# Preface

This is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0330. It is primarily for students who have very little experience or have never used Mathematica before and would like to learn more of the basics for this computer algebra system. As a friendly reminder, don't forget to clear variables in use and/or the kernel.

Finally, the commands in this tutorial are all written in bold black font, while Mathematica output is in normal font. This means that you can copy and paste all commands into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.

# Applications

## Vertical Motion

It is known from elementary physics that, in the absence of air friction, a ball thrown up from the ground into earth's atmosphere with initial speed v0 would attain a maximum altitude of $$v_0^2 /(2g) .$$ In this case the return time is $$2\, v_0 /g,$$ independent of the ball's mass. Here g is the acceleration due to gravity. If the ball is thrown up from altitude y0 (which we later assume to be zero), then the time T0 spent traveling is given by

$\left\{ \begin{array}{c} \mbox{travel time with no air resistance} \\ \mbox{when thrown from height y_0} \end{array} \right\} \ = \ T_0 \ =\ \frac{v_0 + \sqrt{v_0^2 +2\, y_0\, g}}{g}$

The presence of air influences the ball's motion: it experiences two forces acting on it---the force of gravity and the air resistance force. Let's define the symbol T as follows:

$\left\{ \begin{array}{c} \mbox{travel time {\it with} air resistance} \\ \mbox{when thrown from height y_0} \end{array} \right\} \ = \ T$

Without air resistance, the object travels farther up than with air resistance. On the way down, without air resistance the object travels a larger distance, but it also gathers more speed. A natural question is, which travel time (with air resistance vs. no air resistance) is larger? Also, it is of interest to find the maximum altitude $$y_{\max}$$ of the ball, the time Tmax to reach maximum altitude, and the time $$T_{\text{down}}$$ to return back from ymax. Therefore, $$T_{\max} + T_{\text{down}} = T_{\text{total}}$$ (the total time the ball spent in the air). The landing velocity is denoted by $$v_{\ell} .$$

Example: model for air resistance. Air resistance is the force that acts in the direction opposite to the motion of an object through air. Air resistance depends on the shape, material, and orientation of the object, the density of the air, and the object's relative speed.

We would like to think that there is a nice formula for the air resistance in terms of speed and other variables. Such a formula would help in making calculations and predicting various quantities. A starting point for obtaining such a formula is our everyday experience. Based on our experience, a reasonable assumption to make\footnote{Our intuition based on everyday experience is limited to a small range of conditions. This may lead to erroneous assumptions.

It has been observed that, under suitable conditions, the magnitude of the air resistance is proportional to a power of the speed s=|v|:

$$\tag{M} F \propto s^p, \qquad \mbox{which may be written as}\qquad F(v) = k \, |v|^p.$$
Here v is velocity, and both k and p are positive constants. For very small objects, such as a speck of dust (about 1 micrometer or 0.001mm), p=1 seems to give a reasonable formula for the air resistance. For larger, human scale objects moving at relatively large speed, p=2 works better. Therefore, the magnitude of the air resistance F as a function of velocity v is assumed be given by formula (M).

The air resistance force depends on the velocity (v) of the object at time t, so let us denote this force with the symbol F(v). Note that the air resistance, force F(v), always acts in the direction opposite to the motion. Therefore, F(v) acts in the down (negative) direction when the ball is moving up, and it acts in the up (positive) direction when the ball is moving down. If we measure the displacement y = y(t) vertically upwards from the ground, then $$v= {\text d}y/{\text d}t = \dot{y}$$ is the velocity of the object. Newton's law of motion for the ball on the way up gives the differential equation

$$\label{vert.1} m\, \dot{v} = -m\,g -F(v) \ , \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = - g - \frac{1}{m}\, F(v),$$
and on the way down,
$$\label{vert.2} m\, \dot{v} = -m\,g +F(v)\ , \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = - g + \frac{1}{m}\, F(v).$$
Since we assume $$F(v) = k \, |v|^p ,$$ the equation of motion on way up becomes
$$\label{vert.3} m\, \dot{v} = -m\,g -k\,|v|^p \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = -g-\frac{k}{m}\,|v|^p ,$$
and on the way down,
$$\label{vert.4} m\, \dot{v} = -m\, g +k\, |v|^p \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = -g+\frac{k}{m}\,|v|^p .$$

