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Math Help - Brain Teasers

  1. #1
    Newbie EightballLock's Avatar
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    Brain Teasers

    Frankly, I'm not sure what some of these problems have to do with Calculus, but since they were assigned by a Calculus instructor, I'll assume they do.


    The first one doesn't seem so bad at first.
    For what values of c does the equation \ln{x}=cx^2 have exactly one solution?
    I tried graphing to compare the two functions, and all I've been able to ascertain from the graphs is that c must be less than one. I don't know any way to solve this equation analytically. Ideas?


    The other one, oh boy.
    When a flexible cable of uniform density is suspended between two fixed points and hangs of its own weight, the shape y=f(x) of the cable must satisfy a differential equation of the form \frac{d^2y}{dx^2}=k\sqrt{1 + \frac{dy}{dx}^2} where k is a positive constant. Let z=\frac{dy}{dx} in the differential equation. Solve the resulting first-order differential equation (in z) and then integrate to find y.
    Maybe it's because I've been ill this weekend, but I'm completely stumped. I don't even know where to start.

    Thanks for the help!
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  2. #2
    Newbie EightballLock's Avatar
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    Hahaha, I realized that for the first problem that we have exactly one solution if C = 0. However, I don't think that's the answer we're looking for.

    EDIT: Additionally, I've also been able to determine that whatever the other value of C is, it lies between .183 and .184, as if that helps at all.
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  3. #3
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    Quote Originally Posted by EightballLock View Post
    Hahaha, I realized that for the first problem that we have exactly one solution if C = 0. However, I don't think that's the answer we're looking for.

    EDIT: Additionally, I've also been able to determine that whatever the other value of C is, it lies between .183 and .184, as if that helps at all.
    What about C=\frac{1}{2e}?

    At the point where the curves meet, the functions are equal: cx^2=\ln x, and their derivatives are equal: 2cx=\frac{1}{x}. Why is it so? Because otherwise there would be a second intersection. It is clear from the picture or, to be more rigorous: Consider indeed the function f(x)=cx^2-\ln x. You have f(x)\to+\infty when x\to\infty (except if c=0...) or when x\to 0^+. If f(x)=0 and f'(x)>0, then f(y)<f(x)=0 for some y close to x on the left, and by the intermediate value theorem there must be a zero between 0 and y. You can do the same on the right of x if f'(x)<0.

    So you get two equations, that you can solve.
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  4. #4
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    Quote Originally Posted by EightballLock View Post
    The other one, oh boy.
    When a flexible cable of uniform density is suspended between two fixed points and hangs of its own weight, the shape y=f(x) of the cable must satisfy a differential equation of the form \frac{d^2y}{dx^2}=k\sqrt{1 + \frac{dy}{dx}^2} where k is a positive constant. Let z=\frac{dy}{dx} in the differential equation. Solve the resulting first-order differential equation (in z) and then integrate to find y.
    Maybe it's because I've been ill this weekend, but I'm completely stumped. I don't even know where to start.

    Thanks for the help!
    For this one: the function z satisfies: z'=k\sqrt{1+z^2}, or: \frac{z'}{\sqrt{1+z^2}}=k (the denominator is positive). The big hint is that: \frac{z'}{\sqrt{1+z^2}}=({\rm arcsinh }\, z)' where {\rm arcsinh} is the inverse hyperbolic sinus. Now you can integrate both side of the equality ({\rm arcsinh }\, z)'=k, deduce z, and finally deduce y by integrating.
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