# Thread: Integration Problem, sector of a circle.

1. ## 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.

2. ## Re: Integration Problem, sector of a circle.

We know that the arc length is $rx$.

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

3. ## 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?

4. ## 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.

5. ## 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.

6. ## Re: Integration Problem, sector of a circle.

We're told that $\frac{dV}{dt}=-2$, 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 $S=4\pi r^2$ for a sphere, where S is the surface area.

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

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

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

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

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

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

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

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

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

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

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

And then you can just substitute in $r=12$ 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.

7. ## Re: Integration Problem, sector of a circle.

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?
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.