Question4 in Application Problems





In this page question4 in application problems we are going to see solution of first question

Question 4:

The radius of a spherical balloon is increasing at the rate of 4 cm/sec. Find  the rate of increases of the volume and surface area when the radius is 10 cm.

Solution :

Let "V' be volume of spherical balloon and "S" be the surface area.

Here we need to find "dV/dt" and "dS/dt"

dr/dt = 4 cm/sec and r = 10 cm

Volume of the spherical balloon (V) = (4/3) Π r³

Differentiate with respect to t

dV/dt = (4/3) Π 3 r² (dr/dt)

dV/dt = (4/3) Π 3 (10)² (4)

dV/dt = 16 Π  (10)²

dV/dt = 16 Π  (100)

dV/dt = 1600 Π cm³/sec

Surface area of the spherical balloon S = 4 Π r²

differentiate with respect to t

   dS/dt = 4 Π 2r (dr/dt)

   dS/dt = 8 Π r (dr/dt)

   dS/dt = 8 Π (10) (4)

   dS/dt = 80 Π  (4)

   dS/dt = 320 Π cm²/sec

Questions

Solution

(1) The radius of a circular plate is increasing in length at 0.01 cm per second. What is the rate at which the area is increasing when the radius is 13 cm?

Solution

(2) A square plate is expanding uniformly each side is increasing at the constant rate of 1.5 cm/min. Find the rate at which  the area is increasing when the side is 9 cm.

Solution

(3) A stone thrown into still water causes a series of concentric ripples. If the radius of outer ripple is increasing at the rate of 5 cm/sec,how fast  is the area of the distributed water increasing when the outer most ripple has the radius of 12 cm/sec.

Solution

(5)  A balloon which remains spherical is being inflated be pumping in 90 cm³/sec. Find the rate at which the surface area of the balloon is increasing when the radius is 20 cm.

Solution






Question4 in application problems to Rate of Change