To find an equation for $v_{\ell}$, the landing velocity, we rewrite \refeq{vert.3} as $${\text d}\,t = - {\text d}v/(g +F(v)/m)$$ and integrate both sides from t=0 and v = v0 to $$t=T_{\max}$$ and v=0. Here $$T_{\max}$$ is the time to reach the maximum altitude $$y_{max} ,$$ which is also the time to have velocity v=0. We obtain,

$$\label{Tmax.1} \int_{0}^{T_{max}}\,{\text d}t = - \int_{v_0}^{0} \frac{{\text d}\,v}{g+\frac{k}{m}\,|v|^p} \, , \quad \mbox{that is,} \quad T_{max} = - \int_{v_0}^{0} \frac{{\text d}\,v}{g+\frac{k}{m}\,|v|^p}$$
A similar formula is valid for time T. From \refeq{vert.4}, we get
$$\label{Tmax.2} \int_{T_{max}}^T \, {\text d}t = \int_0^{v_\ell} \frac{ \,{\text d}v}{-g+\frac{k}{m}\,|v|^p} \quad\Longrightarrow \quad T - T_{\max} = \int_0^{v_\ell} \frac{ \,{\text d}v}{-g+\frac{k}{m}\,|v|^p}$$
Equating the time $$T_{max}$$ in the above equations, we have
$$\label{T landing} T \ = \ - \int_{v_0}^{0} \frac{{\text d}\,v}{g+\frac{k}{m}\,|v|^p} + \int_0^{v_\ell} \frac{{\text d}\,v}{-g+\frac{k}{m}\,|v|^p}$$
This equation gives T as a function of $$v_\ell .$$ The next step is to find an equation for $$v_\ell .$$

To find an equation for $v_\ell$, we rewrite \refeq{vert.3} as $$v\,{\text d}t = -v\, {\text d}v/(g +F(v)/m)$$ and integrate both sides from t=0 and v = v0 to $$t=T_{\max}$$ and v=0. Here $$T_{\max}$$ is the time to reach the maximum altitude $$y_{\max} ,$$ which is also the time to have velocity v=0. Using the fact that the integral of the velocity is the displacement, we obtain,

$$\label{ymax.1} \int_{0}^{T_{max}} v\,{\text d}t = - \int_{v_0}^{0} \frac{v\,{\text d}v}{g+\frac{k}{m}\,|v|^p} \, , \quad \mbox{that is,} \quad y_{max} - y_0 = - \int_{v_0}^{0} \frac{v\,{\text d}v}{g+\frac{k}{m}\,|v|^p}$$
A similar formula is valid for the distance traveled down. From \refeq{vert.4} we get
$$\label{ymax.2} \int_{T_{max}}^T v\, {\text d}t = \int_0^{v_\ell} \frac{v\,{\text d}v}{-g+\frac{k}{m}\,|v|^p} \quad\Longrightarrow \quad - y_{\max} = \int_0^{v_\ell} \frac{v\,{\text d}v}{-g+\frac{k}{m}\,|v|^p}$$
Equating the maximum distance $$y_{\max}$$ traveled in both ways, we get an equation involving the landing velocity $$v_{\ell} :$$
$$\label{v landing} \int_0^{v_\ell} \frac{v\,{\text d}v}{-g+\frac{k}{m}\,|v|^p} \ = \ - y_0 + \int_{v_0}^{0} \frac{v\,{\text d}v}{g+\frac{k}{m}\,|v|^p}$$
This equation is an equation where the unknown is $$v_{\ell} ,$$ which does not appear explicitly solved for.

g := 9.806
Solve[Integrate[v/(-g + 0.01*v^2), {v, 0, vl}] == Integrate[v/(g + 0.01*v^2), {v, 50, 0}], vl]
Out[2]= {{vl -> -26.5393}, {vl -> 26.5393}}

We would like to determine the ratio: $\gamma = T/T_0 = Tg/ \left( v_0 + \sqrt{v_0^2 +2 x_0 g}\right),$ where T time in air with air resistance and T0 is the time in air without air resistance.

For a tennis ball thrown upward with the initial velocity v0 =10, it is possible to find x0 that γ > 1 when p=0.9. In general, it is unknown for what values of p< 1, x0, and v0 we can achieve γ > 1.

