Problem 30
Question
A stone dropped into a calm lake causes a series of circular ripples. The radius of the outer one increases at \(2.00 \mathrm{ft} / \mathrm{s} .\) How rapidly is the disturbed area changing at the end of 3.00 s?
Step-by-Step Solution
Verified Answer
The disturbed area is changing at a rate of \(24.00 \pi \text{ ft}^2/\text{s}\) at the end of 3.00 s.
1Step 1: Identify Given Information
The problem provides the rate at which the radius of the circular ripple increases, which is \(2.00 \text{ft/s}\). It also asks for the rate at which the area is changing after 3.00 seconds.
2Step 2: Write Down the Formula for the Area of a Circle
The area, \(A\), of a circle is given by the formula \(A = \pi r^2\), where \(r\) is the radius of the circle.
3Step 3: Apply the Chain Rule to Find the Rate of Change of the Area
To find how rapidly the area is changing, we differentiate both sides of the area formula with respect to time \(t\), obtaining \(\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}\).
4Step 4: Substitute Known Values
Substitute \( r = 2.00 \text{ ft/s} \times 3.00 \text{ s} = 6.00 \text{ ft}\) and \(\frac{dr}{dt} = 2.00 \text{ ft/s}\) into the expression for \(\frac{dA}{dt}\) to find the rate at which the area is changing after 3 seconds.
5Step 5: Calculate the Rate of Change of the Area
After substituting the values, we get \(\frac{dA}{dt} = 2 \pi \times 6.00 \text{ ft} \times 2.00 \text{ ft/s} = 24.00 \pi \text{ ft}^2/\text{s}\), which is the rate of change of the area after 3 seconds.
Key Concepts
Circular Ripple AreaCalculus in PhysicsChain Rule Application
Circular Ripple Area
Imagine a pebble dropping into a still pond and forming a circular ripple that expands over time. Understanding how the area of this ripple changes as it grows is a fascinating application of calculus.
Firstly, to grasp the concept of the changing area, recall the fundamental formula for the area of a circle: the area, denoted as \( A \), is given by \( A = \text{\pi} r^2 \), where \( r \) is the radius. In our ripple scenario, the radius is not constant; it is growing over time due to the constant spread of the ripples. The key here is to realize that as time progresses, the radius—and consequently, the area of the ripple—increases.
During the initial moments after the pebble impacts the water, the ripple is small and its area is minimal. But as the ripple radiates outward, the area increases at a rate determined by how fast the radius is expanding.
Firstly, to grasp the concept of the changing area, recall the fundamental formula for the area of a circle: the area, denoted as \( A \), is given by \( A = \text{\pi} r^2 \), where \( r \) is the radius. In our ripple scenario, the radius is not constant; it is growing over time due to the constant spread of the ripples. The key here is to realize that as time progresses, the radius—and consequently, the area of the ripple—increases.
During the initial moments after the pebble impacts the water, the ripple is small and its area is minimal. But as the ripple radiates outward, the area increases at a rate determined by how fast the radius is expanding.
Calculus in Physics
Calculus plays a pivotal role in physics, particularly when dealing with dynamic systems where quantities change over time. One such application is the study of motion and change, exemplified by our example of circular ripples on water.
Mathematically, when we want to know how quickly something like the ripple's area is changing, we seek the rate of change. This rate can be conceptualized as the derivative of the area with respect to time, denoted as \( \frac{dA}{dt} \). Derivatives form the core of differential calculus, enabling us to calculate instantaneous rates of change -- just as the velocity of an object is the instantaneous rate of change of its position.
By bringing calculus into physics, we can describe the behavior of physical systems with precision. From the acceleration of a car to the expansion of a ripple on a lake, calculus allows us to quantify and represent these changes with equations that model our observations with remarkable accuracy.
Mathematically, when we want to know how quickly something like the ripple's area is changing, we seek the rate of change. This rate can be conceptualized as the derivative of the area with respect to time, denoted as \( \frac{dA}{dt} \). Derivatives form the core of differential calculus, enabling us to calculate instantaneous rates of change -- just as the velocity of an object is the instantaneous rate of change of its position.
By bringing calculus into physics, we can describe the behavior of physical systems with precision. From the acceleration of a car to the expansion of a ripple on a lake, calculus allows us to quantify and represent these changes with equations that model our observations with remarkable accuracy.
Chain Rule Application
The chain rule is a powerful tool in calculus, especially when one quantity's rate of change depends on another quantity that is also changing with respect to time. In our example of the circular ripple, the rate of change of the area of the ripple depends on the radius, which in turn depends on time.
Following the chain rule, to find the rate of change of the area, we compute the derivative of the area with respect to the radius and then multiply by the derivative of the radius with respect to time. The formula for the area's rate of change is \( \frac{dA}{dt} = 2 \text{\pi} r \frac{dr}{dt} \). This relationship simplifies a seemingly complex problem down to a straightforward calculation once we know the rate at which the radius is changing.
In our circular ripple example, by knowing the constant rate at which the radius increases (\( \frac{dr}{dt} \)), we can easily compute the ever-changing rate at which the circle's area expands. The chain rule allows us to link these rates together, giving us a complete picture of the dynamics at play.
Following the chain rule, to find the rate of change of the area, we compute the derivative of the area with respect to the radius and then multiply by the derivative of the radius with respect to time. The formula for the area's rate of change is \( \frac{dA}{dt} = 2 \text{\pi} r \frac{dr}{dt} \). This relationship simplifies a seemingly complex problem down to a straightforward calculation once we know the rate at which the radius is changing.
In our circular ripple example, by knowing the constant rate at which the radius increases (\( \frac{dr}{dt} \)), we can easily compute the ever-changing rate at which the circle's area expands. The chain rule allows us to link these rates together, giving us a complete picture of the dynamics at play.
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