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Math Help - Linear Density

  1. #1
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    Linear Density

    I came across this problem just now:

    The mass of the part of a rod that lies between its left end and a point x meters to the right is 5x^{2} kg. The linear density of the rod at 8 meters is ___kg/meter and at 4 meters the density is ___kg/meter.

    Now, I understand that meter is x while kg is f(x), since density is kg/meter, the density of of the rod at 8m should be f(8)/8, which is 320/8 = 40 kg/m. But the answer turns out to be incorrect, would anyone please explain how to get the right answer?

    Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader
    I came across this problem just now:

    The mass of the part of a rod that lies between its left end and a point x meters to the right is 5x^{2} kg. The linear density of the rod at 8 meters is ___kg/meter and at 4 meters the density is ___kg/meter.

    Now, I understand that meter is x while kg is f(x), since density is kg/meter, the density of of the rod at 8m should be f(8)/8, which is 320/8 = 40 kg/m. But the answer turns out to be incorrect, would anyone please explain how to get the right answer?

    Thank you.
    Linear density means the strings/rods mass divided by it length.
    Thus, it mass after x meters is 5x^2 and its length is x thus, its linear density is,
    \frac{5x^2}{x}=5x thus, for x=8 its linear density is 40\frac{\mbox{kg}}{\mbox{m}}. Thus, for 8 meters the density is 40, I do not see what the problem is?
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  3. #3
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    I don't know, my prof assigns homework online via the webwork system, so you can excatly input and answer and see if it is correct or not right away.

    My answer was 40 and then 20, and they are incorrect.
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  4. #4
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    ThePerfectHacker is almost right. What you have been given is the "mass function," m(x)=5x^2 from one end of the rod. Since it is not constant you have to figure out the function. What TPH gave you is the average mass density, not the instantaneous. The instantaneous mass density then, is simply:
    \lambda = \frac{dm}{dx}=10x
    So I get that at x = 8 m, lambda = 80 kg/m, and at x = 4 m, lambda = 40 kg/m.

    -Dan
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  5. #5
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    Thanks for the help, I now understand how to solve the problem. But the next problem is even worse...

    1/f=1/p+1/q

    Find the rate of change of p in respect to q if q=7 and f=3.

    Now I have tried simply plug in the numbers, and got -4p+21, then the derivative is -4, but it isn't the right answer.

    I understand that rate of change of p in respect of q, (dp/dq?), but I cannot figure out how to use it in this problem.

    Any ideas?
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