gibbenergy

I don't know how to solve this one. Can anyone help me?

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks from the pole with a speed of 5ft/s along the straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

[math]\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x}[/math]

It can be solved quickly if you know derevative. I can post my sols quickly.

Let [math]f(x)=(x+27)^{\frac{1}{3}} - 3[/math] so [math]f(0)=0[/math] then

[math]f'(0)=\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x-0}[/math]

So you know how to do next?

It's all about the n'th root of 2' date=' and proving its irrationality (where n > 2).

 

Suppose [imath']\sqrt[n]{2}[/imath] is rational. Then [imath]\sqrt[n]{2} = \frac{a}{b}[/imath] for [imath]a, b \in \mathbb{Z}[/imath]. So we must have that [imath]2 = \frac{a^n}{b^n} \Rightarrow a^n = b^n + b^n[/imath].

 

However, this is a contradiction of Fermat's Last Theorem, so it's irrational for n > 2. Just thought it was really neat.

Sorry but I don't think it is nice. Because, in order to prove that ,you use Fermar's last theorem which is too strong. In fact, we can prove it very simple by number theory . For example:

Suppose it is rational then we can write it as

[imath] a^n= 2b^n[/imath] so [imath] a^n [/imath] must be even so

[imath] a [/imath] must be even .Put [imath] a=2k[/imath] then we get [imath] (2k)^n=2b^n[/imath]

[imath]2^{n-1}k^n=b^n [/imath]

so [imath]b [/imath] must be even.Then [imath]b=2k1 [/imath] .... And we can countinue it .So it is only ended when [imath] a=b=0[/imath] with is never happen. So [imath] \sqrt[n]{2} [/imath] can't be rational:D