Problem 83 - Algebra and Trigonometry Real Mathematics, Real People

Question

A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius (in feet) of the outermost 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 \(A(r)=\pi r t^{2} .\) Find and interpret \((A \circ r)(t).\)

Step-by-Step Solution

Answer

The composition of functions \((A \circ r)(t) = 0.36 \pi t^{2}\) represents how the area of the ripple, caused by a pebble being dropped into a pond, increases as time passes.

Step 1: Substitution

First, we're going to substitute \(r(t)\) into \(A(r)\). This means that every occurrence of \(r\) in \(A(r)\) must be replaced with \(0.6t\). So, \(A(r) = \pi r^{2}\) becomes \(A(r(t)) = \pi (0.6t)^{2}\).

Step 2: Simplification

Now, we simplify the resulting expression. \(A(r(t)) = \pi (0.6t)^{2}\) simplifies to \(A(r(t)) = \pi (0.36t^{2})\) and further simplifies to \(A(r(t)) = 0.36 \pi t^{2}\).

Step 3: Interpretation

The resulting expression \(A(r(t)) = 0.36 \pi t^{2}\) is the area of the ripple at a given time \(t\). This equation tells us how the area of circles, created by a pebble that was dropped in the pond, increases over time. Since the multiplier of \(t^{2}\) is positive, we know the area of the ripple enlarges as time passes.

Key Concepts

Substitution in Functions

Substitution is a fundamental concept in algebra, especially when dealing with functions. It involves replacing a variable in an expression with another expression. This technique allows us to evaluate composite functions, which are functions composed of two or more other functions. In the context of our exercise, we are given a composite function \(A \circ r\)(t) where \(r(t)\) is a function defining the radius of a ripple at time \(t\), and \(A(r)\) represents the area as a function of the radius \(r\).

By substituting \(r(t) = 0.6t\) into \(A(r)\), we effectively combine these two separate functions into a single function that directly relates the time \(t\) to the area of the ripple. This is achieved by replacing each instance of \(r\) with \(0.6t\) which results in \(A(r(t)) = \pi (0.6t)^{2}\). This process simplifies the evaluation of the real-world phenomenon described by the functions—how the size of the ripple changes over time after the pebble is dropped into the pond.

Simplifying Expressions

Simplifying algebraic expressions is an essential skill that helps make complex relationships more understandable and calculations more manageable. It involves reducing expressions to their most basic form by performing operations such as multiplication, division, and combining like terms.

In our problem, once the substitution has been made (as described in the previous section), the next step is to simplify the expression \(A(r(t)) = \pi (0.6t)^{2}\). To make this expression simpler, we square the coefficient \(0.6\) and then multiply it by \(\pi\), which gives us \(0.36\pi t^{2}\). Simplifying algebraic expressions in this way allows us to see the most concise version of the relationship between variables—allowing us, in this case, to better understand how the area of the ripple grows as a function of time.

Concepts of Area and Circles

The concepts of area and circles are closely linked in geometry. The area of a circle is the amount of two-dimensional space inside the circle's boundary and is calculated using the formula \(A = \pi r^{2}\), where \(A\) is the area and \(r\) is the circle's radius.

In the context of the ripple created by a pebble in a pond, the circle's radius grows as a function of time, as modeled by the function \(r(t)\). To understand how the area of the circle changes over time, we use the area formula in conjunction with the function for the radius. As time \(t\) increases after the pebble strikes the water, the radius—and consequently, the area—expands.

This close relationship between the radius and the area is significant because it helps us understand how the impact of a single event (a pebble hitting water) propagates over time and space. In real-world terms, the area of a ripple—or any circular region—grows quadratically with respect to the radius, which is why the \(t^{2}\) term appears in the final simplified expression for \(A(r(t)) = 0.36\pi t^{2}\).