Verified answer

hey

well read the problem and see..,

now we have a spherical object

the area decreases with the time, so there would be a tangent to a curve showing the curve falling... in the area-time graph..

that tangent is equal to dA/dt

implies dA/dt= - 1 cm^3/min (here negative shows decrasing)

now area of a sphere is 4piR^2

dA/dt= d(4piR^2)/dt

R= D/2 (radius is half of diameter)

= d(pi (D)^2)/dt

= 2pi D dD/dt (chain rule)

where D= 10cm

and dA/dt (on LHS)= 1

therefore dD/dt= - 1/20pi

that means when area decreases at 1cmsq./min then diameter decreases at 1/20pi units...

dA/dt is the rate of decrease of area, and is 1 cm^2/sec.

dr/dt is the rate of decrease of radius.

dd/dt is the rate of decrease of radius and is to be found.

The relations ARE:

A = 4 pi r^2, and d = 2r, so r = d/2

Then A = 4 pi(d/2)^2 = pi d^2

Differentiate implicitly w.r.t. time:

dA/dt = 2 pi d dd/dt

Substitute, using the fact that d = 10:

- 1 = 2 pi (10) dd/dt

Solve:

dr/dt = -1/20pi