Question

Use differentials to approximate the error and relative error in using the given measurement to calculate the value of the function, if the measurement may be off by the specified amount.
(a) Using radius 10 miles for the area of a circular island if the radius may vary by 0.02 miles

-approximate error:
-approximate relative error:

(b) Using volume V = 4 m^3 for the pressure P = 220/V of a gas in a balloon if the volume may be off by 0.2 m^3
-approximate error:
-approximate relative error:
(c) Using an intensity of I = 64 lumens for the distance
-approximate error:
-approximate

Answer 1

how about
area A= \( \pi r^2 \)
d A= \( 2\pi r \ dr \)
now replace dr with 0.02 and r with 10 to find d A

Answer 2

ok sec

Answer 3

2pi(10)(0.02) = 1.25664

Answer 4

@phi what do i do now?

Answer 5

I think the relative error is error divided by area

Answer 6

oh ok let me try that!

Answer 7

You may want to write the relative error as a percent

Answer 8

yay that worked!

Answer 9

@phi thanks soo much... Would u help me with part b too?

Answer 10

can you take the derivative of
P = 220/V
?

Answer 11

-220/v^2?

Answer 12

you should also show the differentials. in other words derivative of P is dP
and the derivative of v^-1 is -V^-2 dV
so you get
\[ dP = - \frac{220}{V^2} dV \]

now find dP using the data they gave you.

Answer 13

i get -2.75 but it says is wrong

Answer 14

(-220/4^2)*0.2=-2.75 but it says the answer is wrong for the approximate error

Answer 15

they may not want the minus sign. they don't say the error is plus or minus

Answer 16

yup that worked!

Answer 17

so now i do the voluma divided by dP?

Answer 18

relative error: error/"truth"

Answer 19

so the other way round dP/P

Answer 20

sweet thank you soo much!!