Problem 29 - Calculus Early Transcendentals

Question

(a) Suppose that a particle moves along an \(s\) -axis in such a way that its velocity \(v(t)\) is always half of \(s(t)\) Find a differential equation whose solution is \(s(t)\) (b) Suppose that an object moves along an \(s\) -axis in such a way that its acceleration \(a(t)\) is always twice the velocity. Find a differential equation whose solution is \(s(t) .\)

Step-by-Step Solution

Answer

For (a), \(\frac{ds}{dt} = \frac{1}{2}s(t)\). For (b), \(\frac{d^2s}{dt^2} = 2\frac{ds}{dt}\).

Step 1: Understanding the Relationship for Part (a)

We are given that the velocity of the particle, denoted as \(v(t)\), is half of its position, \(s(t)\). This can be mathematically expressed as:\[v(t) = \frac{1}{2}s(t)\]

Step 2: Expressing Velocity as a Derivative for Part (a)

Since velocity \(v(t)\) is the derivative of position \(s(t)\) with respect to time, we can write:\[v(t) = \frac{ds}{dt}\]

Step 3: Setting up the Differential Equation for Part (a)

Substitute the expression for velocity into the equation from Step 1:\[\frac{ds}{dt} = \frac{1}{2}s(t)\]This is the required differential equation for Part (a).

Step 4: Understanding the Relationship for Part (b)

For Part (b), we are given that the acceleration \(a(t)\) is twice the velocity. Therefore:\[a(t) = 2v(t)\]

Step 5: Expressing Acceleration as a Second Derivative for Part (b)

Acceleration \(a(t)\) is the derivative of velocity \(v(t)\). We can write this as:\[a(t) = \frac{dv}{dt}\]Considering velocity as \(v(t) = \frac{ds}{dt}\), acceleration can be expressed in terms of \(s(t)\) as:\[a(t) = \frac{d^2s}{dt^2}\]

Step 6: Setting up the Differential Equation for Part (b)

Using the relationship \(a(t) = 2v(t)\), and substituting the expressions for acceleration and velocity, we get:\[\frac{d^2s}{dt^2} = 2\frac{ds}{dt}\]This is the differential equation for Part (b).

Key Concepts

Calculus

Calculus plays a pivotal role in understanding motion and change. It provides the tools to model and solve problems involving rates of change, like velocity and acceleration. Calculus uses derivatives and integrals as its main building blocks. Derivatives represent the rate of change, while integrals handle the accumulation of quantities. When analyzing motion, the position of an object over time can be represented with functions, which can then be differentiated to find velocity or further differentiated to find acceleration.
In this context, differentiation helps us find how quickly position changes over time (velocity) and how velocity itself changes over time (acceleration). Key calculus concepts include:

• Derivatives: Indicate how a quantity changes as the input changes. Velocity is the derivative of position.
• Second Derivatives: Show the rate of change of the rate of change, giving us acceleration when differentiation is applied to velocity.
• Differential Equations: Equations involving derivatives that describe relationships between functions and their rates of change.

Understanding these basic calculus concepts is crucial to solving problems related to motion.

Velocity and Acceleration

Understanding velocity and acceleration is key to analyzing the movement of objects. Velocity refers to the speed of the object in a particular direction, which can change over time as the object speeds up or slows down. Acceleration describes how the velocity of an object changes with time.
In Part (a) of the exercise, velocity is expressed as half of the position function, leading to the equation \( v(t) = \frac{1}{2}s(t) \). By considering the derivative relationship \( v(t) = \frac{ds}{dt} \), we formulate a differential equation \( \frac{ds}{dt} = \frac{1}{2}s(t) \), which ties together velocity and position through calculus.
For Part (b), acceleration is described as twice the velocity, which introduces \( a(t) = 2v(t) \). As acceleration is the derivative of velocity \( a(t) = \frac{d^2s}{dt^2} \), this gives rise to another differential equation \( \frac{d^2s}{dt^2} = 2\frac{ds}{dt} \).
It's important to note:

• Velocity is the first derivative of the position function.
• Acceleration is the second derivative of the position function.
• Both are essential in modeling real-world movements and solving related problems.

Position Function

The position function is a fundamental concept in motion analysis. It describes where exactly a particle or object is located at any given time along a specific axis, typically using a function like \( s(t) \). The mathematical function conveys the object's location as time progresses.
When solving differential equations related to motion, the position function is the primary unknown we seek to solve for. By understanding how it relates to velocity and acceleration, we can derive differential equations representing its motion.
In the exercises given, the position function \( s(t) \) is connected to velocity and acceleration through differential equations:

• In Part (a), since velocity is half of the position function, solving the differential equation \( \frac{ds}{dt} = \frac{1}{2}s(t) \) provides insight into how position evolves over time.
• In Part (b), acceleration being twice the velocity leads to the equation \( \frac{d^2s}{dt^2} = 2\frac{ds}{dt} \), which is used to analyze how the position function changes more deeply.

Mastering the position function enables you to predict accurately where a moving object will be at any given point in time.