I'm stuck on a question and was wondering if someone could explain how it's worked out.
The question is:
"The length of a side of a square is increasing at a rate of 0.2cm s^-1. At what rate is the area increasing when the length of the side is 10cm?"
Any help would be greatly appreciated, thanks in advance xx :)
Differentiation using the chain rule.?
2015-07-12 6:36 pm
回答 (3)
2015-07-12 6:44 pm
A (x) = x²
A ` (x) = 2x [ dx/dt ]
A ` (10) = [ 20 ] [ 0.2 ] = 4 cm² / s
A ` (x) = 2x [ dx/dt ]
A ` (10) = [ 20 ] [ 0.2 ] = 4 cm² / s
2015-07-12 6:44 pm
The area of the square with the side s:
A= s^2
dA/dt = 2 s ds/dt
= 2 ( 10 cm )( 0.2 cm s^-1
dA/dt = 4 cm^2 s^-1
A= s^2
dA/dt = 2 s ds/dt
= 2 ( 10 cm )( 0.2 cm s^-1
dA/dt = 4 cm^2 s^-1
2015-07-12 6:39 pm
so area of a square is A=s^2, where s is the length of a side.
You are now given that ds/dt is 0.2 cms^-1
so da/dt=2s ds/dt
so da/dt= 2(10cm)(0.2cms^-1)=4cm^2s^-1
You are now given that ds/dt is 0.2 cms^-1
so da/dt=2s ds/dt
so da/dt= 2(10cm)(0.2cms^-1)=4cm^2s^-1
收錄日期: 2021-05-01 15:22:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150712103615AAE0MJk