Problem 63 - Algebra and Trigonometry

Question

Area of a Balloon A spherical weather balloon is being inflated. The radius of the balloon is increasing at the rate of 2 \(\mathrm{cm} / \mathrm{s}\) . Express the surface area of the balloon as a function of time \(t(\text { in seconds). }\)

Step-by-Step Solution

Answer

The surface area as a function of time is \( S(t) = 16\pi t^2 \).

Step 1: Recall Surface Area Formula of a Sphere

The surface area of a sphere is given by the formula \( S = 4\pi r^2 \), where \( S \) is the surface area and \( r \) is the radius of the sphere.

Step 2: Express Radius as a Function of Time

The radius \( r \) increases at a constant rate of \( 2 \, \text{cm/s} \). Therefore, we can express \( r \) as a function of time \( t \) as \( r(t) = 2t \).

Step 3: Substitute Radius in Surface Area Equation

Substitute \( r(t) = 2t \) into the surface area formula \( S = 4\pi r^2 \) to express \( S \) as a function of time: \[ S(t) = 4\pi (2t)^2 \].

Step 4: Simplify Surface Area Function

Simplify the equation from Step 3: \[ S(t) = 4\pi (4t^2) = 16\pi t^2 \]. So, the surface area as a function of time is \( S(t) = 16\pi t^2 \).

Key Concepts

Surface Area of a Sphere

Understanding the surface area of a sphere is crucial when dealing with problems involving spherical objects. The surface area is the total area that the surface of the sphere occupies, and it is calculated using the formula \( S = 4 \pi r^2 \). This formula signifies that the surface area \( S \) is directly proportional to the square of the radius \( r \) of the sphere.
The constant \( \pi \) is a mathematical constant approximately equal to 3.14159 and is crucial in calculations involving circles and spheres.

• The formula demonstrates that even a small increase in the radius can result in a significant increase in the surface area.
• This relationship is important to understand when considering how a change in radius influences the entire sphere.

Grasp this concept as it's foundational in solving problems related to the geometry of spheres.

Function of Time

A function of time describes how a given quantity varies as time progresses. In the context of this exercise, the radius of the sphere changes over time because the balloon is being inflated. Therefore, expressing the surface area as a function of time shows how the surface area evolves as time passes.
We express the radius \( r \) as \( r(t) = 2t \) because the radius increases at a constant rate of 2 cm/s.

• This expression links time \( t \) directly with the radius, simplifying calculations and predictions about the balloon's surface area.
• Applying this relationship, we can further explore how rapidly the balloon's surface area is changing at any given moment in time.

Understanding this concept allows us to see the dynamic nature of the problem and perform more complex calculations efficiently.

Radius Expansion

The expansion of the radius is central to this problem. As the balloon inflates, the radius increases at a constant rate of 2 cm/s. This consistent change is crucial because it simplifies how we express the radius as a function of time.
In this sense, radius expansion is not just a physical change but a mathematical concept that results in the expression \( r(t) = 2t \).

• This formula means that at any given time \( t \), the radius \( r \) can be easily calculated by multiplying \( 2 \) by the time in seconds.
• Applying this idea to calculate surface area makes the process straightforward, as the relationship between radius and time is clear and linear.

Understanding how the radius expands allows us to predict how other attributes of the sphere will also change over time, making it an essential concept in related rates problems.