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course MTH 173
5/12 1AM
The rate at which the position of a coasting ball on a constant incline changes is given as a function of clock time by velocity function v(t) = .00086 t^2 + .21 t + .6, with v in meters/sec when t is in seconds. Determine the rate of position change for clock times t = 0, 7 and 14 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 = 10.5 sec and compare with the approximation given by the graph.
By how much does the antiderivative function change between t = 0 and t = 14 seconds, what is the meaning of this change, and what is the graph's approximation to this change?
SOLUTION:
v(t) = .00086t^2 + .21t + .6
v(0) = .6
v(7) = 2.11
v(14) = 3.71
Interval [0,7]
A1 = ((.6 + 2.11)/2)7
A1 = 9.49
Slope = (2.11-.6)/(7-0) = .216
Interval [7,14]
A2 = ((2.11+3.71)/2)7
A2 = 20.4
Slope = (3.71-2.11)/(14-7) = .229
A = A1 + A2
A = 29.89
v'(t) = .00172t + .21
v'(10.5) = .228
10.5 is in the interval [7,14] of the graph.
The average change between 7 and 14:
(3.71+2.11)/7 = .831
The change in the antiderivative function is the area of the function. So, the change in the antiderivative function is near 29.89 (actual value is 29.77) between t = 0 and t = 14."
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Good, but you haven't yet explained, in terms of the behavior of the ball, the meanings of your slopes and areas.
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