course Mth 271
The velocity of an automobile coasting down a hill is given as a function of clock time by v(t) = .00059 t^2 + .28 t + .6, with v in meters/sec when t is in seconds. Determine the velocity of the vehicle for clock times t = 0, 9 and 18 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 = 13.5 sec and compare with the approximation given by the graph.
By how much does the antiderivative function change between t = 0 and t = 18 seconds, what is the meaning of this change, and what is the graph's approximation to this change?
To find the velocity of the vehicle for the clock times given, I plugged the values of t into the given equation:
v(t) = .00059t^2 + .28t + .6
t = [0, 9, 18]
t = 0
v(0) = .00059(0)^2 + .28(0) + .6 = .6
t = 9
v(t) = .00059(9)^2 + .28(9) + .6 = 3.16779
t = 18
v(t) = .00059(18)^2 + .28(18) + .6 = 5.83116
To sketch the graph of the trapezoidal approximation, I graphed the following points: (0, .6), (9, 3.16779), (18, 5.83116)
I then drew trapezoids from these points to the x axis.
Slope of first line:
(0, .6) (9, 3.16779)
(y2 – y1)/(x2 – x1) = (3.16779 - .6)/(9 – 0) = .28531
Slope of the second line:
(y2 – y1)/(x2 – x1) = (5.83116 – 3.16779)/(18 – 9) = .29593
'graph altitudes' represent velocities in m/s, so the rise between two points represents a change in velocity.
The run of a trapezoid represents the corresponding change in clock time, in seconds.
So the slope represents (change in velocity) / (change in clock time) = average rate of change of velocity with respect to clock time. The units will be (m/s) / s = m/s^2.
The areas of the trapezoids are found by using the Fundamental Theorem of Calculus:
integral of f(x) at 0 to 9 = F(9) – F(0)
F(9) = 3.16779
F(0) = .6
F(9) – F(0) = 3.16779 - .6 = 2.56779
Integral of f(x) at 9 to 18 = F(18) – F(9)
F(18) = 5.83116
F(9) = 3.16779
F(18) – F(9) = 2.66337
The areas of the trapezoids are found geometrically as average 'graph altitude' * width.
The average altitudes are respectively about 1.58 and 4.5. Widths are both 9 so areas are about 14 and 40.
The units of altitudes are m/s so the units of average altitudes are m/s; the units of width are seconds. So the units of area are m / s * s = m.
The areas correspond to approximate displacements in meters.
The derivative of velocity function:
v(t) = .00059t^2 + .28t + .6
v’(t) = .00059(2)t + .28
v’(t) = .00118t + .28
This is the acceleration
To find acceleration at time 13.5, I plugged 13.5 in for t.
v’(t) = .00118(13.5) + .28 = .29593
By comparing this value to the graph, this is identical to the slope at t = 13.5
Since the function is quadratic its slope changes at a constant rate; the average slope for a trapezoid will therefore be equal to the slope at the midpoint of the trapezoid.
This is a unique property of quadratic functions, in the sense that if for every possible interval the midpoint slope is equal to the average slope, the function must be quadratic. If the function is not quadratic, then there will be at least one interval (and usually an infinite number of intervals, and for most basic functions all possible intervals) for which this does not hold.
To find the antiderivative, I look the integral of the original function:
v(t) = .00059t^2 + .28t + .6
integral v(t) = (.00059t^3)/2 + .28t^2 + .6t + c
the correct antiderivative is .00059 t^3 / 3 + .28 t^2 / 2 + .6 t + c; if you use this antiderivative you will get results that agree reasonably well with the trapezoidal approximation. However it should be clear from your graph that the total area of the trapezoids differs, though not by much, from the actual area beneath the graph.
Because the values of t are given and c is an arbitrary value, I will leave it out of the calculations.
t = 0
integral v(0) = (.00059(0)^3)/2 + .28(0)^2 + .6(0) = 0
t = 18
integral v(18) = (.00059(18)^3)/2 + .28(18)^2 + .6(18) = 103.24044
‘dv = 103.24044 – 0 = 103.24044
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