Mag|cK
This is quotes from online pdf files from University of Toronto Scarborough.

"the terminal velocity is F/b , where b is drag constant

Let’s solve the equation of motion and see how
this is reflected in the solution. Newton's law

m (dv(t)/dt) = F - b v(t)

which we have written entirely in terms of the
velocity v(t) and its first derivative.

How to solve this? Well, if F were zero, we
would have dv/dt=–(b/m)v, which has its solution
some constant times exp(-bt/m). By inspection,
we find that we can account for nonzero F by
simply adding a constant, F/b. That is,

v(t) = F/b + (constant)exp(-bt/m)
--------------------------------------

Now what i dont understand is the last paragraph, and where the last equation came from. Can anyone explain more clearly? Please help Thank you very much.

Homework Helper
A clue: http://www.sosmath.com/diffeq/first/lineareq/lineareq.html" [Broken].

Last edited by a moderator:
Mag|cK
Hmm, i am still in high school, but already learned differential and integration. Do i need to learn first about something i don't learn in high school?
If so what is the topic title? Or maybe u can teach me if its possible?
thx

Staff Emeritus
Gold Member
Hmm, i am still in high school, but already learned differential and integration. Do i need to learn first about something i don't learn in high school?
If so what is the topic title? Or maybe u can teach me if its possible?
thx

Have you learnt the method of substitution for integration?

Mag|cK
i dont think so. Is there any websites that teach that?
thx

Staff Emeritus
Gold Member
i dont think so. Is there any websites that teach that?
thx

Another question: do you know how to find

$$\int \frac{1}{x} dx?$$

Homework Helper
Gold Member
Dearly Missed
It seems you already know the solution to the simplest type of diff.equations, that is the one of the type: du/dt=ku, having u(t)=Ae^(kt) as its general solution.

What would be the next simplest type of a solution to a diff.eq?
At least my intuition says this would be a function of the form:
$$U(t)=A+Be^{kt}$$, where A,B and k are constants.

Let us see what sort of differential equation such a function satisfies!
1. We have that $\frac{dU}{dt}=kBe^{kt}$
2. We can convert the right-hand side of 1. into an expression involving U itself as follows:
$$U=A+Be^{kt}\to{B}e^{kt}=U-A$$
3. Thus, we may insert 2. into 1, and gain:
$$\frac{dU}{dt}=k(U-A)=C+kU, C=-kA$$
4. Note that our equation 3 is a DIFFERENTIAL equation of the general form:
$$\frac{dU}{dt}=A+BU$$
But conversely, this ought to entail that we now can SOLVE such diff.eqs, gaining solutions similar to the function we studied in 1.-3.!
We'll see that this is indeed the case:
We may rewrite 4. as:
$$\frac{dU}{dt}=B(\frac{A}{B}+U)$$
Now, let us define a new function: $$Y(t)=\frac{A}{B}+U$$
Now, note that by differentiation, we have the identity: $\frac{dY}{dt}=\frac{dU}{dt}$

Thus, reconverting our diff.eq in terms of Y, rather than U, we get:
$$\frac{dY}{dt}=BY\to{Y}(t)=Ce^{Bt}\to{U}(t)=Y(t)-\frac{A}{B}=Ce^{Bt}-\frac{A}{B}$$
and we see that this is a solution of the form described in 1.-3.

Mag|cK
Another question: do you know how to find

$$\int \frac{1}{x} dx?$$

yes, ln x + c

Mag|cK
"It seems you already know the solution to the simplest type of diff.equations, that is the one of the type: du/dt=ku, having u(t)=Ae^(kt) as its general solution."

Arildno, No i haven't known that Staff Emeritus
Gold Member

m dv/dt = F - b v.

Proceeding very informally, treat dv as one symbol, and dt as another. First, divide both sides of the above, by the right side of the above, and then multiply both sides dt.

What do you get?

Homework Helper
If you haven't yet done differentiation and integration of exponentials, logs, trig function etc, yet you are getting a little bit ahead of yourself with this equation.

