Question

Toward the end of their lives many stars become much bigger. Assume that they remain spherical in shape and that their masses do not change in this process. If the volume of a star increases by a factor of 872 by what factors do the following change: a) its surface area, b) its circumference, c) its diameter?

Short answer

Answer: The surface area increases by a factor of (872)^(2/3), while circumference and diameter both increase by a factor of (872)^(1/3).

Step-by-step solution

Write down given information and relevant formulas

We are given that the volume of a star increases by a factor of 872. We need to find the factors by which the surface area, circumference, and diameter change. Spherical volume formula: V = (4/3)πr^3 Surface area formula: A = 4πr^2 Circumference formula: C = 2πr Diameter formula: D = 2r

Find the new radius in terms of the old radius

We are given that the new volume is 872 times the old volume. To find the new radius in terms of the old radius, we can use the volume formula: V1 = (4/3)πr1^3 V2 = (4/3)πr2^3 Since V2 = 872V1, we have: (4/3)πr2^3 = 872(4/3)πr1^3 Dividing both sides by (4/3)π, we get: r2^3 = 872r1^3 To find r2, take the cube root of both sides: r2 = r1·(872)^(1/3)

Find the factor by which the surface area changed

Now we can find the factor by which surface area changed using the new radius, r2, in terms of the old radius, r1: A1 = 4πr1^2 A2 = 4πr2^2 = 4π(r1·(872)^(1/3))^2 Dividing A2 by A1 to find the factor by which the surface area changed, we get: Factor(A) = A2/A1 = [4π(r1·(872)^(1/3))^2] / [4πr1^2] = (872)^(2/3) The surface area increases by a factor of (872)^(2/3).

Calculate the factors for circumference and diameter changes

We will do the same for circumference and diameter: C2/C1 = [2πr1·(872)^(1/3)] / [2πr1] = (872)^(1/3) D2/D1 = [2r1·(872)^(1/3)] / [2r1] = (872)^(1/3) The circumference and diameter both increase by a factor of (872)^(1/3). In conclusion: a) The surface area changes by a factor of (872)^(2/3). b) The circumference changes by a factor of (872)^(1/3). c) The diameter changes by a factor of (872)^(1/3).

Key concepts

Spherical Volume Calculation

Understanding the calculation of spherical volume is fundamental when examining changes to celestial bodies like stars. The spherical volume helps us measure how much three-dimensional space is occupied by a sphere. The formula for the volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius of the sphere.

The existing solution for the textbook exercise used this formula to determine how an increase in volume by a factor of 872 affects other properties of the sphere. Specifically, by understanding the relationship between volume and radius, one can deduce the impact on surface area, circumference, and diameter.

For students to truly grasp this concept, it's important to remember that volume scales with the cube of the radius. This means that if you increase the radius by a certain factor, the volume increases by that factor cubed. When a star's volume increases, as per the given factor, we apply the cube root to find the change in radius. This is key to calculating the subsequent changes in the sphere's geometric properties.

Surface Area Scaling

Surface area scaling can seem complex, but it becomes clearer once you understand the relation between surface area and radius. The formula for a sphere's surface area is \( A = 4\pi r^2 \). When dealing with scaling, it's crucial to recognize that the surface area is proportional to the square of the radius.

When the exercise states that a star expands and its volume increases by a factor of 872 without mass change, its radius also changes. The surface area changes according to the square of the ratio of the new radius to the original radius. This is where the solution used the exponent \( (872)^{\frac{2}{3}} \) to express the factor by which the surface area increases. This scaling effect is not just crucial for understanding stars but is also widely applicable in other spheres of science and mathematics.

In order to improve comprehension, visualize the relationship by considering a balloon being inflated. As the balloon expands, the surface area increases, but not as quickly as the volume does. That's a practical way to remember how scaling works for surface area relative to volume.

Geometric Properties of Spheres

When discussing geometric properties of spheres, we usually refer to volume, surface area, diameter, circumference, and sometimes other derived measurements. These properties are interrelated through the radius of the sphere, which is the distance from the center of the sphere to any point on its surface.

In the given example, after determining the new radius as per the expanded volume, we can figure out how other properties, like circumference given by \( C = 2\pi r \), and diameter, indicated by \( D = 2r \), are affected by the expansion. As the radius increases by a certain factor, the circumference and diameter scale linearly with that factor. In the solutions provided, this linear scaling shows both properties increasing by the cube root of 872, which is \( (872)^{\frac{1}{3}} \).

To aid in understanding, it may be helpful to visualize a rubber band being stretched. As it is pulled, both the length (analogous to the diameter) and the overall 'edge' path (analogous to the circumference) increase in direct proportion to the extent of the stretch. This analogy can help students internalize how sphere expansion impacts its geometric properties.