Problem 66 - Calculus Early Transcendentals

Question

Skydiving A skydiver in free fall subject to gravitational acceleration and air resistance has a velocity given by \(v(t)=v_{T}\left(\frac{e^{a t}-1}{e^{a t}+1}\right),\) where \(v_{T}\) is the terminal velocity and \(a>0\) is a physical constant. Find the distance that the skydiver falls after \(t\) seconds, which is \(d(t)=\int_{0}^{t} v(y) d y.\)

Step-by-Step Solution

Answer

Answer: The distance fallen by the skydiver after t seconds is given by the formula: \(d(t)=\frac{v_{T}}{2a}(-at+\ln(e^{at}+1)-\ln(2))\).

Step 1: Identify the given information

We know the velocity equation for the skydiver as a function of time: \(v(t)=v_{T}\left(\frac{e^{a t}-1}{e^{a t}+1}\right)\). The goal is to find the distance fallen after \(t\) seconds, which can be found by integrating the velocity function with respect to time: \(d(t)=\int_{0}^{t} v(y) d y.\)

Step 2: Rewrite the integral with given velocity function

Replace \(v(y)\) with the given function in the integral: \(d(t)=\int_{0}^{t} v_{T}\left(\frac{e^{a y}-1}{e^{a y}+1}\right) dy.\)

Step 3: Factor out constants

We can take the constant term \(v_{T}\) out of the integral: \(d(t)=v_{T}\int_{0}^{t} \left(\frac{e^{a y}-1}{e^{a y}+1}\right) dy.\)

Step 4: Perform the integration

Now, we need to evaluate the integral: \(d(t)=v_{T}\int_{0}^{t} \left(\frac{e^{a y}-1}{e^{a y}+1}\right) dy.\) Make a substitution: Let \(u=e^{ay}\). Then, \(du = a e^{ay} dy\). So, \(dy = \frac{du}{a e^{ay}}=\frac{du}{a u}\). The integral becomes: \(d(t) = \frac{v_{T}}{a} \int_{1}^{e^{at}} \frac{u-1}{u(u+1)} du\).

Step 5: Use partial fraction decomposition

We can rewrite the integrand as a sum of simpler fractions: \(\frac{u-1}{u(u+1)}=\frac{A}{u}+\frac{B}{u+1}\). Solving for \(A\) and \(B\), we find \(A=-\frac{1}{2}\) and \(B=\frac{1}{2}\). So the integral becomes: \(d(t)=\frac{v_{T}}{a}\left(-\frac{1}{2}\int_{1}^{e^{at}}\frac{1}{u}du+\frac{1}{2}\int_{1}^{e^{at}}\frac{1}{u+1}du\right)\).

Step 6: Evaluate the integrals

We now evaluate the integrals separately: \(d(t)=\frac{v_{T}}{a}\left(-\frac{1}{2}(\ln u)|_{1}^{e^{at}}+\frac{1}{2}(\ln(u+1))|_{1}^{e^{at}}\right)\). Now, plug in the limits of integration: \(d(t)=\frac{v_{T}}{a}\left(-\frac{1}{2}(\ln e^{at}- \ln 1)+\frac{1}{2}(\ln(e^{at}+1)-\ln(2))\right)\).

Step 7: Simplify the expression

Finally, simplify the expression: \(d(t)=\frac{v_{T}}{a}\left(-\frac{1}{2}(at-0)+\frac{1}{2}(\ln(e^{at}+1)-\ln(2))\right)\). Combine terms and express the result as: \(d(t)=\frac{v_{T}}{2a}(-at+\ln(e^{at}+1)-\ln(2))\). Now we have the distance fallen by the skydiver after \(t\) seconds, given by the formula: \(d(t)=\frac{v_{T}}{2a}(-at+\ln(e^{at}+1)-\ln(2))\).

Key Concepts

Velocity Function

When exploring the movement of objects under gravity, like a skydiver, the velocity function helps describe how velocity changes over time. In this case, our velocity function given is:
\[ v(t) = v_{T}\left(\frac{e^{at} - 1}{e^{at} + 1}\right) \]
Here:

• \(v_{T}\) is the terminal velocity. This is the constant speed the skydiver eventually reaches when the force of air resistance equals the force of gravity.
• \(a\) represents a positive constant that affects how quickly the skydiver approaches terminal velocity.

This function expresses velocity in terms of the exponential constant \(e\), reflecting the impact of air resistance which varies over time. Understanding this function is fundamental in calculating how far the skydiver falls, as distance is the integral of velocity over time.

Partial Fraction Decomposition

Before you can integrate certain rational functions like \(\frac{u-1}{u(u+1)}\), one useful technique is called partial fraction decomposition. This method allows us to express a complicated fraction as the sum of simpler fractions.
In our exercise, we decompose the fraction:
\[ \frac{u-1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \]
This equation needs to be solved to find the values of \(A\) and \(B\). By equating coefficients, we find:

• \(A = -\frac{1}{2}\)
• \(B = \frac{1}{2}\)

This decomposition breaks down the expression into parts that are easier to integrate. Once decomposed, the integration can be executed separately for each component, leading to more straightforward calculations.

Integration by Substitution

Integration by substitution is a method that simplifies integration by transforming the variable of integration. It's particularly useful when dealing with functions that contain nested operations.
For the given exercise, this method involves substituting \(u = e^{ay}\). Then, differentiate to find \(du\):
\[ du = ae^{ay} dy \quad \Rightarrow \quad dy = \frac{du}{au} \]
This substitution changes the original variable \(y\) to \(u\), transforming the integral into:
\[ \int_{1}^{e^{at}} \frac{u-1}{u(u+1)} du \]
With this new integral, we simplify the function using partial fraction decomposition and integrate each term independently. Substitution makes the integral manageable and leads to solving complex problems with higher efficiency.