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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?
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The second term in y vs. x is x, and the second terms is the independent variable, which is graphed on the horizontal axis.
So you have that right.
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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.
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The average rate is not the average of the two derivatives, but (change in y) / (change in x).
That would correspond to the slope of your v vs. t trapezoid.
It seems that below, however, you have constructed a graph for v ' vs. t, not v vs. t.
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In any case, if you multiply the average rate by the interval you do get the change in v, and adding this to the initial value of v for the given interval does give you the next value of v.
Fortunately this function is quadratic, so that the average of the two rates for a given interval is in fact the average rate. For any other type of function it would not have been so.
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the graph requested is the chart of the derivative i believe, so I will map it.
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The requested graph would correspond to the table of values, which you expressed as
v(0)=1.4
v(6)=5.86016
v(12)= 10.36064
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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.
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Your graph of v ' (t) would have two trapezoids, one representing the interval from t = 0 to t = 6, the other the interval from t = 6 to t = 12.
However since v ' (t) is linear, the area of the single big trapezoid is identical to the sum of the areas of the two trapezoids.
This would not be the case for the trapezoidal graph of the quadratic velocity function.
However note that in this case the area is the change in the velocity from t = 0 to t = 12. This illustrates is the importance of the area. If all you knew was the v ' (t) function, the area of a trapezoid would allow you to approximate the change in the v(t) function for the given interval. If the trapezoids are thin enough, corresponding to small t intervals, the approximations will be sufficiently accurate for your purposes.
In this case, the linearity of the v ' (t) function results in approximations that are exact (i.e., the region of the graph is actually trapezoidal, so the area of the trapezoid is identical to the area beneath the graph).
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(.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?
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A constant rate of slope change implies, and is exhibited by, a quadratic function.
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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.
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This relationship is the very heart of the calculus. What you've discovered, within the limited context of this problem, is the fundamental theorem of calculus.
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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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Had you graphed the velocity function instead of the v ' (t) function, the graph would not have been linear, and the trapezoids would have only approximated the actual changes in the antiderivative function.
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