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course mth 173
7/19 3
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.
The three points of our table are (0, 1.8), (15, 4.125), (30, 6.9)
slope = dy/dx
appox. Slope from t = 0 to t = 15
(4.125-1.8)/ 15 = .144 m/s^2
approx. slope from t = 15 to t = 30
(6.9-4.125)/(30-15) = .185 m/s^2
Area = avg. altitude * dt
Area t = 0 to t = 15
(1.8 + 4.125)/2 = 2.96
2.96 * 15 = 44.4375
Area t = 15 to t = 30
(4.125 + 6.9)/2 = 5.51
5.51 * 15 = 82.69
I'm not sure what these areas will give us.
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What does an altitude on a v vs. t graph represent?
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What do the two altitudes of a single trapezoid on this graph represent?
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What is represented by the average of those two altitudes?
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What is represented by the width of the trapezoid?
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What therefore is represented by the product of the average altitude and the width?
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Areas under the graph of a rate function will give us the change in quantity but we were not asked to graph the rate function.
If we had been asked to graph the rate function this is how I would do it:
Area = [(f'(x) + f'(x + dx))/2] * dx
v'(x) = .002t + .14
v'(0) = .14
v'(15) = .17
v'(30) = .2
(.14+ .17)/ 2 = .155
.155*15 = 2.325 m
(.2 + .17)/2 = .185
.185* 15 = 2.775 m
Evaluate the derivative of the velocity function for t = 22.5 sec and compare with the approximation given by the graph.
v(t) = .001 t^2 + .14 t + 1.8
v'(t) = 2(.001)t + .14 = .002t + .14
v'(22.5) = .002(22.5) + .14 = .185 m/s^2
This line lies directly on the slope of the trapezoid formed by (15, 4.125) and (30, 6.9)
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?
v(30)- v(0) = 6.9 - 1.8 = 5.1 m
By finding the area under each trapezoid and adding them together we can find the graph's approximation to this change.
Area = [(f'(x) + f'(x + dx))/2] * dx
v'(x) = .002t + .14
v'(0) = .14
v'(15) = .17
v'(30) = .2
(.14+ .17)/ 2 = .155
.155*15 = 2.325 m
(.2 + .17)/2 = .185
.185* 15 = 2.775 m
2.775 + 2.325 = 5.1 m so our approximation is very accurate.
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Good overall.
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