Example: We consider a model of falling object, say a tennis ball, to a flat surface that moves up and down periodically. Using vertical axis directed upward, we denote v(t) as the velocity of the ball and y(t) as its position/height at time t. It has been observed that, under suitable conditions, the magnitude of the air resistance is proportional to the power of speeds=|v|:

$$\tag{M} F \propto s^p, \qquad \mbox{which may be written as}\qquad F(v) = k \, |v|^p.$$
Note that the air resistance force F(v) always acts in the direction opposite to the motion. Therefore, F(v) acts in the down (negative) direction when the ball is moving up, and it acts in the up (positive) direction when the ball is moving down.

Suppose that initially at t=0, the ball of mass m is dropped from the altitude $$y=h > 1$$ without initial velocity. At the same time, it is assumed that the floor starts moving according to the formula $$z= \sin \omega t.$$ When elastic ball hits a hard flat surface, it bounces back with the same velocity. It is assumed that the collision is totally elastic, so the ball loses no kinetic energy in the collision, and its speed after collision is the same as before the collision. At this point, ignore the time needed for the ball to be deformed during collision before fully rebounded and has lifted off from the surface instantly. Hence the ball can be treated as a rigid body with negligible deformation during impact.

After collision, the ball climbs up until its velocity becomes zero, and then the ball falls vertically downward under the influence of gravity, hits the the moving floor, and bounces back.

## Derivation of a differential equation

Newton's law of motion for the ball on the way down is

\begin{equation*} %\label{vert.1} m\, \dot{v} = -m\,g +F(v)\ , \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = - g + \frac{1}{m}\, F(v), \end{equation*}
and on the way up
\begin{equation*} %\label{vert.2} m\, \dot{v} = -m\,g -F(v) \ , \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = - g - \frac{1}{m}\, F(v), \end{equation*}
where g is the acceleration due to gravity. Since we assume $$F(v) = k \, |v|^p ,$$ the equation of motion on the way down becomes
\begin{equation*} %\label{vert.3} m\, \dot{v} = -m\, g +k\, |v|^p \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = -g+\frac{k}{m}\,|v|^p . \end{equation*}
and on way up
$$%\label{vert.4} m\, \dot{v} = -m\,g -k\,|v|^p \qquad\mbox{or}\qquad \frac{{\text d}v}{{\text d}t} = -g-\frac{k}{m}\,|v|^p ,$$
Assuming that p=2 for a tennis ball, the above differential equations can be integrated using separation of variables:
$$%\label{vert.1} \int \frac{{\text d}v}{-g+\frac{k}{m}\,|v|^2} = \int {\text d}t \qquad\Longrightarrow \qquad \sqrt{\frac{m}{g\,k}}\, \mbox{Arctanh} \left( v\,\sqrt{\frac{k}{g\,m}} \right) = t ,$$
and on way up
$$%\label{vert.2} \int \frac{{\text d}v}{g+\frac{k}{m}\,|v|^2} = -\int {\text d}t \qquad\Longrightarrow \qquad \sqrt{\frac{m}{g\,k}}\, \mbox{Arctan} \left( v\,\sqrt{\frac{k}{g\,m}} \right) = -t ,$$
where
$\mbox{Arctanh} \, (x) = \frac{1}{2} \, \ln \frac{1+x}{1-x} \quad \mbox{for} \quad |x| < 1.$
Since the velocity v(t) is the derivative of ball's position y(t), you may need to integrate
$\int \arctan (\alpha v)\, {\text d}v = v\,\arctan (\alpha v) - \frac{1}{2\alpha} \, \ln \left( 1 + \alpha^2 v^2 \right) , \quad \int \mbox{arctanh} (\alpha v)\, {\text d}v = v\,\mbox{arctanh} (\alpha v) - \frac{1}{2\alpha} \, \ln \left\vert 1 - \alpha^2 v^2 \right\vert .$
For example, at the initial stage, you need to solve two IVPs:
$\frac{{\text d}v}{{\text d}t} = -g+\frac{k}{m}\,|v|^2, \quad v(0) =0; \qquad \frac{{\text d}y}{{\text d}t} = v, \quad y(0) =h .$

## Input parameters

$$g \approx 9.806$$ m/sec2  the acceleration due to gravity near sea level at 45 deg. latitude;
m=0.08  mass of the object, in kg because a tennis ball is about 80 grams;
k=0.02   drag coefficient, positive;
p=2   power of the speed term in the resistance force;
ω = π   frequency of the oscillating floor;
$$y_0 =h > 1$$  initial altitude, positive, in meters.

