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HMMT Calculus Problem

  1. Nov 25, 2007 #1
    A plane curve is parameterized by x(t) = [tex]\int^{\infty}_{t}[/tex] [tex]\frac{cos u}{u}[/tex] du and y(t) = [tex]\int^{\infty}_{t}[/tex] [tex]\frac{sin u}{u}[/tex] for 1 [tex]\leq[/tex] t [tex]\leq[/tex] 2. What is the length of the curve?

    I know the formula for the length of the curve so I know you need to find dy/dt and dx/dt and integrate the square root of the sum of their squares and integrate that over the interval [1,2].

    But my question is finding each derivative. I understand that you could apply the fundamental theorem of calc that in essences states the derivative of the integral is the original function. So you just flip the limits of integration, put a negative out front, and simply write dx/dt = -cos u / u. But in the actual statement of this FTC, the lower limit of integration is a constant (commonly denoted a or c), so I'm second guessing myself. Should it matter that the lower limit is infinity? I guess I was thrown a bit by the infinity and our AP class just covered the FTC before I attempted this problem, it seems like the infinity wouldn't affect the differentiability of an already integrated function.
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  3. Nov 25, 2007 #2


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  4. Nov 25, 2007 #3
    Ah thank you, that makes a lot of sense. So as long as one limit of integration is a variable and the other is a quantity other than a variable, then we can directly apply this FTC if we want the derivative?

    Also is the equation x^a = e^(a ln(x)) pretty common?
  5. Nov 26, 2007 #4


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    I don't know what you mean by "pretty common" but that equation is certainly true and shows that an exponential with any base can be written in terms of "the" exponential function ex.
  6. Nov 26, 2007 #5
    By the way, both limits of integration can be variable and you can still do the FTC.
  7. Nov 26, 2007 #6
    I see, rewriting in terms of e certainly is useful for evaluating limits such as x^x.

    Yes, I think I meant to say that if both are variable, it's just a bit more work.

    OK, I fully understand how to do problems 1-5 of HMMT 2005 and they all seem very straightforward with just a basic understanding of calculus. Knowing that I'm probably getting ahead of my problem-solving abilities and/or my calculus knowledge, I attempted #10:

    10. Let f: R -> R be a smooth function such that f '(x) = f(1-x) for all x and f(0) = 1. Find f(1).

    Seeing the potential for recursion and no other approach, I differentiated both sides to get f''(x) = -f'(1-x) = -f(x). Here I tried plugging in values, ended up with zero, and realized I was using circular reasoning... So now I'm stuck. I took a peek at the solution and it involved lots of trig. Any suggestions on how I should proceed or what I should consider?
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