Question

If each side s of a square is increased by 1%, then the area is increased approximately:
0.02s
0.02s^2
0.01s^2
1%
0.01s
I dont get this one either

Answer 1

sorry but can you help?

Answer 2

0.01s^2 because its a square ?

Answer 3

its an AP calc question i dont think thats it

Answer 4

area = s^2
new area = [s(1.01])^2 = (s^2)(1.01)^2

difference is ... I am not supposed to give answers.

Answer 5

@douglaswinslowcooper don't worry, he is an AP cal student

Answer 6

oh i see kinda what you did

Answer 7

so its 0.02s^2

Answer 8

I think you're dealing with "differentials" here. If the area of a square of side s is A=s^2, then the derivative w. r. t. s is dA/ds = 2s, and the resulting differential is dA=2sds.
Supposing that the initial side of the square was s and the increase in s is 1%, then the increase in s can be expressed as 0.01s.
Then dA = 2 s ds = 2 s (0.01s). Simplify this expression. That's the approx. increase in the area of the square.

Answer 9

Yes, 2%.

In general, for small percentage changes
if f(x) = x^n,
a
p percentage change in x will produce
an np percentage change in f.

1% change in length produces 3% change in volume etc.

Answer 10

take length as 100 then %1 of it is 1 then new lentgh become 101
at the beginnnig area is 10.000 then it becomes 10.201 so there is a 201 m2 increasing
so 201/10.000*100 =2.01

Answer 11

yes this is a differentials question thank you @douglaswinslowcooper