#$&*
course Mth 173
2/11 10:21 PM
The velocity of an automobile coasting down a hill is given as a function of clock time by v(t) = .00056 t^2 + .74 t + 1.4, with v in meters/sec when t is in seconds. Determine the velocity of the vehicle for clock times t = 0, 6 and 12 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 = 9 sec and compare with the approximation given by the graph.
By how much does the antiderivative function change between t = 0 and t = 12 seconds, what is the meaning of this change, and what is the graph's approximation to this change?
???just checking to make sure, when you say term 1 v term 2, you consider term 2 the x-axis, right?
v(0)=1.4
v(6)=5.86016
v(12)= 10.36064
rate is derivative which for this instance is
v'=.00112t+.74
v'(0)= .74
v'(6)=.74672
v'(12)=.75344
between 0 and 6 the average rate would be .74336 and from 0 to 6, is 6 seconds, so *6 is 4.46016. + 1.4 is 5.86016, so does check out.
between 6 and 12 the average rate would be .75008, 6 seconds have elapsed, so * 6 is 4.50048, + 5.86016 is 10.36064, checks out.
the graph requested is the chart of the derivative i believe, so I will map it.
the two altitudes of the trapezoid, are .74 and .75344, with an average of .74672, in this case meters/sec
the width spans from t=0 to t=12, so a width of 12, making the width 12. so 12seconds* .74672m/sec is 8.96064, which is the area, or 8.96064m/sec
????what is the purpose for knowing the area. i seem to have trouble catching onto this with trapezoids.
(.75344-.74672)/(12-6)
.00672/6= .00112 (the constant in our derivative function)
(.74672-.74)/(6-0)
.00672/6=.00112 (same as previous), this shows a constant rate of slope change. just trying to pull in extra information, this proves that whatever function this rate was for, it would not be exponential, correct?
the change from 0 to 12 is 8.96064 for the antiderivative
!!!!I believe I have answered my own question, the trapezoid's area (at least from the derivative) gives the amount of change from two points. this fascinates me. the meaning of this change is the actual value velocity has changed from the two times, and knowing that change you can better guess changes for times around these points, or for these points know a more actual value.
what is meant by asking the graph's approximation to this change?
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