Problem 57 - Algebra and Trigonometry

Question

Area of a Ripple A stone is dropped in a lake, creating a circular ripple that travels outward at a speed of 60 \(\mathrm{cm} / \mathrm{s}\) . (a) Find a function \(g\) that models the radius as a function of time. (b) Find a function \(f\) that models the area of the circle as a function of the radius. (c) Find \(f \circ g .\) What does this function represent?

Step-by-Step Solution

Answer

(a) \( g(t) = 60t \); (b) \( f(r) = \pi r^2 \); (c) \( f \circ g(t) = 3600\pi t^2 \), representing area as a function of time.

Step 1: Model the Radius Function

The ripple travels outward at 60 cm/s. We need to express the radius as a function of time. If time is denoted by \( t \), then the radius of the circle at time \( t \) seconds is given by the linear equation: \( r(t) = 60t \). So, the function representing this is \( g(t) = 60t \).

Step 2: Model the Area Function

The area of a circle is given by the formula \( A = \pi r^2 \). We need to express the area as a function of the radius \( r \). The function that models the area is \( f(r) = \pi r^2 \).

Step 3: Find the Composite Function

The function \( f \circ g \) represents the composition of \( f \) and \( g \), which describes the area in terms of time. Substitute \( g(t) = 60t \) into \( f(r) = \pi r^2 \), giving \( f(g(t)) = \pi (60t)^2 = 3600\pi t^2 \). Thus, \( f \circ g(t) = 3600\pi t^2 \).

Step 4: Interpret the Composite Function

The function \( f \circ g(t) = 3600\pi t^2 \) describes how the area of the circular ripple changes over time. It shows that the area increases with the square of time, scaled by the factor \( 3600\pi \).

Key Concepts

Radius Function

In calculus, a radius function can describe how the radius of a circular object changes with time. In the example of a ripple in a lake, the radius function is particularly useful. When a stone creates a ripple, it moves outward at a constant speed. Here, this speed is 60 cm per second.

To find the radius function, we look at how far the ripple spreads in a given time—measured as time in seconds. We describe this relationship with the function \( g(t) = 60t \), where \( t \) is time in seconds. This equation is based on simple linear growth since the speed is constant. The radius grows linearly over time, meaning every second, the ripple increases by 60 cm in radius. Therefore, understanding the radius function helps us calculate how large the ripple grows with time.

This concept forms the basis for understanding more complex relationships, as we can estimate the size of circular objects as they change over time based on a steady rate of change.

Area of a Circle

The area of a circle is a fundamental concept in calculus and geometry. To calculate the area of a circle, you use the formula \( A = \pi r^2 \), with \( r \) representing the radius. This area function, expressed as \( f(r) = \pi r^2 \), links the size of a circle directly to its radius.

Understanding this formula is crucial because it allows you to find how much space, or area, the circle occupies in a plane. In practical situations, such as when determining the area affected by the ripple in water, this function can be a powerful tool. By squaring the radius and multiplying by the constant \( \pi \), you can predict how the circle's area grows with a changing radius.

It's essential to recall that \( \pi \) is a constant approximately equal to 3.14159. This constant ensures the formula accurately represents the perfect nature of a circle.

• Radius \( r \) directly influences area; double the radius, quadruple the area.
• This function is quadratic, meaning the area increases much faster as the radius grows.

Therefore, the area function is significant for connecting the geometric properties of the circle to its radius.

Composite Function

A composite function combines two functions to create a new function. In this case, we explore the composite function \( f \circ g \), which can describe changes in a system over time. To form this composite function, we take \( f(r) = \pi r^2 \) for area and \( g(t) = 60t \) for radius and substitute one into the other.

The resulting expression \( f(g(t)) = \pi (60t)^2 \) gives us \( f \circ g(t) = 3600\pi t^2 \). This new function provides a comprehensive view of how the area of a ripple changes with time. It tells us that the area increases with the square of time, which is logical since both speed (given as a constant) and growth (as seen with radius) play into how quickly the area expands.

• Compositing functions can simplify complex problems by merging two simpler ideas.
• The relationship between components meant for radius and area offers an elegant solution for changes over time.
• This approach aids in visualizing how dynamic systems evolve across different dimensions.

Composite functions are thus vital in calculus for addressing challenges where multiple variable transformations occur in sequence.