a circular ink-blot grows at a rate of 2 sq cm per second. find the rate at which the radius is increasing after (28/11)seconds.

i tried this way but answer not coming..... :/

ans is 0.25

assuming it had zero area at t=0,

a = pi r^2
da/dt = 2pi r dr/dt
2 = 2pi (28/11) dr/dt
11/(28pi) = dr/dt

nope. I don't buy it. I think the answer lost a factor of 2 somewhere. 11/(28pi) is 0.125

Can you explain it please

To find the rate at which the radius is increasing, we can use the formula for the area of a circle.

Let's denote the radius of the circular ink-blot as r and the area as A.

We know that the rate at which the ink-blot grows is 2 square centimeters per second. This means that the rate at which the area changes with respect to time (dA/dt) is 2 cm²/s.

The formula for the area of a circle is A = πr². To find the rate at which the radius is increasing (dr/dt), we need to differentiate this equation with respect to time.

dA/dt = d/dt (πr²)
2 = π * (2r) * (dr/dt) (using the chain rule)

Now we can solve for dr/dt:

2 = 2πr * (dr/dt)
dr/dt = 2 / (2πr)
dr/dt = 1 / πr

To find the rate at which the radius is increasing after (28/11) seconds, we need to substitute this value into the equation for dr/dt:

dr/dt = 1 / πr
dr/dt = 1 / (π * r), when t = 28/11

Hence, the rate at which the radius is increasing after (28/11) seconds is 1 / (π * r).