Spherical Balloon Inflation Rate Exploring Radius Change

In this article, we delve into a classic calculus problem involving a spherical balloon being inflated with gas. We will explore the rate of change of the radius of the balloon as it inflates, a concept that beautifully illustrates the power of related rates in calculus. Our main focus will be on determining how quickly the radius changes when the balloon reaches specific sizes. We will investigate the scenario where a spherical balloon is inflated with gas at a constant rate of 700 cubic centimeters per minute. This means the volume of the balloon is increasing at this steady pace. Our goal is to calculate the rate of change of the radius, denoted as dr/dt, at two specific instances: when the radius (r) is 20 centimeters and when it is 45 centimeters. This problem is a prime example of related rates, a concept in calculus that deals with the rates of change of two or more variables that are related to each other. In this case, the volume and the radius of the sphere are related, and we're given the rate of change of the volume (dV/dt) and asked to find the rate of change of the radius (dr/dt).

Establishing the Foundation: Volume and Differentiation

The volume of a sphere is a fundamental concept in geometry, and it's crucial for solving this problem. The formula for the volume (V) of a sphere with radius (r) is given by:

$$V = \frac{4}{3}πr^3$$

This formula establishes the relationship between the volume and the radius of the balloon. As the balloon inflates, both its volume and radius increase. The rate at which these quantities change is what we're interested in.

To find the relationship between the rates of change of the volume and the radius, we use the technique of implicit differentiation. This involves differentiating both sides of the volume formula with respect to time (t). Remember, both V and r are functions of time, as they change as the balloon inflates. Differentiating both sides of the equation with respect to t, we get:

$$\frac{dV}{dt} = \frac{d}{dt} (\frac{4}{3}πr^3)$$

Applying the chain rule, which is essential for differentiating composite functions, we treat r as a function of t. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the outer function is ${\frac{4}{3}πr^3}$ and the inner function is r(t). Applying the chain rule, we get:

$$\frac{dV}{dt} = \frac{4}{3}π * 3r^2 * \frac{dr}{dt}$$

Simplifying the equation, we obtain a crucial relationship between the rates of change:

$$\frac{dV}{dt} = 4πr^2 \frac{dr}{dt}$$

This equation is the cornerstone of our solution. It connects the rate of change of the volume (dV/dt), which is given, to the rate of change of the radius (dr/dt), which we want to find. The equation also includes the radius (r), highlighting that the rate of change of the radius depends on the current size of the balloon.

Solving for the Rate of Change of the Radius

Now that we have the equation relating the rates of change, our next step is to isolate the rate of change of the radius, ${\frac{dr}{dt}}$. We can do this by dividing both sides of the equation by ${4πr^2}$:

$$\frac{dr}{dt} = \frac{1}{4πr^2} \frac{dV}{dt}$$

This equation provides a direct formula for calculating the rate of change of the radius (dr/dt) given the rate of change of the volume (dV/dt) and the radius (r). It clearly shows that the rate at which the radius changes is inversely proportional to the square of the radius. This makes intuitive sense: as the balloon gets larger, the same amount of added gas will result in a smaller increase in the radius.

We are given that the balloon is inflated at a rate of 700 cubic centimeters per minute. This means ${\frac{dV}{dt} = 700}$ cm³/min. We will now use this information and the formula we derived to calculate ${\frac{dr}{dt}}$ at the specified radii.

Case 1: Radius (r) = 20 centimeters

Let's first consider the case when the radius of the balloon is 20 centimeters. We substitute the given values, ${r = 20}$ cm and ${\frac{dV}{dt} = 700}$ cm³/min, into our formula:

$$\frac{dr}{dt} = \frac{1}{4π(20)^2} (700)$$

Simplifying the expression, we get:

$$\frac{dr}{dt} = \frac{700}{1600π}$$

$$\frac{dr}{dt} = \frac{7}{16π}$$

This gives us the rate of change of the radius in centimeters per minute when the radius is 20 cm. We can approximate this value to get a better sense of the magnitude of the change. Using a calculator, we find:

$$\frac{dr}{dt} ≈ 0.139\ \text{cm/min}$$

Therefore, when the radius of the balloon is 20 centimeters, the radius is increasing at a rate of approximately 0.139 centimeters per minute. This means that for every minute that passes, the radius of the balloon grows by about 0.139 cm.

Case 2: Radius (r) = 45 centimeters

Now, let's analyze the scenario when the radius of the balloon is 45 centimeters. We follow the same procedure as before, substituting ${r = 45}$ cm and ${\frac{dV}{dt} = 700}$ cm³/min into our formula:

$$\frac{dr}{dt} = \frac{1}{4π(45)^2} (700)$$

Simplifying the expression, we have:

$$\frac{dr}{dt} = \frac{700}{8100π}$$

$$\frac{dr}{dt} = \frac{7}{81π}$$

This gives us the rate of change of the radius when the radius is 45 cm. To get a numerical approximation, we calculate:

$$\frac{dr}{dt} ≈ 0.0276\ \text{cm/min}$$

Thus, when the radius of the balloon is 45 centimeters, the radius is increasing at a rate of approximately 0.0276 centimeters per minute. This is a significantly smaller rate compared to when the radius was 20 cm. This demonstrates the inverse relationship between the rate of change of the radius and the square of the radius: as the balloon gets larger, the rate at which the radius increases slows down.

Insights and Implications

The results we obtained highlight an important concept in related rates problems: the rate of change of one variable with respect to time depends on the instantaneous values of the other variables. In this case, the rate at which the radius of the balloon increases depends on the current radius. When the balloon is small (r = 20 cm), the radius increases relatively quickly because the same volume of gas is being distributed over a smaller surface area. However, when the balloon is larger (r = 45 cm), the same volume of gas is distributed over a much larger surface area, leading to a slower increase in the radius.

This problem also illustrates the power of calculus in modeling real-world phenomena. By using the concept of related rates and the formula for the volume of a sphere, we were able to quantitatively analyze the inflation of a balloon and determine how the rate of change of its radius varies with its size. This type of analysis has applications in various fields, including engineering, physics, and even economics, where understanding how different variables change in relation to each other is crucial.

Conclusion

In summary, we have successfully determined the rate of change of the radius of a spherical balloon being inflated with gas at a constant rate. We found that when the radius is 20 cm, the radius is increasing at approximately 0.139 cm/min, and when the radius is 45 cm, the radius is increasing at approximately 0.0276 cm/min. These results demonstrate the inverse relationship between the rate of change of the radius and the square of the radius. This problem provides a clear example of how related rates can be used to analyze dynamic systems and understand how different variables influence each other. The principles and techniques used in this problem can be applied to a wide range of other related rates problems, making it a valuable exercise in understanding calculus and its applications.