Question

A sphere is expanding in such a way that the area of any circular cross section through the sphere’s center is increasing at a constant rate of 2cm³/sec sec. At the instant when the radius of the sphere is 4 centimeters, what is the rate of change of the sphere's volume? (The volume of a sphere with radius r is given by V= 4/3 π r³.) Hint: Use the equations A=π r³ and V= 4/3 π r³.

a, 8cm³/sec sec
b. 16cm³/sec
c, 8π cm³/sec
d. 64π cm³/sec
e, 128π cm³/sec

Answer 1

The rate of change of the sphere's volume at the instant when the radius of the sphere is 4 centimeters is 32π cm³/sec. This was calculated using related rates and the fact that the area of a cross-section of the sphere is increasing at a rate of 2 cm³/sec.

Step-by-step explanation:

To find the rate of change of the sphere's volume, we can use the related rates technique in calculus. We are given that the area of a cross-section A is increasing at a constant rate of 2 cm³/sec and the volume V of a sphere with radius r is given by V = 4/3 π r³. We first express the area A of a cross-section in terms of r, which is A = π r². Since we know dA/dt = 2 cm³/sec, we can find dr/dt by differentiating A with respect to t and solving for dr/dt.

Now, using dr/dt, we can find the rate of change of volume, dV/dt, by differentiating the volume formula V = 4/3 π r³ with respect to t and substituting the value of dr/dt into this equation:

1. First, differentiate A with respect to t to find dr/dt:
   dA/dt = d(π r²)/dt
   dA/dt = 2π r (dr/dt)
   2 = 2π (4) (dr/dt)
   (dr/dt) = 1/(2π) cm/sec
2. Then, differentiate V with respect to t:
   dV/dt = d(4/3 π r³)/dt
   dV/dt = 4π r² (dr/dt)
3. Now, replace dr/dt with the value found in step 1 and r with 4 cm:
   dV/dt = 4π (4²) (1/(2π))
   dV/dt = (64π/2) cm³/sec
   dV/dt = 32π cm³/sec

Thus, the rate of change of the volume of the sphere at the instant when the radius is 4 centimeters is 32π cm³/sec.