Related Rates A spherical balloon is being inflated at a rate of 10cm^3/sec. How
ID: 2879205 • Letter: R
Question
Related Rates A spherical balloon is being inflated at a rate of 10cm^3/sec. How fast is the radius of the balloon increasing at the instant when the radius is 10cm? [Volume of sphere is V = 4/3 pi r^3] A 24 ft ladder is leaning against a house while the base of the ladder is pulled away from the house at a constant rate of 1 ft/sec. At what rate is the top of the ladder sliding down the side of the house at the instant when the base is 10 feet from the house? The radius of a sphere increases at a rate of 3 ft/sec. How fast is the volume increasing at the instant when the diameter is 24 ft? (See (a) for volume formula] As a snowball melts its surface area decreases at a rate of 1 cm^2/min. Find the rate at which the diameter decreases at the instant when the diameter is 6cm. [Surface area of sphere is A = 4 h r^2] The radius of a cone [i.e.. the radius of the circular base of a cone] is increasing at a rate of 3 inches/sec. The height of the cone is always 3 times the radius. Find the rate at which the volume of the cone is increasing at the instant when the radius is 7 inches. [Volume of a cone is pi/3 hr^2] Pouring sand forms a conical [i.e., shaped like a cone] pile whose height is always equal to the base diameter. The height of the pile is increasing at a rate of 5 feet/hour. Find the rate of change of the volume of sand in the conical pile at the moment when its height is 4 feet. A cylindrical tank is being filled with water at a rate of 2 m^3/min. The radius of the tank is 8 m. How fast is the water level inside the tank rising? [Volume of cylinder V = pi r^2 h]
Explanation / Answer
a)
V=(4(pi)R^3)/3
dV/dR=(4pi*3*R^2)/3
dV/dR=4piR^2
Given,
dV/dt=10cm^3/sec
R=10 cm
dV/dt=dR/dV*dV/dt
=1/(dV/dR)*dV/dt
=10/(4pi*10^2)
=1/40pi
dR/dt=1/40pi=0.007957cm/sec
Rate of change of radius is 1/40pi cm/sec or 0.007957 cm/sec.
b)
Given,
rate of change of ladder dX/dt=1 ft/sec
we have to find dY/dt at X=10 feet
b pythagorean theroem,
X^2+Y^2=24^2
100+Y^2=576
Y^2=476
Y=21.81 feet
X^2+Y^2=576
2X*dX/dt+2Y*dY/dt=0
dY/dt=-(X/Y)*dX/dt
substituting values,
dY/dt=-(10/21.81)*1
dY/dt=-0.04583
At the rate of (-0.04583) ft/sec the ladder will sliding down the side.
c)
Given
d=24 feet
so r=d/2=12 feet
rate of change of radius dr/dt=3 ft/sec
find dV/dt
V=4pir^3/3
dV/dt=4pir^2dr/dt
substituting value,
dV/dt=4pi(12)^2*3
dV/dt=5428.6721 ft^3/sec
At diameter 24 feet the volume increasing is 5428.6721 ft^3/sec.
d)
given,
d=6cm
r=d/2=3 cm
dA/dt=1 cm^3/sec
surface area A=4pir^2
taking derivative,
dA/dt=4pi*2*r dr/dt
dA/dt=8pi*r*dr/dt
substituting values,
-1=8*pi*3*dr/dt
-1=24*pi*dr/dt
dr/dt=(-1)/24pi
but we have to find the rate of change in diameter
d=2r
2dr/dt=2*(-1)/24pi
2dr/dt=(-1)/12pi
2dr/dt=-0.261799 cm/sec
Therefore our diameter decerases at a rate of 1/12pi cm/sec or 0.261799 cm/sec.