Examples

Catenary

Catenary

Problem: A catenary is the shape formed by a hanging chain, string, or cable, and also the shape formed by a stable arch. A cable can only resist expansion forces directed along its length, while an arch can only resist compression forces directed along its length. What is the shape of a catenary?

Solution: Consider a hanging chain. Let $F_T$ be the tension force tangent to the chain at a point $p=(x,y)$, and let $F_H$ be the horizontal component of the tension, so $F_H=F_T\cos \theta$, where $\theta$ is the angle formed by the line tangent to the chain at $p$.

The vertical balance of forces is more complicated. Set $x=0$ at the lowest point on the chain, and let $s(x)$ measure the arclength of the chain with $s(0)=0$. At $x=0$, the chain has a horizontal slope, so the vertical component of tension force there is zero. At $p=(x,y)$, the vertical component of tension force is $F_T\sin \theta$. The total vertical force on the portion of chain from the bottom up to $p$ is therefore $F_T\sin \theta$, and this must balance the total weight of the chain from the bottom up to $p$, which is $\rho s(x)$, where $\rho$ is the chain density. So $F_T\sin \theta =\rho s(x)$.

By dividing our two equations, we have $\rho s(x)=F_H\tan \theta$. We may write $\frac{dy}{dx}=\tan \theta$ for the slope of the tangent at $p$. This gives us the differential equation $y^{\prime }=\frac{\rho }{F_H}s(x)$. To solve this, we need to relate the arclength function $s(x)$ to $y(x)$. Recall the arclength formula:

$$s(x)={\int }_{u=0}^x\sqrt{1+(y^{\prime }(u){)}^2}du.$$

Plugging this in gives a differential equation that can be solved algebraically. The solution (family of curves) is:

$$y(x)=\frac{F_H}{\rho }\cosh \left(\frac{\rho }{F_H}x\right)+D.$$

If we set $h_0=y(0)$, i.e. the vertical position of the bottom of the chain, we can solve for $D$ and obtain:

$$y(x)=\frac{F_H}{\rho }\cosh \left(\frac{\rho }{F_H}x\right)+\left(h_0-\frac{F_H}{\rho }\right).$$

https://www.desmos.com/calculator/ngjjfg5yxj

Tractrix

Tractrix

A tractrix is a curve given by dragging a point behind a string of fixed length.

Problem: Suppose we drag the point $p$ behind a string of length $a$. Suppose $p$ starts at $(a,0)$, and we start at $(0,0)$ and walk in the $+y$ direction. What is the shape of the path of $p$?

Solution: Write $p=(x,y)$. The tangent line to the string at $p$ has slope given by

$$\frac{dy}{dx}=-\frac{\sqrt{a^2-x^2}}{x}.$$

To see this, draw the triangle with vertices $(x,y)$ and $(0,y)$ and $(0,y+\sqrt{a^2-x^2})$. The hypotenuse of this triangle is the string itself, and we are pulling from the top vertex.

This equation is separable, and the solution is:

$$y(x)=a\ln \left(\frac{a+\sqrt{a^2-x^2}}{x}\right)-\sqrt{a^2-x^2}.$$

https://www.desmos.com/calculator/9mu1stojhl

Pursuit curve

Pursuit curve

Problem: A gazelle runs up the $y$-axis at speed $v_1$ starting at $(0,0)$, and a cheetah runs towards the gazelle at speed $v_2$ starting at $(a,0)$. What is the shape of the path taken by the cheetah?

Solution: The gazelle’s path can be given parametrically by the curve $\mathbf{r}(t)=(0,v_{1}t)$. Suppose the cheetah is at point $p=(x,y(x))$. The vertical distance from cheetah to gazelle is $y-v_{1}t$, and the horizontal distance is $x$.

We assume the cheetah is always running straight toward the gazelle, so the slope of the tangent line to the cheetah’s path is $(y-v_{1}t)/x$, and we obtain the differential equation:

$$\frac{dy}{dx}=\frac{y-v_{1}t}{x}.$$

This equation involves two independent variables, $t$ and $x$, so we cannot immediately solve it. Of course they are related by the fact that $x(t)$ is the $x$-coordinate of the cheetah’s path. We would like to eliminate $t$ from the equation by substituting a formula for $t$ in terms of $x$, $y$, and $y^{\prime }$ quantities. This is the hard part.

