# Brain Teasers

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• Dec 1st 2008, 06:49 AM
EightballLock
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 $\displaystyle \ln{x}=cx^2$ 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 $\displaystyle y=f(x)$ of the cable must satisfy a differential equation of the form $\displaystyle \frac{d^2y}{dx^2}=k\sqrt{1 + \frac{dy}{dx}^2}$ where k is a positive constant. Let $\displaystyle z=\frac{dy}{dx}$ 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!
• Dec 1st 2008, 08:16 AM
EightballLock
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.
• Dec 1st 2008, 08:43 AM
Laurent
Quote:

Originally Posted by EightballLock
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 $\displaystyle C=\frac{1}{2e}$?

At the point where the curves meet, the functions are equal: $\displaystyle cx^2=\ln x$, and their derivatives are equal: $\displaystyle 2cx=\frac{1}{x}$. 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 $\displaystyle f(x)=cx^2-\ln x$. You have $\displaystyle f(x)\to+\infty$ when $\displaystyle x\to\infty$ (except if $\displaystyle c=0$...) or when $\displaystyle x\to 0^+$. If $\displaystyle f(x)=0$ and $\displaystyle f'(x)>0$, then $\displaystyle f(y)<f(x)=0$ for some $\displaystyle y$ close to $\displaystyle x$ on the left, and by the intermediate value theorem there must be a zero between $\displaystyle 0$ and $\displaystyle y$. You can do the same on the right of $\displaystyle x$ if $\displaystyle f'(x)<0$.

So you get two equations, that you can solve.
• Dec 1st 2008, 08:54 AM
Laurent
Quote:

Originally Posted by EightballLock
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 $\displaystyle y=f(x)$ of the cable must satisfy a differential equation of the form $\displaystyle \frac{d^2y}{dx^2}=k\sqrt{1 + \frac{dy}{dx}^2}$ where k is a positive constant. Let $\displaystyle z=\frac{dy}{dx}$ 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 $\displaystyle z$ satisfies: $\displaystyle z'=k\sqrt{1+z^2}$, or: $\displaystyle \frac{z'}{\sqrt{1+z^2}}=k$ (the denominator is positive). The big hint is that: $\displaystyle \frac{z'}{\sqrt{1+z^2}}=({\rm arcsinh }\, z)'$ where $\displaystyle {\rm arcsinh}$ is the inverse hyperbolic sinus. Now you can integrate both side of the equality $\displaystyle ({\rm arcsinh }\, z)'=k$, deduce $\displaystyle z$, and finally deduce $\displaystyle y$ by integrating.