A drop of ink spreads over a blotting paper so that the circumferences

1.	A drop of ink spreads over a blotting paper so that the circumferences of the blot is 4πcm and it changes 3cm/sec. Find the rate of increase of its radius and also find the rate of increase of its area?
Answer» Given c = 4π, \(\frac{dc}{dt}\)= 3cm/sec \(\frac{dA}{dt}\)= ? \(\frac{dr}{dt}\)= ? Circumference = c = 2πr 4π = 2πr ⇒ r = 2 Again C = 2πr & A = πr² \(\frac{dc}{dt}\) = 2π \(\frac{dr}{dt}\).\(\frac{dA}{dt}\) = π . 2r.\(\frac{dr}{dt}\) 3 = 2π . \(\frac{dr}{dt}\) = π . 2. 2. \(\frac{3}{2\pi}\) ⇒ \(\frac{dr}{dt}\) = \(\frac{3}{2\pi}\)cm/ sec \(\frac{dA}{dt}\) = 6cm²/sec

A drop of ink spreads over a blotting paper so that the circumferences of the blot is 4πcm and it changes 3cm/sec. Find the rate of increase of its radius and also find the rate of increase of its area?

Answer»

Given c = 4π, \(\frac{dc}{dt}\)= 3cm/sec \(\frac{dA}{dt}\)= ? \(\frac{dr}{dt}\)= ?

Circumference = c = 2πr

4π = 2πr ⇒ r = 2

Again

C = 2πr & A = πr²

\(\frac{dc}{dt}\) = 2π \(\frac{dr}{dt}\).\(\frac{dA}{dt}\) = π . 2r.\(\frac{dr}{dt}\)

3 = 2π . \(\frac{dr}{dt}\) = π . 2. 2. \(\frac{3}{2\pi}\)

⇒ \(\frac{dr}{dt}\) = \(\frac{3}{2\pi}\)cm/ sec \(\frac{dA}{dt}\) = 6cm²/sec