Problem 62 - College Algebra and Calculus: An Applied Approach

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 time in seconds after the pebble strikes the water. The area of the outermost circle is given by the function \(A(r)=\pi r^{2}\) Find and interpret \((A \circ r)(t)\).

Step-by-Step Solution

Answer

The composite function \((A \circ r)(t)=0.36 \pi t^{2}\) represents the area of the outermost ripple as a function of time, which grows with the square of the time after the pebble strikes the water.

Step 1: Understanding the Composite Function

The composite function \(A \circ r(t)\) means that we substitute the function \(r(t)=0.6t\) into \(A(r)\).

Step 2: Substituting \(r(t)\) into \(A(r)\)

By substituting \(r(t)=0.6t\) into \(A(r)=\pi r^{2}\), we get \(A(r(t))=\pi (0.6t)^{2}\).

Step 3: Simplifying the Expression

The expression simplifies to \(A(r(t))=0.36 \pi t^{2}\). This represents the area of the outermost ripple as a function of time.

Step 4: Interpreting the Result

\((A \circ r)(t)=0.36 \pi t^{2}\) means that the area of the outermost ripple grows with the square of the time after the pebble strikes the water, with a multiplier of \(0.36\pi\), which is a consequence of the circle's area formula as well as the linear growth of the radius in relation to time.

Key Concepts

Understanding the Radius Function

When speaking about the radius function, we describe how the radius of a circle changes over time. If you're observing ripples in a pond from a pebble drop, the outermost ripple is an excellent example. The radius function tells us how far the edge of the ripple is from the center at any given moment.

The given function is \(r(t) = 0.6t\).

This means every second, the radius increases by 0.6 feet.

It captures the linear growth of the ripple.

It's essential to understand that this relationship is straightforward: as time doubles, so does the radius. This simple linear function makes calculations easy when predicting the ripple size after several seconds.

Calculating the Area of a Circle

The area of a circle is crucial when discussing ripples, as it helps us understand how much surface the ripple covers. The formula to calculate the area is given by \(A(r)=\pi r^{2}\).

Here, \(\pi\) (approximately 3.14159) is a constant.

\(r\) represents the radius of the circle.

The equation means you multiply \(\pi\) with the square of the radius.

For the ripple, once you know the radius, you plug it into this formula to get the area. Notice that area grows with the square of the radius. This quadratic relationship means that small changes in the radius lead to much larger changes in the area. This is why even as the radius grows linearly over time, the area expands significantly faster.

Understanding Time-Dependent Growth

When considering time-dependent growth, you're looking at how the ripple's area expands as time goes on. This relates directly to the composite function, \((A \circ r)(t)\). Starting with an equation like \(r(t)=0.6t\), you'll calculate the area using \(A(r(t))=\pi (0.6t)^2\), which results in \(0.36\pi t^2\).

• This tells us the area increases with the square of time, \(t^2\).
• The factor of \(0.36\pi\) represents how quickly it expands compared to just time alone.

The growth pattern is quadratic, meaning for every doubling of time, the area grows by four times, not just two. This illustrates why understanding how variables like time affect growth helps predict future changes accurately. This knowledge informs practical concepts, like anticipating how quickly a ripple could reach a certain size depending on the elapsed time.