We know that the cheetah is running at speed $v_2$, so we also have the equation:

$$v_2=\frac{ds}{dt}=\frac{ds}{dx}\frac{dx}{dt}=-\sqrt{1+(y^{\prime }{)}^2}\frac{dx}{dt}.$$

(Here we use $s(t)$ for the arclength of the cheetah’s path.) This equation relates $t$ to the quantities $x$, $y$, and $y^{\prime }$.

Now we seek to combine the equations. We can flip the second equation to obtain

$$\frac{dt}{dx}=-\frac{1}{v_2}\sqrt{1+(y^{\prime }{)}^2}.$$

If we differentiate the first equation in $x$, we can make the $t$ disappear, with $dt/dx$ appearing instead:

$$\begin{array}{rl}{y}^{\prime \prime }(x)& =\frac{x(y^{\prime }-v_1{t}^{\prime })-(y-v_{1}t)}{x^2}\\ & =\frac{y^{\prime }}{x}-\frac{v_1}{x}\frac{dt}{dx}-\frac{y^{\prime }}{x}\\ & =-\frac{v_1}{x}\frac{dt}{dx}\\ x{y}^{\prime \prime }& =\frac{v_1}{v_2}\sqrt{1+(y^{\prime }{)}^2}.\end{array}$$

Set $k=v_1/v_2$. Reduce the order using $p(x)=y^{\prime }(x)$ and obtain:

$$p(x)=\frac{1}{2}\left({\left(\frac{x}{a}\right)}^k-{\left(\frac{a}{x}\right)}^k\right)$$

Integrate this again to find $y$, solving for the constant using the initial condition $y(a)=0$:

$$y(x)=\frac{1}{2}\left(\left(\frac{a}{k+1}\right){\left(\frac{x}{a}\right)}^{k+1}-\left(\frac{a}{1-k}\right){\left(\frac{x}{a}\right)}^{1-k}\right)+\frac{ak}{(1-k^2)}.$$

https://www.desmos.com/calculator/geosk6qjrb

Growth of retirement fund with compound interest

Growth of retirement fund with compound interest

A quantity of money $A(t)$ in a savings account with compound interest rate $r$ grows according to the differential equation $A^{\prime }(t)=rA$. The solution of this equation is an exponential function $A(t)=A_0{e}^{rt}$, where $A_0=A(0)$ is the initial quantity in the account.

Theory: Compound interest with continuous deposit or withdrawal
Suppose you make continuous deposits or withdrawals at a rate of $d$ (in dollars per year, with $d>0$ for deposits, $d<0$ for withdrawals). Now the equation is:

$$\frac{dA}{dt}=rA+d.$$

This equation is linear and separable. The solution can be found algebraically:

$$A(t)=C{e}^{rt}-\frac{d}{r}.$$

If the account starts with $A(0)=0$, then $C=\frac{d}{r}$ and $A(t)=\frac{d}{r}(e^{rt}-1)$.

Achieving a retirement fund
Suppose this situation describes your Roth IRA. How much should you deposit in the account per year to achieve $800,000 after 40 years, with $r=0.05$? Answer:

$$d=\$800,000\cdot \frac{0.05}{e^{0.05\cdot 40}-1}=\$6261.$$

Constant rate of retirement withdrawal, with inflation
Suppose you have a retirement account with $A_0$ dollars, growing at the interest rate $r$, and you plan to withdraw money at a rate which is increasing with inflation at $k=2\%$ per year, and starting at $80,000 per year. What should $A_0$ be to ensure you will have money for 30 years of retirement?

This problem involves two exponential growths, one stemming from interest and the other from inflation. The rate of withdrawal $W(t)$ satisfies $W^{\prime }(t)=kW$, so $W(t)=W_0{e}^{kt}$ per year. Then the account value satisfies:

$$\frac{dA}{dt}=rA-W_0{e}^{kt}.$$

This equation is linear. The solution is:

$$A(t)=C{e}^{rt}-\frac{W_0}{k-r}{e}^{kt}.$$

Setting $A(0)=A_0$, we can solve for $C$. Then plugging in $A(30)=0$, obtain the equation:

$$\left(A_0+\frac{W_0}{k-r}\right)e^{r\cdot 30}=\frac{W_0}{k-r}{e}^{k\cdot 30}.$$

Setting $W_0=$ $80,000, $r=0.05$, and $k=0.02$, solve for $A_0$ and find $1,580,000.

