Carlos is blowing air into a spherical soap bubble at the rate of 5 cm3/sec. How fast is the radius of the bubble changing when the radius is 10 cm? (Round your answer to four decimal places.)

How fast is the surface area of the bubble changing at that time? (Round your answer to one decimal places.)

The rate at which the radius of the bubble is changing at that time is

The rate at which the surface area of the bubble is changing at that time is

The formula for calculating the volume of a sphere is expressed as;

r is the radius of the sphere

Differentiating the volume with respect to time;

Given the following parameters

dV/dt = 5 cm^3/sec.

r = 10 cm

Substitute into the formula to get dr/dt

The surface area of the bubble is expressed as:

Hence the rate at which the surface area of the bubble is changing at that time is

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The answer is below

Step-by-step explanation:

a) dV/dt = 5 cm³/s

The volume (V) of the bubble is:

b) The surface area of the sphere (A) is:

2.56 is the width of the garden

step-by-step explanation:

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