Approximating change

[esloveday]

Q: A cube's side is measured with an error of 1%, what is the approximate percentage error in SA.

A: SA = 6x[sup:30ribe5n]2[/sup:30ribe5n], SA'=12x, x[sub:30ribe5n]0[/sub:30ribe5n]=x, h=0.01x

so f(x+0.01x) = 6x[sup:30ribe5n]2[/sup:30ribe5n]+0.01x(12x)
= 6.12x[sup:30ribe5n]2[/sup:30ribe5n]

Is this accurate? Does it mean the approximate percentage error is 12?

 

[soroban]

Hello, esloveday!

I used a standard approach.

A cube's side is measured with an error of 1%.
What is the approximate percentage error in surface area?

\(\displaystyle \text{Surface area: }\:S \:=\:6x^2\)

\(\displaystyle \text{Then: }\:dS \:=\:12x\,dx\)

\(\displaystyle \text{We have: }\:dS \:=\:12x(0.1x) \:=\:0.12x^2\)

\(\displaystyle \text{Percentage error: }\:\frac{dS}{S} \:=\:\frac{0.12x^2}{6x^2} \:=\:0.02 \:=\:2\%\)

 

[esloveday]

Ah, so my approximate is 6.12x[sup:1sjtnswx]2[/sup:1sjtnswx] whereas the actual is 6x[sup:1sjtnswx]2[/sup:1sjtnswx] so my % error is (6.12x[sup:1sjtnswx]2[/sup:1sjtnswx]-6x[sup:1sjtnswx]2[/sup:1sjtnswx])/6x[sup:1sjtnswx]2[/sup:1sjtnswx]=2 Thanks for the help!

 

[Deleted member 4993]

Ah, so my approximate is 6.12x[sup:1mdq71aa]2[/sup:1mdq71aa] whereas the actual is 6x[sup:1mdq71aa]2[/sup:1mdq71aa] so my % error is (6.12x[sup:1mdq71aa]2[/sup:1mdq71aa]-6x[sup:1mdq71aa]2[/sup:1mdq71aa])/6x[sup:1mdq71aa]2[/sup:1mdq71aa]=2 so is 2% too large Thanks for the help!

It depends on the situation - that is if you are building a box to hold clothes - 2% is quite acceptable.

If you are building a box to generate laser - 2% is not acceptable.