Integration Problem, sector of a circle.

I am currently studying for my calculus 1 final (which is on the 12th) and have reached some questions I cannot answer. For those who might have the book they are in section 8.2.8 of "Modern Engineering Mathematics 4th Edition" by Glyn James.

Question #17 States:

A wire of length L metres is bent so as to form the boundary of a sector of a circle of radius r metres and angle x radians. Show that

x=(L-2r)/r

and prove that the area of the of the sector is greatest when the radius is L/4.

The second part of the question is not too difficult, I have managed to work through it, but I don't know how to approach the first part of the question. I have tried relating the circumference, using triangles, using the sector length formula, and other unsuccessful methods to derive that formula.

Help would be appreciated, thanks.

Re: Integration Problem, sector of a circle.

We know that the arc length is $\displaystyle rx$.

A sector is made up of 2 radii and this length, so $\displaystyle L=2r+rx.$. Solve this for $\displaystyle x$.

Re: Integration Problem, sector of a circle.

I see, I was thinking about it wrong. I interpreted the length as just the arc length and not the perimeter. Thanks, that solves my problem.

I am new here, is it bad form to ask a similar (concept) problem in the same thread?

Re: Integration Problem, sector of a circle.

Not at all, although as a general rule, having a maximum of 2 questions per thread to prevent things being convoluted is advised.

Re: Integration Problem, sector of a circle.

Alright, Question 2 is from the same set of questions:

Gas escapes from a spherical balloon at 2m^3.min^-1. How fast us the surface area shrinking when the radius is 12m.

I worked through this and attained an answer of 1/6m^2.min^-1 which is half the actual answer of 1/3m^2.min^-1.

I re-arranged the volume formula to give me radius then subbed that into the surface area formula, differentiated (I ended up with SA'=2/((3v)/(4pi))^1/3) ), subbed in 12 and solved. The problem is that I don't know how to incorporate the gas escape rate into the calculation.

Re: Integration Problem, sector of a circle.

We're told that $\displaystyle \frac{dV}{dt}=-2$, so we need somehow to bring time into this.

You probably know (and if not, you can calculate by differentiating the volume, as you're probably aware), that $\displaystyle S=4\pi r^2$ for a sphere, where S is the surface area.

We need to find $\displaystyle \frac{dS}{dt}$.

$\displaystyle \frac{dS}{dt}=\frac{dS}{dr}\cdot\frac{dr}{dt}$

$\displaystyle \frac{dS}{dr}=8\pi r$. The problem becomes finding $\displaystyle \frac{dr}{dt}$

We know that $\displaystyle V=\frac{4}{3}\pi r^3$ and so $\displaystyle \frac{dV}{dr}=4\pi r^2$.

We also know that $\displaystyle \frac{dV}{dt}=-2$.

Now we can use the fact that $\displaystyle \frac{dr}{dt}=\frac{dr}{dV}\times\frac{dV}{dt}$to give:

$\displaystyle \frac{dr}{dt}=\frac{1}{4\pi r^2}\cdot -2$

$\displaystyle \frac{dr}{dt}=\frac{-1}{2\pi r^2}$

$\displaystyle \frac{dS}{dt}=\frac{dS}{dr}\cdot\frac{dr}{dt}$

$\displaystyle =8\pi r\cdot\frac{-1}{2\pi r^2}$

$\displaystyle =\frac{-4}{r}$

And then you can just substitute in $\displaystyle r=12$ to finish.

Edit: The negative comes from the fact that the balloon is being depleted/shrinking; if it had been expanding, I would have used a positive. Either way, the result is the magnitude so it isn't relevant here.

Re: Integration Problem, sector of a circle.

Quote:

Originally Posted by

**sdnano** I see, I was thinking about it wrong. I interpreted the length as just the arc length and not the perimeter. Thanks, that solves my problem.

I am new here, is it bad form to ask a similar (concept) problem in the same thread?

Quote:

Originally Posted by

**Quacky** Not at all, although as a general rule, having a maximum of 2 questions per thread to prevent things being convoluted is advised.

Moderator note: Actually, it is better form to introduce all problems in the first post. Forum Rule # 9 indicates that unless you're asking for clarification on the original problem, or are asking the new question in order better to understand *precisely the same concept* as in the OP, you should start a new thread.

Thank you.