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? | |
(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. | |
(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. | |
(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. |