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Math Help - Real-world utility(?) of derivatives to estimate rates of change

  1. #1
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    Real-world utility(?) of derivatives to estimate rates of change

    Pardon a perhaps naive question from someone who is teaching himself calculus.

    I understand that the derivative of a function can be used to estimate things like marginal cost, marginal production etc.

    Yet with spreadsheets, or even function-enabled calculators, available to most people, I wonder -- why not just use the original function with new values, guaranteeing a fully correct answer, rather than using the derivative, which only gives an approximately correct answer?

    Let me give an example from my textbook. (I won't show all the work as I'm not asking how to solve it).

    Suppose z, the total weekly production of widgets, is a function of the number of skilled and unskilled workers (x and y respectively) as follows:

    z = f(x,y) = 12,000 + 5,000y + 10x^2y - 10x^3.

    Assume x = 40 and y = 60.

    I understand (or think I do; as ever, please correct me if I'm wrong) that if you want to estimate the additional number of widgets produced when one more skilled worker is added to the work force, you can determine the partial derivative of x with respect to z, or

    12,000 + 20xy - 30x^2

    and plug in (40,60), which gives you the approximation of 12,000.

    I also understand that if you want more than an estimate, you can "nail" the answer down by simply solving for

    z = f(41,60) = 12,000 + 5,000y + 10x^2y - 10x^3
    = 11,390.

    The approximation provided by the partial derivative is about 5% too high.

    In some cases, a 5% error is no problem. But what if the widgets in question are big-ticket items, meaning that lots of money is at stake? Five percent of a very big number is material.

    So why, when Excel etc makes it so easy to calculate the same function while plugging in different values for the variables, would one even want to bother with the derivative method?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lingyai
    Pardon a perhaps naive question from someone who is teaching himself calculus.

    I understand that the derivative of a function can be used to estimate things like marginal cost, marginal production etc.

    Yet with spreadsheets, or even function-enabled calculators, available to most people, I wonder -- why not just use the original function with new values, guaranteeing a fully correct answer, rather than using the derivative, which only gives an approximately correct answer?

    Let me give an example from my textbook. (I won't show all the work as I'm not asking how to solve it).

    Suppose z, the total weekly production of widgets, is a function of the number of skilled and unskilled workers (x and y respectively) as follows:

    z = f(x,y) = 12,000 + 5,000y + 10x^2y - 10x^3.

    Assume x = 40 and y = 60.

    I understand (or think I do; as ever, please correct me if I'm wrong) that if you want to estimate the additional number of widgets produced when one more skilled worker is added to the work force, you can determine the partial derivative of x with respect to z, or

    12,000 + 20xy - 30x^2

    and plug in (40,60), which gives you the approximation of 12,000.

    I also understand that if you want more than an estimate, you can "nail" the answer down by simply solving for

    z = f(41,60) = 12,000 + 5,000y + 10x^2y - 10x^3
    = 11,390.

    The approximation provided by the partial derivative is about 5% too high.

    In some cases, a 5% error is no problem. But what if the widgets in question are big-ticket items, meaning that lots of money is at stake? Five percent of a very big number is material.

    So why, when Excel etc makes it so easy to calculate the same function while plugging in different values for the variables, would one even want to bother with the derivative method?
    Three reasons come to mind.

    The first is that your function may not be solvable. In that case numerical approximation needs to be done and the derivative method, though slightly incorrect, may be much faster than the numerical solution.

    The second is that you may simply have zillions of variables to consider. Again you might need to resort to numerical calculation with the above comments applying again.

    Finally, a note about Excel's (and many other program's) techniques for solving equations...a number of equation solvers use Calculus in solving equations...almost with certainty if you are doing numerical approximations the program is going to be using Calculus. So theoretically you are using Calculus anyway.

    Basically my point is that you are right as you refer to the type of problem you are looking at. But in solving a general problem you might need to resort to other, less "black-box" like, techniques. And hey, the derivative method gives a simply solved first approximation, so why not use it?

    -Dan
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  3. #3
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    Thanks Dan -- good answers
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    Three reasons come to mind.

    The first is that your function may not be solvable. In that case numerical approximation needs to be done and the derivative method, though slightly incorrect, may be much faster than the numerical solution.

    The second is that you may simply have zillions of variables to consider. Again you might need to resort to numerical calculation with the above comments applying again.

    Finally, a note about Excel's (and many other program's) techniques for solving equations...a number of equation solvers use Calculus in solving equations...almost with certainty if you are doing numerical approximations the program is going to be using Calculus. So theoretically you are using Calculus anyway.

    Basically my point is that you are right as you refer to the type of problem you are looking at. But in solving a general problem you might need to resort to other, less "black-box" like, techniques. And hey, the derivative method gives a simply solved first approximation, so why not use it?

    -Dan
    Also, what you are given are toy problems to motivate the use of calculus.

    If we are to restrict attention to finance - look at the Black-Scholes equation
    for a real application of calculus in finance (for which the the Nobel Prize for Economics in 1997 was awarded).

    RonL
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