Problem 72 - Calculus: Early Transcendental Functions

Question

Suppose that the thrust of a boat's propeller produces a constant acceleration, but that friction with water produces a deceleration that is proportional to the square of the speed of the boat.Write a differential equation for the speed \(v\) of the boat. Find equilibrium points and use a slope diagram to determine the eventual speed of the boat.

Step-by-Step Solution

Answer

The equilibrium speed of the boat is \(\sqrt{\frac{a}{k}}\). As time progresses and the speed of the boat increases, the boat will eventually reach this equilibrium speed.

Step 1: Formulate the Differential Equation

Based on the given conditions, we can set up the following differential equation for the speed \(v\) of the boat: \(\frac{dv}{dt} = a - kv^2\), where \(a\) is the constant acceleration produced by the boat's propeller, \(k\) is a constant of proportionality for the friction, and \(v\) is the speed of the boat.

Step 2: Find Equilibrium Points

The equilibrium points are obtained by solving the equation when \(\frac{dv}{dt} = 0\). This gives us \(a - kv^2 = 0\), which results in \(v = \sqrt{\frac{a}{k}}\).

Step 3: Construct the Slope Diagram

To determine the eventual speed of the boat, a slope diagram can be drawn. As the speed increases, the fraction \(\frac{dv}{dt}\) decreases until it reaches 0 at \(\sqrt{\frac{a}{k}}\). This indicates that over time the speed will approach \(\sqrt{\frac{a}{k}}\) asymptotically, becoming the equilibrium speed of the boat.

Key Concepts

Constant Acceleration

In physics, constant acceleration refers to a scenario where an object's velocity changes at a steady rate over time. It remains unchanged, meaning that the object's speed increases by a fixed amount in each unit of time.
For a boat propelled by its engine, we assume the thrust provides a consistent forward push, leading to a constant acceleration in the direction of motion.
However, when considering factors such as friction, which acts as a decelerating force,the impact combined with constant acceleration results in a net change in the speed of the boat.

• Acceleration due to thrust remains constant: this ensures that the boat speeds up steadily.
• It does not vary over time: strong acceleration ensures constant energy delivery by the engine.

This model, given as \(\frac{dv}{dt} = a - kv^2\),is simple yet insightful in analyzing the boat's motion.
Here, the term \(a\) denotes the constant acceleration, while \(kv^2\) accounts for the decelerating force due to friction.

Equilibrium Points

Equilibrium points in differential equations represent states where the system does not change.In other words, when the differential equation reaches an equilibrium state, the rate of change of variables is zero.For the boat, we set the rate of change of speed to zero,which gives the equilibrium condition as \(a - kv^2 = 0\).
Solving for \(v\), we find \(v = \sqrt{\frac{a}{k}}\).

• This equilibrium speed is crucial because it reveals the boat's eventual maximum steady speed.
• The equilibrium point here represents a balance between the propeller's thrust and the water's friction.

Therefore, this speed \(\sqrt{\frac{a}{k}}\) is the velocity at which the boat will settle after an initial period of acceleration and adjustment.

Slope Diagram

A slope diagram is an effective visual tool for understanding the behavior of solutions to differential equations.In this context, the slope diagram assists in visualizing how the speed of the boat changes over time, eventually stabilizing.
For the given differential equation \(\frac{dv}{dt} = a - kv^2\),
we can sketch a graph indicating where the speed will increase or decrease:

• Below the equilibrium speed \(\sqrt{\frac{a}{k}}\), \(\frac{dv}{dt}\) is positive, suggesting that speed is increasing.
• Above this speed, \(\frac{dv}{dt}\) would be negative if calculated, indicating a decrease, but in reality, the speed never exceeds this point as it is an upper bound.

Therefore, over time, the speed of the boat approaches this equilibrium speed.The diagram confirms that the system is stable around this point, emphasizing that even small deviations in speed will self-correct to reach the equilibrium value.