# Hyperbolic Function: Help Needed

#### chlxowns_

##### New member
Introduction

Hello, I am currently a student in Engineering Degree, and currently I am stuck in this particular question.... Any help is appreciated!

Question

For an object falling through the atmosphere, a good model for air resistance is that the force of air resistance is proportional to the square of the object’s speed. This leads to the following differential equation:

$$\displaystyle m(dv/dt) = mg - kv^2$$

where t is time, v(t) is the speed, m is the mass of the object, g is acceleration due to gravity, and k is a constant.

a) Consider a function of the form:

$$\displaystyle v(t)=b tanh(ct)$$

where b and c are constants.

using $$\displaystyle tanh'(t)=1-tanh^2(t)$$ show that this function has a derivative

$$\displaystyle dv/dt=b(1-tanh^2(ct))c$$

Then substitute these formulas for v(t) and d $$\displaystyle dv/dt$$ into the equation at the start of the problem and find expressions for the values of b and c; they should depend on g ,m, and k, but should not depend on the time t

My Attempt:

Given that

$$\displaystyle v(t)=btanh(ct)$$

Differentiate the following function using chain rule,

$$\displaystyle f(g(x)), dy/dx=f'(g(x))× g'(x)$$

Let $$\displaystyle g(t)=ct , g'(t)=c$$

Let $$\displaystyle f(g(t))=tanh(g(t)) , f'(g(t))=1-tanh^2(g(t))$$

Substitute ct back into g(t)

$$\displaystyle f'(g(t))=1-tanh^2(ct)$$

Apply chain rule:

$$\displaystyle dv/dt=b(1-tanh^2(ct))× c$$

Equate the derived equation with original equation.

$$\displaystyle mbc(1-tanh^2(ct))=mg-kv^2$$

$$\displaystyle mbc(sech^2(ct))=mg-kvg^2$$

Help I need:

I do not know how to continue for this question, as I am very confused on what do the question want, and how do I express them in terms of c and b....

#### tkhunny

##### Moderator
Staff member
It appears to me that in your laudable attempts to be formal and thorough, you are not seeing what it is you are doing.

In particular, you start with a parameter "b", but then we don't hear from "b" again until your sixth entry. Personally, I thought you lost "b", but then it magically returned.

I so wish this made sense: "using tanh′(t)=1−(tanh(t))^2 show that this function has a derivative " (Altered for clarity.) Why does one need to show there is a derivative when one is simply GIVEN the derivative?

#### Dr.Peterson

##### Elite Member
I so wish this made sense: "using tanh′(t)=1−(tanh(t))^2 show that this function has a derivative " (Altered for clarity.) Why does one need to show there is a derivative when one is simply GIVEN the derivative?
I think it's meant to say:

Using the fact that $$\displaystyle \tanh'(t)=1-\tanh^2(t)$$, show that the function $$\displaystyle v(t)$$ has derivative $$\displaystyle dv/dt=b(1−\tanh^2(ct))c$$​

This is just an application of the chain rule. But the work, which I think is too detailed, fails to define what is meant by f, which I think is probably
$$\displaystyle f(t)=\tanh(x)$$, so that $$\displaystyle v(t) = b f(ct) = b f(g(t))$$.

Then substitute these formulas for v(t) and d $$\displaystyle dv/dt$$ into the equation at the start of the problem and find expressions for the values of b and c; they should depend on g ,m, and k, but should not depend on the time t

$$\displaystyle mbc(sech^2(ct))=mg-kvg^2$$

Help I need:

I do not know how to continue for this question, as I am very confused on what do the question want, and how do I express them in terms of c and b....
You didn't replace $$\displaystyle v$$ with its value!

#### tkhunny

##### Moderator
Staff member
I think it's meant to say:

Using the fact that $$\displaystyle \tanh'(t)=1-\tanh^2(t)$$, show that the function $$\displaystyle v(t)$$ has derivative $$\displaystyle dv/dt=b(1−\tanh^2(ct))c$$​
Altered for even more clarity!

#### chlxowns_

##### New member
You didn't replace vv\displaystyle v with its value!
There is a value for $$\displaystyle v$$? And where do I substitute it? I am so sorry, this is my first time encountering this type of question... Please bear with me...

#### HallsofIvy

##### Elite Member
Yes, there is a value for v! It is b tanh(ct). You were told that right at the start.

#### Dr.Peterson

##### Elite Member
There is a value for $$\displaystyle v$$? And where do I substitute it? I am so sorry, this is my first time encountering this type of question... Please bear with me...
Yes, $$\displaystyle v$$ is the function that you want to be a solution of the differential equation, $$\displaystyle m\frac{dv}{dt}=mg−kv^2$$.

You were given the function to try, $$\displaystyle v=b\tanh(ct)$$, and you replaced $$\displaystyle \frac{dv}{dt}$$ with its expression, but didn't replace $$\displaystyle v$$ with its.

#### chlxowns_

##### New member
You were given the function to try, v=btanh(ct)v=btanh⁡(ct)\displaystyle v=b\tanh(ct), and you replaced dvdtdvdt\displaystyle \frac{dv}{dt} with its expression, but didn't replace vv\displaystyle v with its.
So...

$$\displaystyle mbc(1-tanh^2(ct)= -(mgkb tanh^2(ct))/(1-tanh^2(ct)$$

$$\displaystyle c= -( gk tanh^2(ct))/(1-tanh^2(ct))$$

How do i reach the objective of the question from here?

Thank you...

#### chlxowns_

##### New member
You were given the function to try, v=btanh(ct)v=btanh⁡(ct)\displaystyle v=b\tanh(ct), and you replaced dvdtdvdt\displaystyle \frac{dv}{dt} with its expression, but didn't replace vv\displaystyle v with its.
Sorry ignore the above one, this is what I have so far, how do I reach the question's objective from here?

#### Dr.Peterson

##### Elite Member
So...

$$\displaystyle mbc(1-tanh^2(ct)= -(mgkb tanh^2(ct))/(1-tanh^2(ct)$$

$$\displaystyle c= -( gk tanh^2(ct))/(1-tanh^2(ct))$$

How do i reach the objective of the question from here?

Thank you...
The RHS should be a subtraction, not a multiplication. That will make the result much simpler.

You goal is to show that the resulting equation is an identity, for properly chosen values of b and c.

#### Dr.Peterson

##### Elite Member
Sorry ignore the above one, this is what I have so far, how do I reach the question's objective from here?

View attachment 23195
Now distribute the LHS, and, since you want it to be true for any value of t, equate terms:

$$\displaystyle mbc = mg$$​
$$\displaystyle -mbc\tanh^2(ct) = -kb^2\tanh^2(ct)$$​

Divide by common factors, and you'll have two equations in b and c, which you can solve to find b and c.

#### chlxowns_

##### New member
Divide by common factors, and you'll have two equations in b and c, which you can solve to find b and c.
I got it!! Thank you Dr! I really appreciate it!

#### rajputaman04

##### New member
I so wish this made sense: "using tanh′(t)=1−(tanh(t))^2 show that this function has a derivative " (Altered for clarity.) Why does one need to show there is a derivative when one is simply GIVEN the derivative?