# Question5 in Application Problems

In this page question5 in application problems we are going to see solution of first question

Question 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 :

Let "V" be the area and "r" be the radius of the balloon

Now we need to find at which the surface area of the balloon is increasing when the radius is 20 cm

That is find dS/dt and r = 20 cm

dV/dt = 90 cm³/sec.

Surface area of the balloon (S) = 4Π r²

V = (4/3) Π r³

dV/dt = (4/3) Π 3 r² (dr/dt)

90 = (4/3) Π 3 (20)² (dr/dt)

90 = 4 Π (400) (dr/dt)

90 = 1600 Π  (dr/dt)

90/1600 Π = (dr/dt)

dr/dt = 9/160Π

Surface area of the spherical balloon = (S) = 4Π r²

dS/dt = 4 Π (2r) dr/dt

dS/dt = 4 Π 2(20) (9/160Π)

dS/dt = 9 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 (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

Question5 in application problems to Rate of Change