## Derivation of solution

A ball that is dropped from height h> 1 can be described by its velocity v(t) and position y(t):

$\frac{{\text d}v}{{\text d}t} = -g+\frac{k}{m}\,|v|^2, \quad v(0) =0; \qquad \frac{{\text d}y}{{\text d}t} = v, \quad y(0) =h .$
Separation of variables yields
$\int_0^v \frac{{\text d}v}{-g+\frac{k}{m}\,|v|^2} = \int_0^t {\text d}t \qquad \Longrightarrow \qquad \mbox{Arctanh} \left( \sqrt{\frac{k}{gm}} \, v \right) = - \sqrt{\frac{gk}{m}} \,t .$
Therefore, we find the velocity v(t) on first stage of falling from hight h> 1 explicitly:
$v(t) = - \sqrt{\frac{gm}{k}} \,\tanh \left( \sqrt{\frac{gk}{m}} \,t \right) , \qquad k\,v^2 < g\,m.$
Then we find ball's position by integrating v(t):
$y(t) = h- \frac{m}{k}\,\ln \cosh \left( \sqrt{\frac{gk}{m}} \,t \right) .$
This equation is valid until the ball meets the surface $$z= \sin \pi t.$$ Therefore, we need to solve the transcendent equation:
$\sin \pi t = h- \frac{m}{k}\,\ln \cosh \left( \sqrt{\frac{gk}{m}} \,t \right) .$

