Brain Teasers

**EightballLock** · December 1st 2008, 06:49 AM

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!

**EightballLock** · December 1st 2008, 08:16 AM

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.

**Laurent** · December 1st 2008, 08:43 AM

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 $C=\frac{1}{2e}$ ?

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.

**Laurent** · December 1st 2008, 08:54 AM

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 $y=f(x)$ 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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