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course Mth 173
Assignment 7 and 8 quizzes I think
Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant: •If a sand pile 2.1 meters high has a mass of 5000.939 kg, then what would we expect to be the mass of a geometrically similar sand pile 6.1 meters high?
• If there are 1.323 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?
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.
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?
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?
My work for this assignment
M1/m2=h1/h2
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The mass of a pile is not proportional to its height, but to the cube of its height.
You need to work through the assigned Modeling Projects.
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2.1/6.1=5000,939/x
X=5000,939/2.1*6.1=14526.537
S1/s2=pie r square root R2+h1^2/pie R square root R^2+h2^2=s1/s2= square root R2+h1^2/R^2+h2^2=1.323/s^2=square root 1+h1^2/1+h2^2=1.323/s^2= square root 1+4*41/1+37*21=1.323/s^2=square root 0.141=0.375=s^2=1.323/0.375=3.528
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The expressions are much simpler that this. These piles are geometrically similar, so there is no need to consider radii.
Again see the introductory problem sets.
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V(t)=0.001t^2+0.14t+0.18=0.001(0)+0.14(0)+1.8=1.8 V(0)=1.8=V(15)=0.001(15)^2+0.14(15)+1.8=0.225+2.1+1.8=4.125
V(30)=0.01(30)^2t+0.14 (30)+1.8=0.9+4.2+1.8=6.9
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So far so good, but
V(t)=0.001t^2+0.14t+0.18=0.001(0)+0.14(0)+1.8=1.8 V(0)=1.8=V(15)=0.001(15)^2+0.14(15)+1.8=0.225+2.1+1.8=4.125
V(30)=0.01(30)^2t+0.14 (30)+1.8=0.9+4.2+1.8=6.9
takes way longer to read than
V(t)=0.001t^2+0.14t+0.18=0.001(0)+0.14(0)+1.8=1.8
V(0)=1.8
V(15)=0.001(15)^2+0.14 (15)+1.8=0.225+2.1+1.8=4.125
V(30)=0.01(30)^2t+0.14 (30)+1.8=0.9+4.2+1.8=6.9
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0.001(2)t+0.14=0.001(2)(22.5)+0.14=0.185
T=0=0.001t^2+0.14+1.8=0.001(t^3)/3+0.14/(t^2)/2+1.8t+c=0.001
0.001/3(t^3)+0.7(t^2)t+1.8+c=0.001/3(0)+0.7(0)+1.8(0)+c
0.001 13(0)+0.7(0)+1.8(0)+c
0.001/3(30^3)+0.7*30^2+1.8*30+c
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You have expressions for the antiderivative at t = 0 and t = 30 sec, but you haven't evaluated them or found the change.
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300=k^x 2,954
K=300/2954
K=0.10155
P=kp dt=
Given population at t=0 is 2954 so p=k p (0)+c=2954 then c=2954
So p=kpt+2954
At t=1.2 p=0.10155*2954*1.2+2954=p=359.92+2954=3313.97
T=2.4
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It is unclear what you mean by
kpt tc
The equation is
dP/dt = k P.
The given information tells you that dP/dt = 300 when P = 2954, allowing you to evalute k. You do get k = 0.10155.
So the equation is
dP/dt = 0.10155 P.
Now given the value of dP/dt at t = 0, how much change would there be in P between t = 0 and t = 1.2?
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I've inserted a number of notes above.
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