Not that any of those topics are "hard", but there is a far amount of detail to remember, so you have to start somewhere and take it one step at a time. Most differential equations in physics, biology, etc involve exponentials and trig functions, so it makes sense for your courses to leave the topic of "differential equations" till after you have done integration and differentiation of those functions.

Mag|cK

m dv/dt = F - b v.

Proceeding very informally, treat dv as one symbol, and dt as another. First, divide both sides of the above, by the right side of the above, and then multiply both sides dt.

What do you get?

(m dv)/(F - bv)=dt

Btw, is there any program downloads from the internet specially made to write mathematical equations that i can easily put on microsoft word or other applications?

Saketh
Btw, is there any program downloads from the internet specially made to write mathematical equations that i can easily put on microsoft word or other applications?
There is http://www.latex-project.org/" [Broken], the engine this forum uses to format mathematics. You can't put equations into Word with it, but you can generate documents separately.

Last edited by a moderator:
Staff Emeritus
Gold Member
(m dv)/(F - bv)=dt

Let v_0 be the speed of the particle at time t_0, and v be the speed of the particle at time t. Tkake this into account by integrating: (over v) the right side of the above equation from v = v_0 to v = v; the left side from t = t_0 to t = t.

$$\int_{v_0}^v \frac{m}{F - bv} dv = \int_{t_0}^{t} dt$$

Can you do these integrations? Remember, m, F, and b are constants.

Mag|cK
Let v_0 be the speed of the particle at time t_0, and v be the speed of the particle at time t. Tkake this into account by integrating: (over v) the right side of the above equation from v = v_0 to v = v; the left side from t = t_0 to t = t.

$$\int_{v_0}^v \frac{m}{F - bv} dv = \int_{t_0}^{t} dt$$

Can you do these integrations? Remember, m, F, and b are constants.

i havent learnt that kind of integration. is there any website that can teach me that?

Staff Emeritus
Gold Member
i havent learnt that kind of integration. is there any website that can teach me that?

Let's look at

$$\int \frac{m}{F - bv} dv.$$

The denominator is a bit messy, thus set u = F - bv, so u is function v. Find du/dv and rearrange this to find dv. In the the integral, substitute u for the denominator, and also substitute for dv.

What does the integral look like now?

Mag|cK
Let's look at

$$\int \frac{m}{F - bv} dv.$$

The denominator is a bit messy, thus set u = F - bv, so u is function v. Find du/dv and rearrange this to find dv. In the the integral, substitute u for the denominator, and also substitute for dv.

What does the integral look like now?

is F constant? I come up with -(m/b) ln u +c

Staff Emeritus
Gold Member
is F constant?

For consistency, all the terms in

m dv/dt = F - b v

must have the same units. Since the right side has the units of force, both F and bv must have units of force. F is a constant force, like the force of gravity on a particle near the Earth's surface.

I come up with -(m/b) ln u +c

Good.

Now change the u back to F - bv and evaluate the resulting expression at the endpoints of the integration interval.

Mag|cK
Now change the u back to F - bv and evaluate the resulting expression at the endpoints of the integration interval.

I haven't learnt about definite integral or integration intervals. I'll come back after i learned it. Thx a lot!

Staff Emeritus
Gold Member
I haven't learnt about definite integral or integration intervals.

Suppose

$$\int f \left( x \right) dx = F \left( x \right) + c.$$

Then,

$$\int_a^b f \left( x \right) dx = F \left( b \right) - F \left( a \right).$$

Geometrically, this is the area bounded by the curve f(x), the vertical lines x = a and x = b, and the x-axis.

Example: $f \left(x \right) = x^2.$

$$\int_2^5 x^2 dx = \frac{5^3}{3} - \frac{2^3}{3} = 39.$$

Mag|cK
well then how to transform -(m/b) ln u +c into integrals with intervals. We just got rid of the integrals right?

Staff Emeritus
After replacing $u$, we have
$$\int \frac{m}{F - bv} dv = - \frac{m}{b} ln \left( F - bv \right) + c.$$