#$&*
course Mth 271
10/25 830p
Week 3 quiz 3solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
If a sand pile 3.3 meters high has a mass of 21562.2 kg, then what would we expect to be the mass of a geometrically similar sand pile 4.7 meters high?
y=ax^3 y= ax^3
21562.2 = 3.3x^3 y = (4.7)(18.7)^3
x^3 = 6534 y = 30709.8 kg
x = 18.7
@&
The proportionality is between the height and the mass of the pile.
y would be the mass of the pile and x its height.
a is the proportionality constnat.
You have substituted the value of x for the value of the proportionality constant.
*@
@&
You need to substitute the height for x and the mass for y, and solve for a.
*@
If there are 2.2869 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?
y=ax^3 y = (4.7)(.8849)^3
2.2869=3.3x^3 y = 3.257 billion grains of sand
x^3 = .693
x=.8849
@&
The surface of the pile can be covered with tiny squares. The appropriate proportionality is y = a x^2, where in this case y would be the number of grains and x the height of the pile.
*@
Week 4 Quiz 1
If a sand pile 4 meters high has a mass of 144000 kg, then what would we expect to be the mass of a geometrically similar sand pile 15 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.
y = ax^3 y = ax^3
144000 = 4x^3 y = 15(33.019)^3
x= 33.019 y = 540000
y = 4.03(33.019)^3
= 145080
@&
540 000 is more than 3 times as great as 145 000.
It should be clear that an increase of .03 meters in the 4-meter height of the pile would not triple its volume.
*@
@&
Once more you have substituted a value of one of your quantities for the proportionality constant.
You would do well to state the meanings of y, a and x at the beginning of the problem. This would help to reinforce the meaning of this process for you.
*@
@&
You did not use a differential estimate in solving this problem. You should review that process and let me know if you have questions.
*@
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.
????Trapezoid is sketched with t as the horizontal axis, marked at 0, 15, and 30. Vertical axis is V, with points at 1.8, 4.125, and 6.9.
The average altitude is (1.8+6.9)/2 = 4.35
Width is 30-0 = 30 sec.
4.35*30 = 130.5 cm
5.1 rise/30 run = slope of .17 cm????
@&
You have calculated your result only for the single trapezoid from t = 0 to t = 30 seconds.
You have two trapezoids and need to calculate results for both.
*@
@&
You have correctly reasoned out the results for the single trapezoid, so you're doing well, but you haven't included units in your calculations and therefore don't have correct interpretations.
*@
@&
What are the units of your average altitude?
What are the units of your width?
What therefore are the units of the area?
What are the units of the rise?
What are the units of your average run?
What therefore are the units of the slope?
*@
Evaluate the derivative of the velocity function for t = 22.5 sec and compare with the approximation given by the graph.
???? 2at +b 2(.001)(22.5) + .14 = .185 Not consistent with the graph. 22.5 comes halfway between 15 and 30 so the derivative should be approximately halfway between 4.125 and 6.9 but it is much smaller.????
@&
You calculated only one trapezoid, and its midpoint is at t = 15 seconds, not at t = 22.5 seconds. That would be the midpoint of one of the two trapezoids.
*@
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?
????? 6.9 - 1.8 = 5.1 meters/sec halfway between 0 and 30 would be 5.½ = 2.55
1.8+2.55 = 4.35 which matches the average altitude
V at 15 sec. is 4.125 so there is a difference of (4.35-4.125) = .225
@&
The function in this problem is
v(t) = .001 t^2 + .14 t + 1.8
The antiderivative function would be an antiderivative of this function.
You haven't given the antiderivative function, and it appears that you didn't use it.
*@
Week 4 Quiz3
Write the differential equation expressing the hypothesis that the rate of change of a temperature is proportional to the temperature T. Evaluate the proportionality constant if it is known that the when the temperature is 4813 its rate of change is known to be 500. If this is the t=0 state of the temperature, then approximately what will be the temperature at t = 1.3? What then will be the temperature at t = 2.6?
dT/dt = 4813/500 = 9.626 at t=0
@&
The proportionality statement is that the rate of change dT/dt is propotional to T.
That proportionality statement would be
dT/dt = k * T,
where k is the proportionality constant.
You need to use the given information to find the value of k, then rewrite the equation in terms of k.
You would then proceed to use the equation to solve the problem.
*@
t=1.3 t= 2.6
x/500 = 9.626(1.3) = 6256.9 x/500 = 9.626(2.6)
T= 6256.9 T = 12513.8"
@&
You have a good start on most of these problems. If you rethink them and submit your work I believe you'll be in pretty good shape. That will take some time an effort, but hopefully not too much.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@