Question

a pebble is dropped into a calm pond, causing ripples in the form of concentric circles. the radius r of the outer ripple is increasing at a constant rate of 1 ft per sec. when the radius is 4 ft, at what rate is the total area disturbed water changing?

Answer 1

LOLOLOLOLOLOLOL i thought it said "the radius r of the outer nipple is increasing"

Answer 2

lol

Answer 3

well it's calculus

Answer 4

hold on - my answer is right but i havent explained it very well

DA/dt is not correct i should have written it as dA / dr

let A = area of the circle of water
A = pi r^2

dr/dt = 1

and dA / dt = dr / dt * dA / dr

dA/ dr = 2 pi r

so dA / dt = 1 * 2 pi r = 2 pi r

at r = 4 the rate of chamge of the area = dA/dt 2 pi 4 = 8pi

Answer 5

u have soo much time on ur hands^ xD

Answer 6

lol = yes i guess ur right - but its better than working!!

Answer 7

so find= da/dt or da/dr?

Answer 8

da/dt - we know what dr/dt is - its 1 and we worked out da/dr as above

do u see why da / dt = dr/dt * da/dr ? its just a matter of fractions - the dr's cancel out

Answer 9

ok, let me work it out myself and dont go anywhere lol