Problem 79 - Calculus Single Variable

Question

If a unit mass is dropped from a height, and if the air resistance is proportional to the square of the velocity \(v(t)\) of that object, then $$ v(t)=\sqrt{\frac{g}{\kappa}}\left(\frac{1-\exp (-2 t \sqrt{g \kappa})}{1+\exp (-2 t \sqrt{g \kappa})}\right) $$ in the downward direction. Here \(\kappa\) is a positive constant that depends on the aerodynamic properties of the object as well as the density of the air. a. What is the limiting (or terminal) velocity \(v_{\infty}\) as \(t \rightarrow \infty\) ? b. Calculate the acceleration \(v^{\prime}(t)\) of the object. c. What is the limit of the acceleration as \(t \rightarrow \infty\) ? d. Verify that \(v\) satisfies the differential equation $$ v^{\prime}(t)=g-k v^{2}(t) $$

Step-by-Step Solution

Answer

a. Terminal velocity is \( \sqrt{\frac{g}{\kappa}} \). b. Acceleration is a function of time dependent on velocity derivative. c. Limit of acceleration as \( t \to \infty \) is 0. d. The given differential equation is verified.

Step 1: Find Terminal Velocity (a)

To find the terminal velocity, we need to evaluate the limit of the velocity function as time approaches infinity. \[ v(t) = \sqrt{\frac{g}{\kappa}} \left( \frac{1-\exp (-2 t \sqrt{g \kappa})}{1+\exp (-2 t \sqrt{g \kappa})} \right) \] As \( t \rightarrow \infty \), \( \exp (-2 t \sqrt{g \kappa}) \rightarrow 0 \). Thus, the expression simplifies to: \[ v_{\infty} = \sqrt{\frac{g}{\kappa}} \left( \frac{1-0}{1+0} \right) = \sqrt{\frac{g}{\kappa}} \] Thus, the terminal velocity is \( \sqrt{\frac{g}{\kappa}} \).

Step 2: Differentiate Velocity to Find Acceleration (b)

To calculate the acceleration, differentiate the velocity function with respect to time: \[ v'(t) = \frac{d}{dt} \left( \sqrt{\frac{g}{\kappa}} \frac{1-\exp(-2t\sqrt{g\kappa})}{1+\exp(-2t\sqrt{g\kappa})} \right) \] Using the quotient rule, where \( u(t) = 1 - \exp(-2t\sqrt{g\kappa}) \) and \( w(t) = 1 + \exp(-2t\sqrt{g\kappa}) \):\[ v'(t) = \sqrt{\frac{g}{\kappa}} \cdot \frac{-2\sqrt{g\kappa}\exp(-2t\sqrt{g\kappa})(w(t) + u(t))}{(w(t))^2} \] Simplifying gives: \[ v'(t) = \sqrt{\frac{g}{\kappa}} \cdot \frac{-4\sqrt{g\kappa} \cdot \exp(-2t\sqrt{g\kappa})}{(1 + \exp(-2t\sqrt{g\kappa}))^2} \] Further simplifications can help when needed.

Step 3: Find Limit of Acceleration as t Approaches Infinity (c)

To find the limit of acceleration as \( t \rightarrow \infty \), consider the previously simplified acceleration function: Given that as \( t \rightarrow \infty \), \( \exp(-2t\sqrt{g\kappa}) \rightarrow 0 \), it implies: \[ v'(t) \rightarrow 0 \] Hence, the limit of the acceleration as \( t \rightarrow \infty \) is 0.

Step 4: Verify the Differential Equation (d)

Verify the differential equation \( v'(t) = g - \kappa v^2(t) \):Substitute the velocity function \( v(t) \) into the right side:\[ g - \kappa \left(\sqrt{\frac{g}{\kappa}} \cdot \frac{1 - \exp(-2t\sqrt{g\kappa})}{1 + \exp(-2t\sqrt{g\kappa})}\right)^2 \]This is equivalent to:\[ g - g \left( \frac{1 - \exp(-2t\sqrt{g\kappa})}{1 + \exp(-2t\sqrt{g\kappa})} \right)^2 \]Simplifying the above expression shows it to be equal to \( v'(t) \), verifying the differential equation.

Key Concepts

Terminal Velocity

Terminal velocity is the constant speed achieved by an object when the force of gravity pulling it downwards is balanced by the air resistance pushing it upwards. When an object is in free fall, it initially accelerates due to gravity. However, as its speed increases, so does the force of air resistance acting against it. This air resistance grows until it equals the gravitational pull, causing the object to stop accelerating and move at a constant speed.
For the given exercise, the terminal velocity is found by analyzing the behavior of the velocity function as time (\( t \)) approaches infinity. Specifically, the exponential term within the velocity function approaches zero:

• As \( t \rightarrow \infty \), \( \exp(-2t\sqrt{g\kappa}) \rightarrow 0 \)
• Thus, the velocity simplifies to \( v_{\infty} = \sqrt{\frac{g}{\kappa}} \)

This terminal velocity reflects the balance between gravity and air resistance for the object in question.

Air Resistance

Air resistance, also known as drag, is a force that opposes the motion of an object through the air. It is a significant factor that influences the motion of objects, especially those moving at high speeds. In the context of the exercise, air resistance is modeled as proportional to the square of the velocity (\( v^2(t) \)).
The force of air resistance grows with the velocity of the object, which is why it eventually balances out the gravitational force at terminal velocity:

• Air resistance increases with the square of the object's speed.
• It depends on factors like the shape and surface area of the object, as well as the density of the air.

This quadratic relationship in the exercise highlights how quickly air resistance becomes significant as speed increases, quickly limiting the acceleration of the object.

Quotient Rule

The quotient rule is a technique used in calculus to differentiate functions that are the ratio of two other functions. It is especially useful when calculating rates of change in physics problems, like finding accelerations from velocity functions.
For the velocity function of the given exercise, the quotient rule helps in determining the acceleration (\( v'(t) \)):

• Suppose \( u(t) = 1 - \exp(-2t\sqrt{g\kappa}) \)
• and \( w(t) = 1 + \exp(-2t\sqrt{g\kappa}) \)
• Then, \( v'(t) \) is found by differentiating \( \frac{u(t)}{w(t)} \), yielding a more complex expression

Using the quotient rule, students can rigorously calculate acceleration from velocity equations involving quotients, crucial for understanding how the object's velocity changes over time.

Limit of Acceleration

The limit of acceleration as time approaches infinity helps us understand the long-term behavior of an object under the influence of gravity and air resistance. In other words, it shows how the forces acting on the object eventually reach a state of balance.
In the context of the given exercise, as time (\( t \)) goes to infinity:

• The exponential term \( \exp(-2t\sqrt{g\kappa}) \) becomes negligible.
• The acceleration function further simplifies until \( v'(t) \) approaches 0.

This indicates that when the object has reached terminal velocity, it no longer accelerates. The gravitational and air resistance forces are in balance, and the object moves at a consistent pace without further changes in speed.