Mathematica confirms:
Assuming[v > 0 && k > 0 && m > 0 && g > 0, Integrate[1/(-g + k/m*v1^2), {v1, 0, v}]]
Integrate[A*Tanh[B*t], t]
k := 0.02; g := 9.806; m := 0.08; h0 := 2; (* h0 is the initial height *)
FindRoot[Sin[$Pi] t] == h0 - 4*Log[Cosh[1.5657266683556232*t]], {t, 0.4}] vv[t_] = omega*Cos[omega*t] omega = Pi T1 = 0.4716548296910227 vv[T1] y1[t_] = h0 - m/k*Log[Cosh[t*Sqrt[g*k/m]]] kk=0.8 V1 = kk*vv[T1] + Sqrt[g*m/k]*Tanh[T1*Sqrt[g*k/m]] 4.158038876332946 Y1 = Sin[omega*T1] 0.9960377590137912 Since Mathematica provides the approximate value of T1 to be 0.4716548296910227 for h0 = 2, we denote it by T1. The velocity $$v(t) = \omega\,\cos (\omega \,t)$$ of the floor/racket at T1 is $$vv(T1) = V1 \approx 0.279386$$ positive, which indicate that the racquet is moving up. So we add 80% of its velocity to the ball. On the second stage, the ball bounced up with the initial velocity $$V1 = kk*vv(T1) + \sqrt{\frac{gm}{k}} \tanh \left( \sqrt{\frac{gk}{m}}\,T1 \right) \approx 4.158$$ and from the position $$Y1 = \sin \left( \omega\,T1 \right) \approx 0.996038.$$ Here kk is the coefficients of elastic damping of the racket, which we set to be 0.8. Therefore, we need to solve two initial value problems: \[ \frac{{\text d}v}{{\text d}t} = -g-\frac{k}{m}\,|v|^2 , \quad v(T1) = V1, \qquad \frac{{\text d}y}{{\text d}t} = v(t), \quad y(T1) = Y1, \quad \mbox{for }t\ge T1.$
Separating variables and integrating, we obtain
$v(t) = V1 + \sqrt{\frac{gm}{k}} \, \tan \left[ \sqrt{\frac{gk}{m}} \left( T1 -t \right) \right] , \quad T1 \le t \le T2 ,$
where T2 is the value of time when v(T2) =0. Then we find ball's position:
$y(t) = Y1+ V1\left( t- T1 \right) +\frac{m}{k} \,\ln \cos \sqrt{\frac{gk}{m}} \left( T1 -t \right) , \quad T1 \le t \le T2 .$
Mathematica find its value to be $$T2 \approx 0.846 :$$
v2[t_] := V1 + Sqrt[g*m/k]*Tan[Sqrt[g*k/m]*(T1 - t)]
T2 = t /. FindRoot[v2[t] == 0, {t, 0.8}]
0.8459828189925936
The altitude of the ball with zero speed will be
y2[t_] = Y1 + V1*(t - T1) + m/k*Log[Cos[Sqrt[g*k/m]*(T1 - t)]]
v3[t_] = -Sqrt[g*m/k]*Tanh[Sqrt[g*k/m]*(t - T2)]
Y2 = y2[T2]
1.8221341319203344
Then we find time (which we denote as T3) when the ball meets the racket:
y3[t_] = Y2 - m/k*Log[Cosh[Sqrt[g*k/m]*(t - T2)]]
T3 = t /. FindRoot[y3[t] == Sin[omega*t], {t, 2}]
1.6703817057459314
The velocity of the racket at time contact is positive and we transfer 80% of its speed to the ball.
vv[T3]
1.6024422817741188
The position of the ball at t=T2 will be $$y(T2) \approx 1.822 .$$ After T2, the ball starts falling down and we need to solve the initial value problems:
$\frac{{\text d}v}{{\text d}t} = -g+\frac{k}{m}\,|v|^2, \quad v(T2) =0; \qquad \frac{{\text d}y}{{\text d}t} = v(t), \quad y(T2) =Y2 \approx 1.822 .$
We again separate variables and obtain
$v(t) = -\sqrt{\frac{gm}{k}} \, \tanh \left[ \sqrt{\frac{gk}{m}} \left( t- T2 \right) \right] , \quad T2 \le t\le T3 ,$
$y(t) = Y2 - \frac{m}{k} \, \ln \cosh \left[ \sqrt{\frac{gk}{m}} \left( t- T2 \right) \right] , \quad T2 \le t\le T3 .$
To find T3, we need to solve the transcendent equation:
$\sin \left( \omega t \right) = Y2 - \frac{m}{k} \, \ln \cosh \left[ \sqrt{\frac{gk}{m}} \left( t- T2 \right) \right] .$
We plot with Mathematica to be sure that the racket does not hit the ball on this stage:
y[t_] = Piecewise[{{h0 - m/k*Log[Cosh[t*Sqrt[g*k/m]]],
0 < t < T1}, {Y1 + V1*(t - T1) + m/k*Log[Cos[Sqrt[g*k/m]*(T1 - t)]],
T1 < t <= T2}, {Y2 - 4*Log[Cosh[Sqrt[g*k/m]*(t - T2)]], T2 < t <= T3}}]
Plot[{y[t], Sin[omega*t]}, {t, 0, T3}]
At $$t= T3 \approx 1.67 ,$$ the ball meets the racket and starts climbing up. So its velocity should be the solution of the following initial value problem:
$\frac{{\text d}v}{{\text d}t} = -g-\frac{k}{m}\,|v|^2 , \quad v(T3) = V3, \qquad \frac{{\text d}y}{{\text d}t} = v(t), \quad y(T3) = Y3, \quad \mbox{for } T3 \le t \le T4,$
where T4 is time where the velocity is zero, and $$V3 = kk*\sin (\omega\,T3) + \sqrt{\frac{gm}{k}} \tanh \left( \sqrt{\frac{gk}{m}}\,T3 \right) \approx 4.69381 .$$ Mathematica provides the numerical values:
V3 = kk*Sin[omega*T3] + Sqrt[g*m/k]*Tanh[Sqrt[g*k/m]*(T3 - T2)]
4.693808534231868
Then we define ball's velocity and its position:
Y3 = Sin[omega*T3]
-0.8601309841556795
v4[t_] = V3 + Sqrt[g*m/k]*Tan[(T3 - t)*Sqrt[g*k/m]]
y4[t_] = Y3 + V3*(t - T3) + m/k*Log[Cos[(T3 - t)*Sqrt[g*k/m]]]
T4 = t /. FindRoot[v4[t] == 0, {t, 2}]
2.08115
Next, we check whether the racket hits the ball from below
TT4 = t /. FindRoot[y4[t] == Sin[omega*t], {t, 2}]
2.05476
Since this number TT4 is less than previously found T4, we conclude that the racket hits the ball on its way up. At this moment TT4, we define the position and the velocity of the ball:
VV3 = vv[TT4] + v4[TT4]
3.48749
YY4 = Sin[omega*TT4]
0.171185
and then define the velocity and new position
vv4[t_] = VV3 + Sqrt[g*m/k]*Tan[(TT4 - t)*Sqrt[g*k/m]]
yy4[t_] = YY4 + VV3*(t - TT4) + m/k*Log[Cos[(TT4 - t)*Sqrt[g*k/m]]]
y[t_] = Piecewise[{{h0 - m/k*Log[Cosh[t*Sqrt[g*k/m]]],
0 < t < T1}, {Y1 + V1*(t - T1) + m/k*Log[Cos[Sqrt[g*k/m]*(T1 - t)]],
T1 < t <= T2}, {Y2 - 4*Log[Cosh[Sqrt[g*k/m]*(t - T2)]],
T2 < t <= T3}, {Y3 + V3*(t - T3) +
m/k*Log[Cos[(T3 - t)*Sqrt[g*k/m]]], T3 < t < TT4}, {yy4[t], TT4 < t < 2.081}}]
Plot[{y[t], Sin[omega*t]}, {t, 0, 2.081}, PlotStyle -> {Thick, Thick}]
The time interval where the ball oscillates over the racket is not enough to show on the picture because time intervals are too small.

