Find the average rate of change of area of a circle with respect to it's radius r as r changes from 5 to 6

Answer 1

A = (pi)r^2

rate of change of A = derivitive of A = 2(pi)r

rate of change of A when (r=5) = 10(pi)

rate of change of A when (r=6) = 12(pi)

i'm not sure about this == > average rate of change of A = ( 10(pi) + 12(pi) )/2 = 11(pi)

Answer 2

A(r) = Area of circle = pi*r^2
A'(r) = 2*pi*r

A'(6) = 12pi
A'(5) = 10pi

Average rate of change from 5 to 6 = [ A'(6) - A'(5) ] / (6 - 5)

=2pi/1 = 2pi

Answer 3

You seem to be missing another factor for this question to make sense. Like time or something.

A = pi Rsquared
dA/dR = 2pi x R

If R increases by 1 every 5 seconds then, dR/dT = 1/5 = 0.2

hence, dA/dT = dA/dR x dR/dT = 2pi x R x 0.2 = 0.4piR

So the rate of change of area when R = 5 is:
0.4 x 5 x pi = 2pi unitsquare/sec

Answer 4

Rate of change relative to what. You have supplied no units and no criteria. Are you looking for relative percentage change per millimeter? Radius changes from 5 inches or 5 miles?

Answer 5

hmmm ..

d(area)/d(radius) ..... I hope you realize that this question is a bit "fuzzy" in its wording

,,,ave when r=5 to r=6

d(pi*r^2)/dr = 2*pi*r

(D(5) - D(6))/2 = 2*pi*(6-5) / 2 = -pi

loooks like answer = "minus pi"

dunno,I wouldn't bet the house on it

Answer 6

feels like there's something missing.

rate of change usually involves time.

you didn't say how long it would take to change.

i dont get it.