2007
Sun
25
Mar

# Small Increments & Approximations - Sectorian Changes

(12)
Posted at 6:49 pm

Today’s a typical Sunday where Miss Loi’s schedule is ultra-packed with joss sticks sessions (aka tuition). In any case, here’s a really standard question re-occurring in many exams to herald the end of your weekend:

A sector has radius r cm and angle θ radians. Given that its area is fixed, use calculus to find the approximate percentage change in its radius when its angle decreases by 2 %.

Should be easy enough for you this time But still many students tend to be a little ‘lost’ when it comes to these kind of questions involving percentage change in angles.

You really need to know the right approach and equation to start off with. And do note the bolded part – don’t make any careless mistake!

### Revision Exercise

To show that you have understood what Miss Loi just taught you, you must:

1. kiroii says

2007
Sep
26
Wed
3:54pm

1

eh btw first and formost i dun see how does the radius change when its a circle? even if the angle changes the radius shouldnt change-.-"??

anyway let angle be 0

area = (0 r^2) /2
da/dr = 0 r

(delta 0 / 0) x 100 = -2
delta 0 = -0 / 50

delta r= dr/da x delta 0
= -1/ ( 50r)

percent change = (delta r / r) x 100

= -2%

tell me im right? but as per se i find it erratic for radius to actually change-.-

2. 2007
Sep
28
Fri
1:05pm

2

Hey kiroii, you're making the same mistake again.

Remember Miss Loi once told you in an email with regards to such questions, before you start jumping into the solution:

1) note all 'variables' in the question i.e. area, radius r and angle θ in this case).

2) Ask yourself which variables do you need to perform calculus with respect to each other?

Note that the question says that the area is fixed, so is it still a variable?

3. kiroii says

2007
Sep
28
Fri
11:52pm

3

erm k i got what ya mean but can i ask is dr/d0 gonna be easy? i get some weird equation and in de end i cant cancel-.-""

4. kiroii says

2007
Sep
29
Sat
12:04am

4

o~la~la~*grinns solved...

k i went through the question with celerity..not much thought to it..

a = (r2θ) / 2
r = √[(2a)/θ]
dr/dθ
=-r/(2 x θ) [θ is the angle)

delta r = -r/(2 x θ) x -(θ / 50)
= r / 100

percentage = delta r/ r x 100

= 1%

correcto?

5. 2007
Sep
29
Sat
1:25pm

5

Kiroii,

From your workings, it's miracle you got the correct answer even though Miss Loi didn't quite understand what you're doing there!

Yes to find your change in r (i.e. ∂r), from your small change formula i,e,

You'll first need to determine .

But like you said may not be easy to differentiate (esp with that square root), so in this case it may be easier to get instead and invert your final expression. So ...

, A represents the area.

and (quicker and simpler right?)

To find ∂θ, given θ decreased by 2%

⇒ ∂θ/θ x 100% = -2%
θ/θ = -2/100
θ = -2θ/100

So sub in all your expressions,

The percentage change in r = ∂r/r x 100%
= 1%

So your final answer is correcto but you'll need to cultivate the habit of scrutinizing the question properly before you jump in to solve it. There's no second-chance for you in the actual exam!

6. kiroii says

2007
Sep
29
Sat
1:44pm

6

eh
∂θ/θ x 100% = 2%
∂θ/θ = 2/100
∂θ = 2θ/100

er i was tinking delta 0 / 0 x 100 = -2% but from ur workings it's 2

but yet the final working is negative thus error or is my concept wrong?

PS: i tink i dun need to worry abt tis-.- rather i need to worry as to waking up on time missed my a-maths prelims paper 1 the first 40mins+ ..LOL

7. 2007
Sep
29
Sat
2:03pm

7

As for your earlier question, think of yourself as a handsome architect tasked to design a stage shaped as a sector. Given to you is A amount of material that you'll need to use up.

Suddenly your boss tells you there's not enough space in the width to accommodate your θ, and you'll need to reduce your angle by 2%, while still using the same amount the materials.

After some intensive calculation and help from a sexy math tutor, you confidently tell your boss the only way to achieve this is to increase the radius of the sector by 1%, thereby lengthening the stage instead.

... after which your boss is so impressed that he started asking you for the number of your sexy math tutor.

Ahh there you go ... a true real life application for this question

8. kiroii says

2007
Sep
29
Sat
4:29pm

8

-.-" what a remarkable story conceived *no further comments? so what ur saying is radius of circles may change if the angle changes? depending?

9. 2007
Sep
29
Sat
6:10pm

9

Oops forgot to add in the -ve sign there. Not easy to type out this entire chunk of expressions okay!

BTW, stop thinking of a fixed circle. Rather think of an elastic circular sector that can expand/contract but always having its area fixed.

10. kiroii says

2007
Sep
29
Sat
7:29pm

10

-.- expand/contract that sounds so wrong LOL in a lewd manner

11. 123 says

2007
Sep
30
Sun
12:43pm

11

Hi Ms Loi,
Can i use ∂θ = dθ/dr x ∂r instead?

12. 2007
Oct
1
Mon
12:09am

12

123,

At the end of the day, we're looking for ∂r.

So from your formula, ∂r ≈ dr/dθ x ∂θ

Isn't it the same???

But do note Miss Loi's earlier comment on choosing the easier expression to differentiate (i.e. dr/dθ or dθ/dr) and then 'inverting' it if necessary.

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