Problem 257 - Calculus Volume 1

Question

The formula for the volume of a sphere is \(S=\frac{4}{3} \pi r^{3},\) where \(r\) (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun. a. Suppose \(r=\frac{1}{(t+1)^{2}}-\frac{1}{12}\) where \(t\) is time in minutes. Use the chain rule \(\frac{d S}{d t}=\frac{d S}{d r} \cdot \frac{d r}{d t}\) to find the rate at which the snowball is melting. b. Use a. to find the rate at which the volume is changing at \(t=1\) min.

Step-by-Step Solution

Answer

The rate at which the snowball melts at \( t=1 \) minute is \(-\frac{\pi}{36}\) cubic feet per minute.

Step 1: Differentiate Volume S with Respect to Radius r

The formula for the volume of a sphere is given by \( S = \frac{4}{3} \pi r^3 \). We need to differentiate \( S \) with respect to \( r \).\[ \frac{dS}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4\pi r^2 \]

Step 2: Differentiate Radius r with Respect to Time t

Given the relation for the radius as \( r = \frac{1}{(t+1)^2} - \frac{1}{12} \), we need to differentiate \( r \) with respect to \( t \).We have:\[ \frac{dr}{dt} = \frac{d}{dt} \left( \frac{1}{(t+1)^2} - \frac{1}{12} \right) = -2 \cdot \frac{1}{(t+1)^3} \]

Step 3: Apply the Chain Rule

Use the chain rule \( \frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt} \) to find the rate of change of volume with respect to time.Combining the results of Step 1 and Step 2:\[ \frac{dS}{dt} = \left(4\pi r^2\right) \left(-2 \cdot \frac{1}{(t+1)^3}\right) \]Substitute \( r = \frac{1}{(t+1)^2} - \frac{1}{12} \) into the equation:\[ \frac{dS}{dt} = 4\pi \left( \left( \frac{1}{(t+1)^2} - \frac{1}{12} \right)^2 \right) \left(-2 \cdot \frac{1}{(t+1)^3}\right) \]

Step 4: Calculate the Rate at t = 1 min

Substitute \( t = 1 \) into the expression obtained in Step 3 to find the rate at which the volume of the snowball is changing.First, calculate \( r \) at \( t = 1 \):\[ r = \frac{1}{(1+1)^2} - \frac{1}{12} = \frac{1}{4} - \frac{1}{12} = \frac{1}{6} \]Now substitute \( r = \frac{1}{6} \) into the equation for \( \frac{dS}{dt} \):\[ \frac{dS}{dt} = 4\pi \left( \left( \frac{1}{6} \right)^2 \right) \left(-2 \cdot \frac{1}{8}\right) \]Simplify the expression:\[ \frac{dS}{dt} = 4\pi \cdot \frac{1}{36} \cdot -\frac{1}{4} = -\frac{\pi}{36} \]

Step 5: Conclusion

The snowball is melting at the rate of \(-\frac{\pi}{36}\) cubic feet per minute at \( t=1 \) minute.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, used to determine the rate of change of a quantity. In our exercise, we found out how the volume of a sphere changes as its radius changes. Differentiation, in this context, takes the form of deriving an expression that represents the volume of the sphere, specifically \( S = \frac{4}{3} \pi r^3 \), with respect to its radius \( r \). Here are the steps involved in differentiating this equation:

• The power rule is applied to the term \( r^3 \), resulting in \( 3r^2 \).
• Multiplying this by the coefficient \( \frac{4}{3} \pi \) gives us \( 4\pi r^2 \).

Differentiation helps us find how fast the volume is changing as we slightly alter the radius, a critical step in many real-world applications involving calculus.
By understanding differentiation, you can better comprehend how tiny changes affect larger systems and solve complex problems involving rates of change.

Integrals

While the exercise primarily utilized differentiation, understanding integrals can provide a more comprehensive grasp on calculus. Integrals, in a broad sense, help in calculating quantities that accumulate over time. If differentiation is about finding rates of change, integration is about adding infinitesimal pieces to find the whole.

Consider integration as the reverse of differentiation. If the derivative of a function gives the rate of change, the integral of this function represents the total change or accumulation. For instance, you may want to know:

• The total volume of a melting snowball over time.
• The amount of liquid collected from the melted snowball.

Integration provides the answer by summing over the infinitesimally small changes at each point in time. Understanding both differentiation and integration offers a full toolkit for solving accumulation problems and dynamic systems, expanding on what is covered by just the differentiation processes.

Application of Calculus

The exercise beautifully illustrates an application of calculus in the real world: modeling the melting of a snowball over time. Calculus allows us to not just calculate static values, but to explore dynamic systems where quantities constantly change. Here, we used:

• The Chain Rule: This was vital in linking the rate of change of the radius to the rate of change of the volume, showing how changes in one dimension affect others.
• Calculating the Rate: By substituting specific values, like \( t = 1 \) minute, we could determine the exact rate the snowball melts at that moment, \(-\frac{\pi}{36}\) cubic feet per minute.

Such calculations can have practical applications in fields ranging from meteorology to physics to engineering. Understanding how objects interact with their environment and how various quantities change over time gives insight into solution strategies for real-life challenges. Using calculus, we can predict, model, and even optimize processes and conditions for various applications.