56x +55)/Dx
(.5x^2 + x*Dx + .5Dx^2 +56Dx -.5x^2)/Dx
(Dx(x + .5Dx +56)/Dx
therefore
ave. rate of change = x + .5Dx + 56
to check:
y'(t) = t +56
y'((x+x+Dx)/2) = (2x+Dx)/2 + 56 = x +.5Dx +56
then
P(t) = $760(1.1)^t
growth rate = .1 or 10%
growth factor = 1.1 or 110%
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Wk 3 Qz 2
Sketch and completely label a trapezoidal approximation graph for the function y = x^2/ 5 + 1, for x = 0 to 2.7 by increments of .9.
(0, 1), (.9, 1.162), (1.8, 1.648), (2.7, 2.458)
0-.9 interval:
slope: .18
area: (.9)(1.081) = .9729
accum. area: .9729
.9-1.8 interval:
slope: .54
area: (.9)(1.405) = 1.2645
accum. area: 2.2374
1.8 to 2.7 interval:
slope: .9
area: (.9)(2.053) = 1.8477
accum. area: 4.0851
we can check the final accum. area with the integral: x^3/15 + x + c
between 0 and 2.7 we get 2.7^3/15 + 2.7 + c - (0^3/15 + 0 + c) = 2.7^3/15 + 2.7 = 4.0122, which is close enough to assume I've done it correctly
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Wk 3 Qz 3
Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
If a sand pile 4 meters high has a mass of 32000 kg, then what would we expect to be the mass of a geometrically similar sand pile 5.9 meters
high?
If there are 6.24 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on
the surface of the second?
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m = kx^3
32000kg = k(4)^3
k = 500 kg/m^3
m = (500kg/m^3)(5.9m)^3
m = 102,690 kg
or dm/dd = 3(500)(x^2)
@& Good, and your meaning is perfectly clear. However dd is potentially confusing. Better to use another variable name for diameter (maybe just x) to avoid confusion. The letter d has its special use in calculus, and it's generally a good idea to reserve it for that use.*@
so 3(500)(4.95)^2 = 36753.75kg/m
dm/1.9 = 36753.75kg/m
dm = 69,832.13 kg
69,832.13kg + 32,000kg =~ 101,832kg
then
S.A. = kx^2
6.24 bill grains = k(4^2)
k =~ 390,000,000 grains/m^2
grains = (390,000,000gr./m^2)(5.9)^2
grains = 1.3576*10^10 or approx. 13.6 billion grains
or
dgrains/dd = 2(.39 bill)x
so 2(.39 bill)(4.95) = 3,861,000,000 grains/m
and dgrains/1.9 = 3,861,000,000 grains/m
and dgrains = 7,335,900,000 grains, so
7,335,900,000 grains + the orig. 6.24 bill grains =~ 1.3576grains
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Week 4 Qz 1
If a sand pile 4 meters high has a mass of 146000 kg, then what would we expect to be the mass of a geometrically similar sand pile 14 meters
high? Using the differential estimate the mass of sand required to increase the height of the pile from 4 meters to 4.03 meters.
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m = kx^3
146000kg = k(4m)^3
k =~ 2281.25 kg/m^3
m = 2281.25 kg/m^3(4.03)^3
m =~ 149,309.7 kg
differential
dm/dd = 3(2281.25)(x^2)
dm/.03m = 3(2281.25)(4.015)^2
dm = 3309.68 kg
so 146,000kg + 3309.68kg =~ 149,309.7kg
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Week 4 Quiz 2
The velocity of an automobile coasting down a hill is given as a function of clock time by v(t) = .001 t^2 + .14 t + 1.8, with v in meters/sec
when t is in seconds. Determine the velocity of the vehicle for clock times t = 0, 15 and 30 sec and make a table of rate vs. clock time.
Sketch and label the trapezoidal approximation graph corresponding to this table and interpret each of the slopes and areas in terms of the
situation.
Evaluate the derivative of the velocity function for t = 22.5 sec and compare with the approximation given by the graph.
By how much does the antiderivative function change between t = 0 and t = 30 seconds, what is the meaning of this change, and what is the graph's
approximation to this change?
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v(0) = 1.8 m/s
v(15) = 4.125 m/s
v(30) = 6.9 m/s
0-15 interval:
slope: .155 m/s^2
area: 15s(2.9625m/s) =~ 44.44m
accum. area: 44.44m
15-30 interval:
slope: .185 m/s^2
area: 15s(5.5125m/s) = 82.69m
accum. area: 127.13m
v'(t) = .002t + .14
v'(22.5s) = .185 m/s^2
it's equivalent to the slope of the 15-30sec interval (midpt clocktime 22.5sec), which is right since the derivative is the change in velocity
over the change in time, and the slope is the average rate of change in velocity in terms of time; so using the instantaneous rate at the midpt
clock time gives us the same rate as the slope (the acceleration).
then
antiderivative = (.001t^3)/3 + (.14t^2)/2 +1.8t + c
and (.001(30)^3)/3 + (.14(30)^2)/2 +1.8(30) + c - ((.001(0)^3)/3 + (.14(0)^2)/2 +1.8(0) + c) = (.001(30)^3)/3 + (.14(30)^2)/2 +1.8(30) = 126 m,
which is decently close to the accumulated 127.13m area, which is right since we're using t=0 to t=30, which encompasses both intervals.
The antiderivative is the position equation, so 126m is the displacement down the hill.
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Week 4 Qz 3
Write the differential equation expressing the hypothesis that the rate of change of a population is proportional to the population P. Evaluate
the proportionality constant if it is known that the when the population is 2954 its rate of change is known to be 300. If this is the t=0
state of the population, then approximately what will be the population at t = 1.2? What then will be the population at t = 2.4?
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dP/dt = kP
300 = k(2954)
k=~.10156
dP/dt = .10156P
if you assume that dP/dt stays close to 300 over the next interval:
dP/1.2 = 300
and
dP = 360
then
360 + 2954 = 3314 organisms
for 2.4s,
dP/1.2 = .10156(3314)
dP =~ 404
3314 + 404 = 3718 organisms
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&#This looks good. See my notes. Let me know if you have any questions. &#