To solve this problem, we need to find how fast the surface area of the bubble is changing when the radius is 11 cm, given that the volume of air is being blown into the bubble at a rate of $\frac{dV}{dt} = 10 \, \text{cm}^3/\text{sec}$.

Given:

• Rate of change of volume: $\frac{dV}{dt} = 10 \, \text{cm}^3/\text{sec}$
• Radius $r = 11 \, \text{cm}$

Step 1: Relationship between volume and radius

The formula for the volume $V$ of a sphere is: $V = \frac{4}{3}\pi r^3$ Differentiating $V$ with respect to $t$, we get: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ We can solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^2} = \frac{10}{4\pi (11)^2}$

Step 2: Calculate $\frac{dr}{dt}$

Let's compute $\frac{dr}{dt}$.The rate of change of the radius when the radius is 11 cm is approximately $\frac{dr}{dt} = 0.0066 \, \text{cm/sec}$.

Step 3: Find the rate of change of the surface area

The formula for the surface area $A$ of a sphere is: $A = 4\pi r^2$ Differentiating $A$ with respect to $t$, we have: $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$ Let's calculate $\frac{dA}{dt}$ using the values we have.The rate at which the surface area of the bubble is changing when the radius is 11 cm is approximately $1.8 \, \text{cm}^2/\text{sec}$.

Would you like more details or have any other questions?

Related Questions:

1. How does the rate of change of volume affect the surface area rate of change as the radius increases?
2. What would be the rate of change of the surface area when the radius is doubled?
3. How does the formula for surface area change with different shapes, like a cube or cylinder?
4. How would the rate of change in radius differ if the volume change rate increased to 20 cm³/sec?
5. Can you derive the formula for surface area starting from the volume of the sphere?

Tip:

In problems involving rates of change, always identify the relationship between the quantities first and differentiate carefully with respect to time.