Question

A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius (in feet) of the outer ripple is given by \(r(t)=0.6 t,\) where \(t\) is the time in seconds after the pebble strikes the water. The area of the circle is given by the function \(A(r)=\pi r^{2} .\) Find and interpret \((A \circ r)(t)\)

Short answer

The function composition \((A \circ r)(t)\) is \(0.36\pi t^{2}\), which represents the area of the circular ripple at any given time \(t\) after the pebble strikes the water. In other words, the ripple's area grows as the square of the time from when the pebble was dropped.

Step-by-step solution

Understand the Function Composition

The original problem is asking for the output of the function \(A(r(t))\), also known as function composition, or \((A \circ r)(t)\). In simple terms, this means we insert the output of the function \(r(t)\) (radius as a function of time) into the function \(A(r)\) (area as a function of radius).

Compute the Function Composition

To find the function composition, we substitute \(r(t)\) into the function \(A(r)\). Given \(r(t) = 0.6t\), we substitute this into \(A(r)\) to find \((A \circ r)(t)\) as follows: \(A(r(t)) = A(0.6t) = \pi (0.6t)^{2}\).

Simplify the Expression

Simplify the obtained expression \(\pi (0.6t)^{2}\) to give a simplified version of \((A\circ r)(t)\). \((A \circ r)(t) = \pi (0.6t)^{2} = \pi \cdot 0.36 \cdot t^{2} = 0.36\pi t^{2}\).

Interpret the Result

The function \((A\circ r)(t) = 0.36\pi t^{2}\) gives the area of the ripple at any given time \(t\) after the pebble is thrown. This demonstrates the ripple's size increasing based on the square of the time passed. This equation gives an accurate representation of how ripples in a pond grow proportionally to the square of the time since the perturbation - in this case, the thrown pebble - occurred.

Key concepts

Calculus

Calculus is a branch of mathematics that studies continuous change, and it is divided into two major areas: differential calculus and integral calculus. Differential calculus concerns the study of rates at which quantities change, which is exemplified by the derivative. Integral calculus, on the other hand, focuses on the accumulation of quantities, such as areas under curves.

In the context of the exercise given, we use differential calculus to understand how the radius of the ripples in the pond changes over time. In more complex scenarios, calculus would allow us to calculate the velocity and acceleration of the ripples' expansion. By employing the principles of calculus, we can model real-world phenomena, like the ripples caused by a pebble dropped into water.

Area of a Circle

The area of a circle is defined as the number of square units inside that circle. The formula for calculating the area, represented as A, for a circle is given as \( A(r) = \pi r^2 \) where \( r \) is the radius of the circle. The symbol \( \pi \) (pi) is a mathematical constant approximately equal to 3.14159, and it represents the ratio of a circle's circumference to its diameter.

This formula is fundamental in our exercise as it allows us to connect the concept of function composition with a geometric interpretation—specifically, how we can find the area of the concentric circles from their radii and how this area changes over time.

Concentric Circles

Concentric circles are circles with a common center. The term 'concentric' means that one circle is inside the other, with both sharing the same center point. In real-world phenomena like ripples on a water surface, each ripple represents a concentric circle. The distance from the center point to any point on the circle remains constant, and this distance is what we call the radius of the circle.

In our exercise, the pebble creates a series of widening ripples that are concentric circles. Understanding concentric circles is crucial when exploring how the area of these circles expands over time, which is precisely described by the function composition \( (A \circ r)(t) \) introduced in the exercise solution.

Radius-Time Relationship

The radius-time relationship is an essential concept when analyzing motion that involves changes in dimensions over time. In this context, the radius is not a static measurement but a dynamic one that changes as time progresses.

In the exercise, the radius-time relationship is modeled by the function \( r(t) = 0.6t \), where \( t \) represents time in seconds and \( r \) represents the radius in feet. The relationship shows that the radius increases linearly over time at a rate of 0.6 feet per second. Through the function composition \( (A \circ r)(t) \) or \( A(r(t)) \), we apply this relationship to find how the area changes as the radius changes with time, giving us insight into how rapidly the area of the ripples increases as time passes.