12 Drill Questions: Rates of change

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Drill 12. 1 How does a rate of change differ from an ordinary rate? What do you need to measure each?

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A rate needs one snapshot; a rate of change needs one snapshot too, but with different units.

A rate is one quantity divided by another from a single snapshot; a rate of change uses changes in two quantities (numerator and denominator), so you need at least two “frames” (e.g., a short “movie”) to form the ratio of those changes.

A rate of change is always with respect to time; an ordinary rate is not.

There is no real difference; the terms are interchangeable.

Drill 12. 2 On a graph of a function, what geometric quantity is the rate of change of the function with respect to its input? What is the name of the function that outputs that rate for every input?

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The rate of change is the height of the graph; the function that gives it is the integral.

The rate of change is the slope of the graph at that input; the function that gives the slope at every input is the derivative.

The rate of change is the area under the graph; the function is the antiderivative.

The rate of change is the intercept; the function is the marginal function.

Drill 12. 3 We approximate the rate of change of f() at input t by the function \[d\!f(t) \equiv \frac{f(t+h) − f(t)]}{h}\ .\] Why can’t we use h = 0? In practice, how do we choose h?

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If h = 0, the numerator \(f(t + h) - f(t)\) is zero. Likewise, the denominator (\(h\)) is 0. But 0/0 is undefined. We choose h “small enough” so that using an even smaller h doesn’t change the result meaningfully (e.g., we can test on a computer).

h = 0 would make the numerator zero, so the rate would always be zero. We choose h as large as possible to avoid this.

h = 0 is not allowed because the function f(t) might not be defined at t=0. We choose h randomly.

Actually, we can use h = 0; it gives the exact derivative. So we don’t need to choose h.

Drill 12. 4 The derivative of a function is itself a function. What are its inputs and what is the output (in words)?

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Inputs: rates of change; output: the original function value.

Inputs: the same as the original function; output: the instantaneous (that is, “marginal”) rate of change (slope of the graph) of the original function at that input.

Inputs: slopes; output: area under the curve.

Inputs: time only; output: position.

Drill 12. 5 Velocity is the rate of change of position with respect to time. Acceleration is the rate of change of velocity with respect to time. So acceleration is the _____ derivative of position.

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first

second

marginal

zero

Drill 12. 6 Why does the chapter say that “important problems of equilibrium,” like balancing spending between Natal and Cancer, are easier to think about when you use the idea of differentiation (rates of change)?

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Because differentiation gives the total QALYs, which we want to maximize.

Because the best allocation is where the marginal benefit (rate of change of QALYs with respect to dollars) is equal for both programs; thinking in terms of derivatives makes that condition clear and easier to work with.

Because we need the second derivative to find the optimum.

Because rates of change are always positive, so the math is simpler.