Claude Perrault  Sir Isaac Newton  Christiaan Huygens

A tractrix (from the Latin verb trahere "pull, drag"; plural: tractrices) is the curve along which an object moves, under the influence of friction, when pulled on a horizontal plane by a line segment attached to a tractor (pulling) point that moves at a right angle to the initial line between the object and the puller at an infinitesimal speed. It is therefore a curve of pursuit. It was first introduced by Claude Perrault in 1670, and later studied by Isaac Newton (1676) and Christiaan Huygens (1692).

Claude Perrault (1613 -- 1688) was a French architect, best known for his participation in the design of the east façade of the Louvre in Paris. He also achieved success as a physician and anatomist, and as an author, who wrote treatises on physics and natural history. Perrault was born and died in Paris. Aside from his influential architecture, he became well known for his translation of the ten books of Vitruvius, the only surviving Roman work on architecture, into French, written at the instigation of Colbert, and published, with Perrault's annotations, in 1673. His treatise on the five classical orders of architecture followed in 1683. As physician and natural philosopher with a medical degree from the University of Paris, Perrault became one of the first members of the French Academy of Sciences when it was founded in 1666.

Sir Isaac Newton (1642 -- 1726/27) was an English mathematician, astronomer, theologian and physicist (described in his own day as a "natural philosopher") who is widely recognized as one of the most influential scientists of all time and a key figure in the scientific revolution. His book Philosophiae Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), first published in 1687, laid the foundations of classical mechanics. Newton also made pathbreaking contributions to optics, and he shares credit with Gottfried Wilhelm Leibniz for developing the infinitesimal calculus.

Christiaan Huygens (Latin: Hugenius; 1629 -- 1695) was a prominent Dutch mathematician and scientist. He is known particularly as an astronomer, physicist, probabilist and horologist. Huygens was a leading scientist of his time. His work included early telescopic studies of the rings of Saturn and the discovery of its moon Titan, the invention of the pendulum clock and other investigations in timekeeping. He published major studies of mechanics and optics (having been one of the most influential proponents of the wave theory of light), and pioneered work on games of chance.

Example: In order to plot tractrix curve, use the following code:

tractrix[a_][t_] := a*{Sin[t], Cos[t] + Log[Tan[t/2]]};
Manipulate[
ParametricPlot[tractrix[a][t] // Evaluate, {t, 0, .99*\[Pi]},
PlotRange -> {0, 7}], {a, 1, 6}]
Manipulate[
Plot[y'[x] = -Sqrt[a^2 - x^2]/x, {x, 0, 20},
PlotRange -> {-10, 10}], {a, 0, 20}]
sol = DSolve[{y'[x] == -Sqrt[a^2 - x^2]/x, y[a] == 0}, y, x]
Manipulate[
Plot[-Sqrt[a^2 - x^2] + a Log[a] - a Log[a^2] - a Log[x] +
a Log[a^2 + a Sqrt[a^2 - x^2]], {x, 0, 20}, PlotRange -> All], {a,
1, 20}]