Proving Arc Length of a Circular Sector using the Arc Length Formula, + Area

TheNerdyGinger

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I'm having trouble with problem 6 in the screenshot below. We need to choose functions f(x) as well as our bounds a and b. I do not know how to find an appropriate function for these. I tried using sqrt(r^2-x^2) for the Arc length part, but that doesn't seem to be getting me where I want to be. Please help!
IMG_2101.jpgIMG_2103.jpg
 
I'm having trouble with problem 6 in the screenshot below. We need to choose functions f(x) as well as our bounds a and b. I do not know how to find an appropriate function for these. I tried using sqrt(r^2-x^2) for the Arc length part, but that doesn't seem to be getting me where I want to be. Please help!
View attachment 10313View attachment 10312

factor out r2 (from top and bottom)
then substitute x/r=sin(phi),
continue.....
 
I'm having trouble with problem 6 in the screenshot below. We need to choose functions f(x) as well as our bounds a and b. I do not know how to find an appropriate function for these. I tried using sqrt(r^2-x^2) for the Arc length part, but that doesn't seem to be getting me where I want to be. Please help!
Also have a look HERE.
 
factor out r2 (from top and bottom)
then substitute x/r=sin(phi),
continue.....

Thank you, that worked! The area portion is also a little tricky. We have to split up the area into the two portions such that A = A1 + A2, and use the integral formula given.
A1 is the area under the line r (triangle), A2 is the area under the curved part of the sector. A2 seems pretty obvious, I should just be able to use the same equation i used in the previous part (sqrt(r^2=x^2). What about A1 though? When drawing a triangle to represent that area, it looks like it should just be the same function. Is this correct? Would I need to just adjust the boundaries to make sure I get the right answer?
IMG_2104.jpg
 
Thank you, that worked! The area portion is also a little tricky. We have to split up the area into the two portions such that A = A1 + A2, and use the integral formula given.
A1 is the area under the line r (triangle), A2 is the area under the curved part of the sector. A2 seems pretty obvious, I should just be able to use the same equation i used in the previous part (sqrt(r^2=x^2). What about A1 though? When drawing a triangle to represent that area, it looks like it should just be the same function. Is this correct? Would I need to just adjust the boundaries to make sure I get the right answer?
View attachment 10317

if you integrate a general eqn, you should end up with eqn for A= 1/2bh,
where b is rcos(theta), or just x
and ht is rsin(theta), or sqrt(r2-x2).

for the specific question they ask, you do integrate f(x)dx;
for the triangle's area, f(x) will be the ht, dx is the width.
your limits then vary with x.
 
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Thank you, that worked! The area portion is also a little tricky. We have to split up the area into the two portions such that A = A1 + A2, and use the integral formula given.
A1 is the area under the line r (triangle), A2 is the area under the curved part of the sector. A2 seems pretty obvious, I should just be able to use the same equation i used in the previous part (sqrt(r^2=x^2). What about A1 though? When drawing a triangle to represent that area, it looks like it should just be the same function. Is this correct? Would I need to just adjust the boundaries to make sure I get the right answer?
View attachment 10317

you could use f(x)=sqrt(r2-x2) for both, but in the case of the triangle r is not constant, and you will end up with a different form for f(x) to the right of the integral sign.
e.g. if r is on 300, then r=2x, and
f(x)=xsqrt3

note: the area of sector eqn. comes from the equation of a circle, where r is constant, and the area of triangle eqn comes from pathagorean theorem, (where r is the hypotenuse, and is not constant).

for triangle area A1
,
with A1=1/2bh,
or/ A1=(1/2)r2cos(theta)sin(theta)

for the sector, A2 ;
*A2=(1/2)r2(theta)-(1/2)r2cos(theta)sin(theta)

and finally AT=A1+A2= (1/2)r2(theta)

 
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