Question

The radius of a spherical watermelon is growing at a constant rate of 2 centimeters per week. The thickness of the rind is always one-tenth of the radius. How fast is the volume of the rind growing at the end of the fifth week? Assume that the radius is initially 0 .

Short answer

The volume of the rind is growing at a rate of 152\( \pi \) cubic cm/week.

Step-by-step solution

Understand the problem

The watermelon is a sphere whose radius grows at a constant rate of 2 cm per week. We want to find the rate at which the volume of the rind is increasing at the end of the fifth week. Given that the rind's thickness is one-tenth of the radius, we'll calculate the volume of the rind as the difference between the volumes of the whole watermelon and the inner sphere (without the rind).

Express the radius and thickness

Let the radius of the watermelon at any time be denoted as \( r(t) \). At the end of the fifth week, \( r(5) = 2 \times 5 = 10 \) cm. The thickness of the rind, \( s \), is \( s = \frac{r}{10} \) hence at the end of the fifth week, \( s = 1 \) cm.

Calculate the inner radius

The inner radius, \( r_{inner} \), is the radius without the rind: \( r_{inner} = r(t) - s = 10 - 1 = 9 \) cm.

Find the volume of the whole watermelon

The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). The whole watermelon at week 5 has radius 10 cm, thus its volume is \( V_{whole} = \frac{4}{3} \pi (10)^3 = \frac{4000}{3} \pi \) cubic centimeters.

Find the volume of the inner sphere

Using the formula for the volume of a sphere regarding the inner radius, \( V_{inner} = \frac{4}{3} \pi (9)^3 = \frac{2916}{3} \pi = 972\pi \) cubic centimeters.

Determine the volume of the rind

The volume of the rind is the difference between the whole volume and the inner volume: \( V_{rind} = V_{whole} - V_{inner} = \left(\frac{4000}{3} \pi\right) - 972\pi = \frac{4000}{3} \pi - \frac{2916}{3} \pi = \frac{1084}{3}\pi \) cubic centimeters.

Differentiate the volume of the rind with respect to time

To find the rate at which the volume of the rind is increasing, we need to take the derivative of \( V_{rind} \) with respect to time \( t \). Since both \( r \text{ and } r_{inner} \) are functions of \( t \) and similar calculations are required, use the chain rule: \[ \frac{dV_{rind}}{dt} = 4 \pi r^2 \frac{dr}{dt} - 4 \pi r_{inner}^2 \frac{dr_{inner}}{dt} \].

Apply the chain rule

Since \( \frac{dr}{dt} = \frac{dr_{inner}}{dt} = 2 \) cm/week, at \( r = 10 \) and \( r_{inner} = 9 \): \[ \frac{dV_{rind}}{dt} = 4 \pi (10)^2 \times 2 - 4 \pi (9)^2 \times 2 = 800 \pi - 648 \pi = 152 \pi \text{ cubic cm/week}. \]

Key concepts

Spherical Geometry

Spherical geometry focuses on shapes and figures on a spherical surface. Unlike flat surfaces, where geometry is traditionally applied to two-dimensional shapes, spherical geometry applies to three-dimensional figures.
In problems like the growing watermelon, the sphere is the primary shape of focus. A sphere is a perfectly symmetrical geometrical object in three-dimensional space, where all points on its surface are equidistant from a single point known as the center. This uniformity makes it possible to use calculus to model changes in its size over time.

• The radius of a sphere extends from its center to any point on its surface.
• The volume of a sphere is calculated with the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius.
• When a sphere grows in size, its geometric measurements such as radius, surface area, and volume change proportionally.

Understanding spherical geometry allows us to calculate the volumes of spherical objects, such as the watermelon described in the exercise.

Differentiation

Differentiation is a key concept in calculus used to determine how a function changes at any given point. It plays a vital role in understanding how various quantities vary over time.
When we talk about the rate at which the volume of the rind grows, we are referring to differentiation. Since the volume changes with time, differentiation helps compute the velocity of this change.

• The derivative essentially represents the rate of change or slope of a function at a given point.
• Given the formulas for volumes of spheres, differentiation can help find how these volumes change as the radius changes.
• Using the chain rule helps differentiate complex functions by breaking them down into simpler parts.

In the example, to find out how fast the rind of the watermelon is growing, we differentiated the volume with respect to time. Here, both the outer and inner radii were functions of time, making the problem a perfect candidate for the application of the chain rule in differentiation.

Mathematical Modeling

Mathematical modeling translates real-life problems into mathematical language, allowing for analysis and predictions.
In this particular problem, the growing watermelon serves as a practical example of mathematical modeling. By understanding the relationship between the radius and the thickness of the rind, we can express these dynamics using mathematical equations.

• Mathematical models use known relationships to simulate the behavior of real-world phenomena.
• The model includes equations representing physical constraints like growth rates and geometric properties.
• This enables us to compute future changes, such as how fast the rind's volume will increase over a defined period.

Through this model, students can predict the growth in volume of the rind based solely on initial conditions and derived formulas. Essentially, mathematical modeling takes theoretical knowledge and applies it to practical scenarios.