Problem 67 - Calculus Single Variable

Question

If the air resistance to a falling object is proportional to the velocity of that object, then the velocity of that object when dropped from a height \(H\) is $$v(t)=-k g\left(1-e^{-t / k}\right)$$ Here \(k\) is a positive constant and \(g\) is the constant acceleration due to gravity. Calculate the height \(y\) of the object as a function of \(t\).

Step-by-Step Solution

Answer

The object's height as a function of time is \(y(t) = H - kgt - kgk e^{-t/k}\).

Step 1: Understand the Problem

We are given the velocity function of a falling object with air resistance as \(v(t) = -kg(1-e^{-t/k})\). We need to find the height \(y\) of the object over time, given its starting height is \(H\).

Step 2: Relate Velocity and Position

Velocity is the derivative of position with respect to time. Thus, \(v(t) = \frac{dy}{dt}\). We need to integrate the velocity function to determine the position function \(y(t)\).

Step 3: Integrate the Velocity Function

Integrate the velocity function \(v(t) = -kg(1-e^{-t/k})\) to find the position function. The integral is given by:\[y(t) = \int -kg(1-e^{-t/k}) \,dt = -kg \int (1-e^{-t/k}) \, dt\]

Step 4: Simplify the Integral into Two Parts

Split the integral into two parts:1. \(-kg \int 1 \, dt = -kgt\)2. \(-(-kg) \int e^{-t/k} \, dt = kg \int e^{-t/k} \, dt\)

Step 5: Calculate the First Part of the Integral

The integral of a constant \(1\) is straightforward:\[-kg \int 1 \, dt = -kgt + C\]where \(C\) is the constant of integration.

Step 6: Calculate the Second Part of the Integral

The integral of the exponential function is:\[kg \int e^{-t/k} \, dt = -kgk e^{-t/k}\]Combine this with the first part of the integral.

Step 7: Combine the Integrals

Add up the results from Step 5 and Step 6:\[y(t) = -kgt + kgk e^{-t/k} + C\]Simplify to get:\[y(t) = -kgt - kgk e^{-t/k} + C\]

Step 8: Apply Initial Condition

Since the object is dropped from height \(H\), at \(t = 0, y(0) = H\). Substitute \(t = 0\) in the equation to find \(C\):- \(y(0) = -kgt + kgk \times 1 + C = H\)- Solving for \(C\), we find \(C = H + kgk\).

Step 9: Determine Final Position Function

Substitute \(C\) into the position equation:\[y(t) = -kgt + kgk e^{-t/k} + H + kgk\]This simplifies to:\[y(t) = H - kgt - kgk e^{-t/k}\]This is the height of the object as a function of time.

Key Concepts

Differential Equations

Differential equations are equations that relate a function with its derivatives. In calculus, they play a central role in modeling situations where there is change, such as with motion, growth, or decay. In our exercise, the velocity function is derived solving a differential equation, reflecting how the speed of a falling object changes over time due to forces like gravity and air resistance.

• A differential equation connects rates of change (derivatives) with the function itself.
• Solve them to understand complex dynamics like our falling object scenario.

When dealing with such problems, we often start with known rates, like the velocity function, and work backward to understand the overall behavior, such as position.

Integration

Integration is a fundamental calculus concept used to find quantities where the rate of change is known, like finding area under curves or, more pertinent here, deriving position from velocity.

• Integration can reverse differentiation, moving from a derivative (like velocity) back to the original function (like position).
• In this exercise, we integrated the velocity equation to discover the position formula: \(y(t) = -kgt + kgk e^{-t/k} + C\).

By breaking down the velocity function into simpler parts, we were able to calculate the position over time, helping us track the height of the object as it descends due to gravity.

Velocity Function

A velocity function in calculus describes how speed (or rate of movement) alters over time. For our exercise, the velocity function \(v(t) = -kg(1-e^{-t/k})\) was provided to represent the velocity of a falling object with air resistance.

• The function shows how velocity is not constant, influenced by dependent variables like time \(t\).
• Here, the negative sign indicates downward motion, typical in problems involving gravity.

Understanding this function is key before integrating, as it tells us how fast the object is moving each moment, while integration will reveal its position over time.

Position Function

Position function gives us the location of an object at any given time. It is derived by integrating a velocity function. In this case, integrating \(v(t) = -kg(1-e^{-t/k})\) allows us to find the height of a falling object over time.

• The initial condition \(y(0) = H\) shows the starting height and is crucial for solving the constant of integration.
• The final position function \(y(t) = H - kgt - kgk e^{-t/k}\) provides a complete view of the object's descent.

This function not only shows how far the object travels from its initial position but also helps predict future positions at different times \(t\), making it indispensable in motion-related analyses.