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 \(d(t)=\int_{0}^{t} v(y) d y\)

Short answer

Question: Calculate the distance fallen by a skydiver after time t, considering gravitational acceleration and air resistance, given the velocity function \(v(t)=v_{T}\left(\frac{e^{at}-1}{e^{at}+1}\right),\) where \(v_{T}\) is the terminal velocity and \(a>0\) is a physical constant. Answer: The distance fallen by the skydiver after time t can be found through the formula \(d(t)=\frac{v_T}{a}\left[ ((e^{at}+1)\ln (e^{at}+1) - (e^{at}+1) -2\ln 2 +2)-\ln\left(\frac{e^{at}+1}{2}\right)\right]\).

Step-by-step solution

Write down the given velocity function

We are given the velocity function \(v(t)=v_{T}\left(\frac{e^{at}-1}{e^{at}+1}\right),\) where \(v_{T}\) is the terminal velocity and \(a>0\) is a physical constant.

Integrate the velocity function

To find the distance fallen, we need to integrate the velocity function with respect to time, as \(d(t)=\int_{0}^{t} v(y) d y\). So, the task is to find the integral: \(\int_{0}^{t} v_{T}\left(\frac{e^{ay}-1}{e^{ay}+1}\right) dy\).

Apply the substitution method

To solve this integral, we can apply the substitution method. Let \(u = e^{ay}\). Then, the derivative of \(u\) with respect to y will be \(\frac{du}{dy} = ae^{ay} = au\). From this, we can find \(\frac{dy}{du} = \frac{1}{au}\). Now, the integral will become \(\int_{0}^{t} v_{T}\left(\frac{u-1}{u+1}\right) \frac{1}{au} du\). The limits of the integral will change accordingly, as \(u=1\) when \(y=0\), and \(u=e^{at}\) when \(y = t\). So, the integral now is: \(\int_{1}^{e^{at}} v_{T}\left(\frac{u-1}{u+1}\right) \frac{1}{au} du\).

Calculate the integral

Now we can calculate the integral. To do that, we can notice that, after multiplying the \(v_{T}\) and the \(\frac{1}{au}\), we get \(\frac{v_T}{au}\). We can also split the integral in two terms, in order to work better with the substitution: \( d(t) = \int_{1}^{e^{at}} \frac{v_T(u-1)}{au(u+1)} du = \frac{v_T}{a} \left[ \int_{1}^{e^{at}} \frac{u}{u+1} du -\int_{1}^{e^{at}} \frac{1}{u+1} du \right] \) The first integral can be solved by another substitution: Let \(w=u+1\), then \(dw=du\). As limits, when \(u=1\), \(w=2\), and when \(u=e^{at}\), \(w=e^{at}+1\). Then: \(\int_{1}^{e^{at}} \frac{u}{u+1} du = \int_{2}^{e^{at}+1} \frac{w-1}{w} dw = \int_{2}^{e^{at}+1} \left(1 - \frac{1}{w}\right) dw\) The second integral can be noticed by inspection that it is a part of the natural logarithm function: \(\int_{1}^{e^{at}} \frac{1}{u+1} du = \ln(u+1)\Big |_{1}^{e^{at}}= \ln(e^{at}+1) - \ln(1+1)\)

Combine the results to find the distance function

Now, we can combine the results of the two integrals to find the distance fallen after time t: \( d(t) = \frac{v_T}{a} \left[\left(\int_{2}^{e^{at}+1} \left(1 - \frac{1}{w}\right) dw\right)- \left(\ln(e^{at}+1) - \ln(1+1)\right)\right]\) So, \(d(t)=\frac{v_T}{a}\left[ (w-w \ln w)\Big|_{2}^{e^{at}+1} - \left(\ln(e^{at}+1) - \ln 2\right)\right]\) \(d(t)=\frac{v_T}{a}\left[ ((e^{at}+1)\ln (e^{at}+1) - (e^{at}+1) -2\ln 2 +2)-\ln\left(\frac{e^{at}+1}{2}\right)\right]\) This is the distance fallen by the skydiver after time t, considering gravitational acceleration and air resistance.

Key concepts

Integration

Integration in calculus is a fundamental technique used to find areas under curves, among many other applications. In the context of motion, such as a skydiver's free fall, integrating the velocity function over time gives us the total displacement or distance traveled.

In simpler terms, if you think of velocity as the speed at which you're moving, integration can be likened to adding up all the tiny distances you've covered over each moment in time, giving you the total path length. This is precisely what we do when we want to calculate how far the skydiver falls after a certain time t, by evaluating the integral of the velocity function with respect to time from 0 to t, denoted as \(d(t)=\int_{0}^{t} v(y) dy\).

Interpreting the Integral

It's helpful to picture the integral as the summation of infinitesimally small ‘slices’ of distance over an interval of time. The result of this sum gives us a finite value representing the overall distance fallen by the skydiver during the time interval from 0 to t seconds.

Velocity Function

A velocity function describes how an object’s velocity changes over time. In the case of the skydiver, it is given by \(v(t) = v_T\left(\frac{e^{at}-1}{e^{at}+1}\right)\), where \(v_T\) represents the terminal velocity and \(a>0\) is a constant related to acceleration due to gravity and air resistance factors.

Understanding the velocity function is crucial for this calculus problem. The function provided implies that the skydiver's velocity begins at zero and approaches \(v_T\) as time increases. The exponential function within it suggests that the change in velocity is not linear but instead depends on the factor \(a\), which could represent the combined effect of gravity and air resistance on the skydiver's acceleration.

Significance of the Velocity Function's Form

The hyperbolic structure of \(\frac{e^{at}-1}{e^{at}+1}\) is important because it inherently defines the asymptotic behavior towards the terminal velocity, \(v_T\), indicating that no matter how long the skydiver falls, their speed will not exceed this maximum value due to the air resistance.

Substitution Method

The substitution method is a technique to solve more complex integrals. It involves replacing a part of the integral with a new variable to simplify the integrand and make integration possible. In our exercise, we substituted \(e^{ay}\) with \(u\) to transform the integral into a more manageable form.

By letting \(u = e^{ay}\), we could differentiate to find \(du\) in terms of \(dy\), which in this case was \(\frac{du}{dy} = ae^{ay}\). With this substitution, the integral became easier to evaluate, changing not just the integrand's appearance but also the limits of integration.

Why Use Substitution?

Without substitution, the integral of the skydiver’s velocity function would be much more difficult to solve directly. The process of substitution effectively ‘translates’ the integral into a form where known integration techniques and rules apply, enabling us to find the solution and, consequently, the distance the skydiver falls over time t.