Problem 77 - Thomas Calculus

Question

Skydiving If a body of mass \(m\) falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body's velocity \(t\) sec into the fall satisfies the differential equation $$ m \frac{d v}{d t}=m g-k v^{2} $$ where \(k\) is a constant that depends on the body's aerodynamic properties and the density of the air. (We assume that the fall is short enough so that the variation in the air's density will not affect the outcome significantly.) a. Show that $$ v=\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{g k}{m}} t\right) $$ satisfies the differential equation and the initial condition that \(v=0\) when \(t=0\) b. Find the body's limiting velocity, \(\lim _{t \rightarrow \infty} v\) c. For a 160 -lb skydiver \((m g=160),\) with time in seconds and distance in feet, a typical value for \(k\) is \(0.005 .\) What is the diver's limiting velocity?

Step-by-Step Solution

Answer

The limiting velocity is approximately 178.89 feet per second.

Step 1: Verify Initial Condition

We need to check if the provided solution satisfies the initial condition. Substitute \(t = 0\) into the proposed solution: \[ v = \sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{g k}{m}} \cdot 0\right) = \sqrt{\frac{m g}{k}} \cdot 0 = 0. \] This confirms that the initial condition is satisfied since it results in \(v = 0\) at \(t = 0\).

Step 2: Plug Function Into Differential Equation

Substitute \(v = \sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{g k}{m}} t\right)\) into the given differential equation. We need to calculate \(\frac{dv}{dt}\) and check if \(m \frac{d v}{d t} = m g - k v^2\) holds.

Step 3: Differentiate the Velocity Function

Calculate the derivative of the velocity with respect to time:\[\frac{dv}{dt} = \frac{d}{dt}\left(\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{g k}{m}} t\right)\right) = \sqrt{\frac{m g}{k}} \cdot \frac{g k}{m} \text{sech}^2 \left(\sqrt{\frac{g k}{m}} t\right).\]

Step 4: Substitute in the Differential Equation

Substitute the derivative back into the differential equation:\[m \cdot \left(\sqrt{\frac{m g}{k}} \cdot \frac{g k}{m} \text{sech}^2 \left(\sqrt{\frac{g k}{m}} t\right)\right) = m g - k \left(\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{g k}{m}} t\right)\right)^2.\]Simplifying both sides confirms that they are equal.

Step 5: Find Limiting Velocity

To find the limiting velocity as \(t\) approaches infinity, evaluate:\[\lim_{t \rightarrow \infty} v = \sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{g k}{m}} t\right). \]Since \(\tanh(x)\) approaches 1 as \(x\) approaches infinity, the limiting velocity is \[\sqrt{\frac{m g}{k}}.\]

Step 6: Calculate Limiting Velocity for the Skydiver

We know \(m g = 160\) and \(k = 0.005\). Calculate the limiting velocity:\[\sqrt{\frac{m g}{k}} = \sqrt{\frac{160}{0.005}} = \sqrt{32000} = 178.89.\]

Key Concepts

Air Resistance

When a body moves through the air, it encounters resistance from the air molecules. This force is known as air resistance, and it acts opposite to the direction of motion, slowing the body down. Unlike other simple models of resistance—like the one proportional to velocity—our exercise involves air resistance proportional to the square of the velocity. This means that as the velocity of the object increases, air resistance increases even more rapidly, significantly affecting the motion.

• Air resistance in our exercise is modeled by the differential equation:
• \[ m \frac{dv}{dt} = mg - kv^2 \]
• where k is a constant that depends on factors such as the shape and surface properties of the body, as well as the air's density.

Understanding this concept helps explain why objects with different shapes and masses fall at different rates. The more streamlined an object, the lower its air resistance coefficient k, and the faster it can potentially fall.

Limiting Velocity

Limiting velocity, also known as terminal velocity, is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. As the object continues to fall, the forces acting on it—gravity and air resistance—come to an equilibrium.
In the context of our exercise, the limiting velocity can be calculated using the formula:

• \[ v_{lim} = \sqrt{\frac{mg}{k}} \]

This formula implies that the limiting velocity is determined by the gravitational force and the resistance constant. The limiting velocity is reached when the net force on the object is zero, which means that the downward gravitational force is exactly balanced by the upward force of air resistance. For the given skydiver, the limiting velocity comes out to approximately 178.89 feet per second after substituting the mass and air resistance coefficient.
Once this velocity is reached, the object will no longer accelerate and will continue to fall at this constant speed.

Initial Condition Verification

Initial conditions are crucial when solving a differential equation because they specify the starting point of the system being modeled. In our exercise, we're given that the velocity of the body at the start, or when time \( t = 0 \), is zero. This is because the body is initially at rest.
Verification of the initial condition involves checking whether our proposed solution for velocity satisfies this condition. By substituting \( t = 0 \) into our solution:

• \[ v = \sqrt{\frac{mg}{k}}\tanh\left(\sqrt{\frac{gk}{m}} \cdot 0\right) = 0 \]

We confirm that \( v = 0 \) when \( t = 0 \). Verification like this is a critical step that ensures the solution is applicable and meaningful in real-world scenarios. Initial condition checks ensure that all variables align correctly from the outset, confirming the integrity of the solution generated by the differential equation.