# Calculus based physics(work) and find f(x) given arc length

• February 3rd 2008, 06:49 PM
ebonyscythe
Calculus based physics(work) and find f(x) given arc length
I've got a test coming up tomorrow and these to problems were giving me some trouble. If anyone could help me along, it'd be greatly appreciated.

1.) Find a function f(x) whose arc length L(a) from (1,2) to (a,f(a)), a>1, is (1/2)a^2+(1/4)*ln(a). Hint: You will need to use the fundamental theorem of calculus at some point. Also, there are 2 possible answers and you only need to find one.

2.) A 20 lb monkey hangs at the end of a 20 ft chain that weighs a total of 10 lb. The end of the chain is attached to the monkey's foot. How much work does the monkey do in climbing the end of the chain to the top of the the chain?

For the first one I've no idea where to begin. We've learned that arc length = $\int_a^b \sqrt{1+(f'(x))^2}*dx$ but I don't know how to connect it to the information given.

As for the second problem, I really never understood how the teacher was explaining work problems relating to chain's or fluids. We were given that
http://us.metamath.org/symbols/cdelta.gifWork=(force increment)(distance)...

Edit: For the second problem, I think I've come up with the work that the monkey would do... correct me if I'm wrong:

$\int_0^{20} 20+(1/2)y dy$

Where y is the distance traveled and (20+.5y) is the weight of the monkey and chain attached to it's foot as it climbs.
• February 3rd 2008, 07:52 PM
ThePerfectHacker
Quote:

Originally Posted by ebonyscythe
1.) Find a function f(x) whose arc length L(a) from (1,2) to (a,f(a)), a>1, is (1/2)a^2+(1/4)*ln(a). Hint: You will need to use the fundamental theorem of calculus at some point. Also, there are 2 possible answers and you only need to find one.

Suppose there is a continous function $f$ on $[1,\infty)$ which is twice countinously differenciable on $(1,\infty)$ such that for any $a>1$ we have that $\int\limits_1^a \sqrt{1+f'(x)^2}~ dx = \frac{1}{2}a^2 + \frac{1}{4}\ln a$ passing through $(1,2)$, i.e. $f(1)=2$. Then by the fundamental theorem for $a>1$ it means (differenciate both sides) $\sqrt{1 + f'(a)^2} = a + \frac{1}{4a}$. Thus, $1+f'(a)^2 = \left( a + \frac{1}{4a} \right)^2 = a^2 + \frac{1}{2} + \frac1{16a^2}$. Thus, $f'(a)^2 = a^2 - \frac{1}{2}+\frac{1}{16a^2} = \left( a - \frac{1}{4a} \right)^2$. Thus, $f'(a) = \pm \left( a - \frac{1}{4a} \right)$. This means the twice continously differencial functions which solve the integral equation above are: $f(a) = \frac{1}{2}a^2 - \frac{1}{4}\ln a + k$ and $f(a) = - \frac{1}{2}a^2 + \frac{1}{4}\ln a + k$. Using the initial condition (and continuity) that $f(1) = 2$ we get that $k=3/2$ in the first class and $k=5/2$ in the second class. Thus, the two solutions are: $\frac{1}{2}x^2 - \ln x + \frac{3}{2} \mbox{ and } - \frac{1}{2}x^2 + \ln x + \frac{5}{2}$ for $x>1$.