RC Circuits

RC Circuits

An electrical circuit with a single resistor and a single capacitor is called an RC circuit:

Suppose the voltage source $E$ produces a voltage $V(t)$ between its terminals, and write $V_R$ and $V_C$ for the voltage drops across the resistor and capacitor.

By Ohm’s Law, $V_R=IR$, where $I$ is the current flowing through the resistor, and $R$ is the resistance of the resistor (a constant). The capacitor voltage drop satisfies $V_C=\frac{q}{C}$, where $q$ is the charge stored in the capacitor, and $C$ is the capacitance of the capacitor (a constant). Current is the rate of flow of charge per unit time: $I=\frac{dq}{dt}$.

By Kerchoff’s Voltage Law, we have an equation for the circuit:

$$\begin{array}{rl}V(t)& =V_R+V_C\\ & =IR+\frac{q}{C}.\end{array}$$

So we find that charge satisfies:

$$\begin{array}{rl}\frac{dq}{dt}& =\frac{V(t)-q/C}{R},\\ q^{\prime }(t)+\frac{1}{RC}q(t)& =\frac{1}{R}V(t),\end{array}$$

which is a linear ODE. Solving this ODE and dividing by $C$ gives us a formula for $V_C(t)$:

$$V_C(t)=\frac{1}{C}q(t)=V_0{e}^{-t/RC}+e^{-t/RC}\frac{1}{RC}{\int }_0^t{e}^{s/RC}V(s)ds.$$

Let us study this solution. The first term, $V_0{e}^{-t/RC}$, represents exponential decay from $V_0$ to $0$. The second term depends on the source function $V(t)$. For example, if $V(t)=V$ a constant, then the entire second term reduces to $V$ itself, and the solution $V_C(t)$ will exponentially decay from $V_0+V$ down to $V$.

Here is a Mathlet that shows the response $V_C(t)$ when the source $V(t)$ is a rectangular wave:
https://mathlets.org/mathlets/periodic-box/

Logistic curve - population dynamics

Logistic curve: population dynamics

Modified exponential growth with a growth rate given by the linear function $h(y)=r-ay$ gives the logistic equation:

$$\frac{dy}{dt}=(r-ay)y.$$

Using the theory of phase lines, we can study solutions to this equation without finding an explicit formula for them. Write $K=r/a$, so $f(y)=r\left(1-\frac{y}{K}\right)y$. This number $K$ is sometimes called the carrying capacity of the population’s environment, because the growth rate $r\left(1-\frac{y}{K}\right)$ will go negative when $y$ exceeds $K$.

Now let us classify equilibrium points. We have $f(0)=f(K)=0$. Compute

$$f^{\prime }(y)=r-2ay=r\left(1-2\frac{y}{K}\right),$$

so $f^{\prime }(0)=r>0$ and $f^{\prime }(K)=-r<0$. Therefore $y=0$ is an unstable equilibrium, while $y=K$ is a stable equilibrium.

The logistic equation is separable, and it can be solved exactly and explicitly. Solutions are called logistic curves. These are given by:

$$y(t)=\frac{y_{0}K}{y_0+(K-y_0)e^{-rt}}.$$

Consider the interpretation of this formula. The initial value is $y(0)=y_0$. As $t\to \infty$, the second term in the denominator goes to zero. So $y(t)\to K$ as $t\to \infty$. This is true whether $y_0>K$ or $y_0<K$.

It is useful to observe that the concavity of the solution trajectories can be ascertained from the formula for $f(y)$ without actually solving the logistic equation. This relies on the chain rule:

$$\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}f(y)=f^{\prime }(y)\frac{dy}{dt}=f^{\prime }(y)f(y).$$

For the logistic equation we are considering, this product is:

$$f^{\prime }(y)f(y)=r^2\left(1-2\frac{y}{K}\right)\left(1-\frac{y}{K}\right)y.$$

The inflection point is where this function changes sign, which is at $y=\frac{1}{2}K$, so the rate of population increase starts slowing down when the population is halfway to the carrying capacity of the environment.