Question

Use a differential to approximate the change in the area of a circle when the radius decreases from 3.0 cm to 2.9 cm. A=(pi)r^2, where r is the radius, in centimeters.

Answer 1

i have the formula \[dy=f \prime(x)dx\] and dx=0.1 so far

Answer 2

\[\pi\times 3^2-\pi\times 2.9^2=\pi(9-8.41)\]

Answer 3

so (pi)0.59?

Answer 4

the formula i have above, is it correct? can it be done that way or was i way off track?

Answer 5

i think you can use "differentials" but that is not what i did, i subtracted

Answer 6

how would i do it using differentials? i'm sure i'll have to show work on my final

Answer 7

\[A(r)=\pi r^2\]
\[\frac{dA}{dr}=2\pi r\]
\[dA=2\pi r dr\]

Answer 8

and as you said, your \(dr=.1\) and your \(r\) is 3

Answer 9

so use the original radius? and where dx/dr is decreasing would it be negative or is it always positive?

Answer 10

so you would get
\[2\times \pi \times 3\times .1=.6\pi\]which is close to what i got exactly by subtracting

Answer 11

oh good point!

Answer 12

i guess it depends on how you interpret the question. change is definitely negative (it decreases) but the area the change represents is positive (because it is an area)

so yes, you might want to write \(-.6\pi\)

Answer 13

ok. i think i understand it now. have to keep practicing. Thank You @satellite73

Answer 14

yw, my pleasure