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Math Help - Derivatives of Inverse Functions

  1. #1
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    Derivatives of Inverse Functions

    Let f(v) be the gas consumption (L/km) of a car going at velocity v (km/h). In other words, f(v) tells you how many liters of gas the car uses to go one kilometer, if it is going at velocity v. You are told that: f(80) = 0.05 and f`(80) = 0.0005.

    a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) and g(v)? Find g(80) and g`(80).


    Okay, I'm pretty sure their relationship is: g(v) = f^-1(v). I tried to find the value of g(80) and came out with the pretty ridiculous value of 159980km/L. I am almost certain I did this wrong, but I can't find any relevant info in my textbook. For g`(80) I know I have to use the inverse rule to find the derivative (which I know), but I can't find this until I have found the value of g(80).

    b) Let h(v) be the gas consumption in L/h at velocity v. What is the relationship between h(v) and f(v)? Find h(80) and h`(80).

    Okay, now I'm just lost. I don't even know where to start here.

    c) How could you explain the practical meaning of the values of these function and their derivatives to a driver who knows no calculus?

    I think I could get this on my own once I get help with the previous problems.
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  2. #2
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    Quote Originally Posted by Jessem View Post
    Let f(v) be the gas consumption (L/km) of a car going at velocity v (km/h). In other words, f(v) tells you how many liters of gas the car uses to go one kilometer, if it is going at velocity v. You are told that: f(80) = 0.05 and f`(80) = 0.0005.

    a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) and g(v)? Find g(80) and g`(80).


    Okay, I'm pretty sure their relationship is: g(v) = f^-1(v).
    NO, they are not inverse functions. Since f takes velocity to gas consumption, its inverse function would take gas consumption to velocity.
    Here both f and g take velocity, v, to some other value. Take a look at the units of the result of both f and g. f(v) gives "liters per kilometer" while g(v) gives "kilometers per liter"- they are reciprocals.

    I tried to find the value of g(80) and came out with the pretty ridiculous value of 159980km/L. I am almost certain I did this wrong, but I can't find any relevant info in my textbook. For g`(80) I know I have to use the inverse rule to find the derivative (which I know), but I can't find this until I have found the value of g(80).

    b) Let h(v) be the gas consumption in L/h at velocity v. What is the relationship between h(v) and f(v)? Find h(80) and h`(80).

    Okay, now I'm just lost. I don't even know where to start here.
    Again, look at the units. f(v) gives "liters per kilometer" and h(v) must be in "liters per hour". Clearly to go from f(v) to h(v), you need to "multiply" by kilometers and "divide" by hours: you need to multiply by kilometers per hour. And what parameter tells you how many kilometers you go in one hour?

    c) How could you explain the practical meaning of the values of these function and their derivatives to a driver who knows no calculus?

    I think I could get this on my own once I get help with the previous problems.
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  3. #3
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    Thank you HallsofIvy!

    a) Since they are reciprocals, g(80) = (f(80))^1 = 20km/L

    I am still having trouble finding g`(80). Would this just be the reciprocal of f`(80) = 0.0005 which would be 2000km/L/L? This seems much too high to make sense.

    b) Like you said, we must multiply h(v) by km/h (or v) so we get: f(v) = h(v)*v

    However, when I tried to apply this to solve for h(80) I got:
    h(80) = f(80)*80^-1 = 0.05*0.0125 = some very small, unreasonable number.
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  4. #4
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    Quote Originally Posted by Jessem View Post
    Thank you HallsofIvy!

    a) Since they are reciprocals, g(80) = (f(80))^1 = 20km/L

    I am still having trouble finding g`(80). Would this just be the reciprocal of f`(80) = 0.0005 which would be 2000km/L/L? This seems much too high to make sense.
    No, it is not just the reciprocal of the derivative of f. For one thing, notice that if f is increasing, its derivative is positive. But its reciprocal would be decreasing so the dervative of its reciprocal must be negative! If g(x)= 1/f(x) then use the quotient rule: g'(x)= (0(f(x))- 1(f'(x))/f^2(x)= -f'(x)/f^2(x).
    Equivalently, write g(x)= (f(x))^{-1} (I have used parentheses there to indicate that this is the -1 power of f, not its inverse function) and use the chain rule: g'(x)= -f(x)^{-2}f'(x).

    b) Like you said, we must multiply h(v) by km/h (or v) so we get: f(v) = h(v)*v
    No, I did not say that! f(v) already has units of "liters per kilometer" and g(v) must have units of "liters per hour": liters/hour= (liters/kilometer)(liters/hour) so h(x)= v*f(x).

    However, when I tried to apply this to solve for h(80) I got:
    h(80) = f(80)*80^-1 = 0.05*0.0125 = some very small, unreasonable number.
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