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MTH 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Inquiry
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http://hyperphysics.phy-astr.gsu.edu/hbase/deriv.html#c1
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The URL provided is a website that I frequent in order to tinker with various forms of math and science.
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If the URL is complete, you'll find information that I've located in order to further illustrate what I couldn't communicate. Truthfully, I normally think Okay, what do I do first? when approaching math and science-related. The best way I learn is to first be exposed to something generic and simple then research my own methods and ways to take it two and three steps further; hence my brief introduction to physics, then later that night studying various forms of astrological measurements of length, namely parsecs.
In closing, I addressed you with a question concerning the rate of change between time 17.7 and 17.701, and I what I was hypothesizing (and cursing for what it's worth) was utilizing a derivative to find the average rate of change.
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In short, for a linear or quadratic function you can easily assess the average rate of change by evaluating the derivative at the appropriate point. For a linear function the derivative is constant so any point would do. For a quadratic function the derivative changes in a linear fashion so you can get the average rate by evaluating the derivative at the midpoint of the interval.
For any function except a linear or quadratic it is pointless to try to use the derivative to find the average rate of change on an interval. For most functions there are ways to predict the point on the given interval at which you would evaluate the derivative to get the average rate, but to do so would be complicated, typically requiring techniques well beyond the level of a first-year calculus course.
The page you quote looks good. The first box explains the relationship between the slope or average rate of change and the derivative. The last two boxes are also useful and relatively accessible. The second box, on partial derivatives, is something usually not encountered until a multivariable calculus course.
More detail:
The reason we deal so much with rates of change in precalculus is that rate of change is central to calculus. The derviative of a function is its 'rate-of-change' function.
A linear function has a constant rate of change. Its graph is a straight line. So for example the function y = 2 t + 7 has a graph with constant slope 2, and the derivative of this function is just 2. If you calculate the rate of change of y with respect to t between any two points you get 2, a result that's identical to the derivative.
A quadratic function has form y = a t^2 + b t + c. Its derivative with respect to t is 2 a t + b. If you were to calculate the average rate of change of a quadratic function between two values of t, the result would be equal to the value of the derivative evaluated at the t value halfway between those two values.
For an exponential function y = A b^t the derivative is A ln(b) * b^t, where ln(b) is the natural log of b. If you were to calculate the average rate of change of this function on some t interval, the result would be equal to the derivative evaluated at a t value somewhere between the two, but not at the midpoint.
The question you asked about concerned the function
y = .026t^2 - 1.2t + 53
which you said had values of 39.90554 and 39.90554 at respective clock times 17.7 and 17.701. Your y values are identical and do not therefore reveal the change in y. Using your values the average rate of change of this function on this interval would be
ave rate = change in y / change in t = 0 / .001 = 0.
The problem with this calculation is that the y value at 17.701 is not 39.90554, but rather 39.905260426. The change in y is thus -0.000279574, and the average rate of change on this interval is
ave rate = -0.000279574 / (.001) = -0.279574
The derivative of y = .026 t^2 - 1.2 t + 53 is .052 t - 1.2. Evaluating this at t = 17.7005, which is halfway between 17.7 and 17.701 we get -0.279574, identical to the average rate of change on this interval.
Had we evaluated the derivative at 17.7 and 17.701 we would have obtained derivatives -.2796 and -0.279548. Averaging these values we would obtain -0.279574, identical to the above midpoint value of the derivative and to the average rate of change on this interval.
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