# Apostol question about the differential equations of a falling object

• zenterix
zenterix
Homework Statement
Refer to example 2 of section 8.6. Use the chain rule to write

$$\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=v\frac{dv}{ds}$$

and thus show that the differential equation in the example can be expressed as follows

$$\frac{ds}{dv}=\frac{bv}{c-v}$$

where ##b=m/k## and ##c=gm/k##. Integrate this equation to express ##s## in terms of ##v##. Check your result with the formulas for ##v## and ##s## derived in the example
Relevant Equations
The cited example 2 is a bit large to be shown here in all its steps.

Here are the main points and equations.

A body of mass ##m## is dropped from rest from a great height in the earth's atmosphere. Assume it falls in a straight line and that the only forces acting on it are the earth's gravitational attraction and a resisting force due to air resistance which is proportional to its velocity.

Newton's second law tells us

$$ma=mg-kv$$

where ##k## is some positive constant and ##-kv## is the force due to air resistance.

$$mv'=mg-kv$$

is a first-order equation in velocity ##v##.

We can write this in the form

$$v'+\frac{k}{m}v=g$$

which we can solve using an integrating factor to obtain (assuming v(0)=0)

$$v(t)=e^{-kt/m}\int_0^t ge^{ku/m}du=\frac{mg}{k}(1-e^{-kt/m})$$

we can differentiate to find acceleration

$$a(t)=ge^{-kt/m}$$

We can also integrate to obtain position

$$s(t)=\frac{mg}{k}t+\frac{gm^2}{k^2}e^{-kt/m}+C$$

and if ##s(0)=0## we have

$$s(t)=\frac{mg}{k}t+\frac{gm^2}{k^2}(e^{-kt/m}-1)$$
Here is my solution to this problem. Unfortunately, I can't check it because it is not contained in the solution manual.

$$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}$$

$$\frac{ds}{dv}=\frac{v}{v'}=\frac{v}{ge^{-kt/m}}$$

$$=\frac{\frac{m}{k}v}{\frac{gm}{k}e^{-kt/m}}$$

$$=\frac{\frac{m}{k}v}{\frac{gm}{k}-\frac{gm}{k}+\frac{gm}{k}e^{-kt/m}}$$

$$=\frac{\frac{m}{k}v}{\frac{gm}{k}-\frac{gm}{k}(e^{-kt/m}-1)}$$

$$=\frac{bv}{c-v}$$

My main question is about the integration of this expression to obtain ##s## in terms of ##v##.

$$\int_0^v \frac{ds}{dv}dv=\int_{s(0)}^{s(v)} ds = s(v)-s(0)=\int_0^v\frac{bv}{c-v}dv$$

$$=bc(\ln{(c)}-\ln{(c-v)})-bv$$

$$s(v)=s(0)+bc(\ln{(c)}-\ln{(c-v)})-bv$$

I don't recall seeing this relationship very often and so I am not sure if this is correct. The problem says to check this result with the equations derived in the cited Example 2. But that example derived equations for ##v## and ##s## relative to ##t##. How would I go about using those equations to check my result of ##s## as a function of ##v##?

I think the intention is to derive the ODE as $$ma = mv\frac{dv}{ds} = mg - kv$$ so that $$\frac{ds}{dv} = \frac{mv}{mg - kv} = \frac{(m/k)v}{(mg/k) - v} = \frac{bv}{c - v}.$$ Note that the solution to the earlier exercise can be written in the form $$\begin{split} v(t) &= c(1 - e^{-b/t}) \\ s(t) - s_0 &= ct - bv(t).\end{split}$$ To show that $$s(v) - s_0 = bc\ln|c| - bv - bc\ln|c - v| = ct - bv = s(t) - s_0$$ the easiest way is to solve $v(t)$ for $t$ to obtain $$\begin{split} e^{-t/b} &= \frac{c - v}{c} \\ t &= b\ln|c| - b\ln|c - v|.\end{split}$$

Last edited:
zenterix
Attached are the full calculations leaving ##v_0## as a variable (just for added suffering with the algebra). Unfortunately, I couldn't figure out a way to post it directly here (file either too large or too low quality).

#### Attachments

• 8.7 - Physical Problems Leading to 1st Order LDE.pdf
1 MB · Views: 53

## What is Apostol's question about the differential equations of a falling object?

Apostol's question typically refers to understanding the differential equations that describe the motion of a falling object, taking into account forces such as gravity and air resistance. This involves setting up and solving equations that model the velocity and position of the object over time.

## How do you derive the differential equation for a falling object with no air resistance?

For a falling object with no air resistance, the only force acting on it is gravity. Using Newton's second law, $$F = ma$$, and knowing that the gravitational force is $$mg$$, we get $$ma = mg$$. This simplifies to $$a = g$$, where $$a$$ is the acceleration and $$g$$ is the acceleration due to gravity. The differential equation is $$\frac{d^2y}{dt^2} = g$$, where $$y$$ is the position of the object.

## How does air resistance affect the differential equation of a falling object?

Air resistance introduces an additional force that opposes the motion of the falling object. This force is often modeled as proportional to the velocity of the object, $$F_{drag} = -kv$$, where $$k$$ is a constant. The differential equation then becomes $$ma = mg - kv$$, or $$m \frac{dv}{dt} = mg - kv$$. This can be rewritten as $$\frac{dv}{dt} = g - \frac{k}{m}v$$.

## What is the solution to the differential equation with air resistance?

The differential equation $$\frac{dv}{dt} = g - \frac{k}{m}v$$ is a first-order linear differential equation. Its solution is $$v(t) = \frac{mg}{k} (1 - e^{-\frac{k}{m}t})$$. This shows that the velocity approaches a terminal velocity $$v_t = \frac{mg}{k}$$ as $$t$$ increases.

## How do you find the position of a falling object with air resistance?

To find the position $$y(t)$$, integrate the velocity function $$v(t) = \frac{mg}{k} (1 - e^{-\frac{k}{m}t})$$. The integral of $$v(t)$$ with respect to time gives $$y(t) = \frac{mg}{k}t + \frac{m^2g}{k^2}(e^{-\frac{k}{m}t} - 1) + C$$, where $$C$$ is the constant of integration determined by initial conditions.

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