Problem 30 - Calculus Early Transcendentals

Question

Suppose that a body moves along an \(s\) -axis through a resistive medium in such a way that the velocity \(v=v(t)\) decreases at a rate that is twice the square of the velocity. (a) Find a differential equation whose solution is the velocity \(v(t)\). (b) Find a differential equation whose solution is the position \(s(t)\).

Step-by-Step Solution

Answer

(a) \( \frac{dv}{dt} = -2v^2 \); (b) \( \frac{dv}{dt} = -2 \left( \frac{ds}{dt} \right)^2 \)

Step 1: Understand the Problem

The problem describes a situation where velocity decreases at a rate twice the square of the velocity. This indicates that the rate of change of velocity with respect to time is related to the velocity itself. We need to find a differential equation for this scenario.

Step 2: Define the Velocity Differential Equation

We know from the problem statement that the rate of change of velocity, \( \frac{dv}{dt} \), is twice the square of the velocity. Hence, we can express this relationship as a differential equation: \[ \frac{dv}{dt} = -2v^2 \]Here, the negative sign indicates that the velocity is decreasing.

Step 3: Solve for Position Differential Equation

To find the position function, we relate velocity to position. Since velocity is the derivative of position with respect to time, \( v = \frac{ds}{dt} \), we substitute \( v \) in the differential equation to get:\[ \frac{dv}{dt} = -2 \left( \frac{ds}{dt} \right)^2 \]This is the differential equation relating the position \( s(t) \).

Key Concepts

Velocity and Its Differential Equation

Velocity describes how fast an object is moving and in which direction. In this exercise, the velocity of a body moving along an axis decreases at a rate described by a specific mathematical relationship. We are told that the rate of change of velocity, \( \frac{dv}{dt} \), is negative and is equal to twice the square of the velocity itself. This can be written as the differential equation: \[ \frac{dv}{dt} = -2v^2 \]

• \(v\) is the velocity
• \(t\) is the time
• The negative sign indicates that velocity is decreasing over time

Such an equation is nonlinear and specifically a first-order differential equation due to its structure. Solving this will give us the function \(v(t)\), the velocity as a function of time, which tells us exactly how the speed and direction of the body change over time through a resistive medium.

Position and Its Relation to Velocity

Position tells us where an object is along an axis. To find the position of the body, we need to relate it to the velocity since velocity is essentially the rate of change of position. Mathematically, this is expressed as \(v = \frac{ds}{dt}\), where \(s\) is the position and \(\frac{ds}{dt}\) represents the rate of change of position with respect to time. If we substitute this relationship into our velocity differential equation, we have \[ \frac{dv}{dt} = -2 \left( \frac{ds}{dt} \right)^2 \] This new differential equation links the position \(s(t)\) with its subsequent derivative, which is crucial when you need to calculate the exact position at any given time as the body moves through a resistive medium.

• This step involves calculus to relate position to velocity
• Understanding the derivative \(\frac{ds}{dt}\) is crucial
• Finding \(s(t)\) involves integrating the function derived from velocity

The Role of a Resistive Medium

A resistive medium is a type of environment through which an object moves and experiences resistance to its motion. This resistance typically results in a decrease in velocity, as described in the exercise. For example, air resistance is a well-known resistive force that acts opposite to the direction of motion, slowing down moving objects. In our context, the resistive medium leads to the specific rate of decrease in velocity as given by the differential equation, \( \frac{dv}{dt} = -2v^2 \). Here's why the resistive medium is important:

• It affects the motion directly by altering the velocity
• It causes non-linear changes in velocity, shown by the \(v^2\) term
• Understanding resistance helps predict how long an object can maintain its speed

By comprehending how resistive forces influence motion, we can better understand real-world scenarios such as vehicle deceleration, the motion of projectiles, and much more.