Hyperbolic Function: Help Needed

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chlxowns_

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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:

[MATH]m(dv/dt) = mg - kv^2[/MATH]
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:

[MATH]v(t)=b tanh(ct)[/MATH]
where b and c are constants.

using [MATH]tanh'(t)=1-tanh^2(t)[/MATH] show that this function has a derivative

[MATH]dv/dt=b(1-tanh^2(ct))c[/MATH]
Then substitute these formulas for v(t) and d [MATH]dv/dt[/MATH] 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

[MATH]v(t)=btanh(ct)[/MATH]
Differentiate the following function using chain rule,

[MATH]f(g(x)), dy/dx=f'(g(x))× g'(x)[/MATH]
Let [MATH]g(t)=ct , g'(t)=c[/MATH]
Let [MATH]f(g(t))=tanh(g(t)) , f'(g(t))=1-tanh^2(g(t)) [/MATH]
Substitute ct back into g(t)

[MATH]f'(g(t))=1-tanh^2(ct) [/MATH]
Apply chain rule:

[MATH]dv/dt=b(1-tanh^2(ct))× c[/MATH]
Equate the derived equation with original equation.

[MATH]mbc(1-tanh^2(ct))=mg-kv^2[/MATH]
[MATH]mbc(sech^2(ct))=mg-kvg^2[/MATH]
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....
 
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?
 
tkhunny said:
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?
Click to expand...

I think it's meant to say:

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

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
[MATH]f(t)=\tanh(x)[/MATH], so that [MATH]v(t) = b f(ct) = b f(g(t))[/MATH].

chlxowns_ said:
Then substitute these formulas for v(t) and d [MATH]dv/dt[/MATH] 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

[MATH]mbc(sech^2(ct))=mg-kvg^2[/MATH]
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....
Click to expand...
You didn't replace [MATH]v[/MATH] with its value!
 
Dr.Peterson said:
I think it's meant to say:

Using the fact that [MATH]\tanh'(t)=1-\tanh^2(t)[/MATH], show that the function [MATH]v(t)[/MATH] has derivative [MATH]dv/dt=b(1−\tanh^2(ct))c[/MATH]​
Click to expand...
Altered for even more clarity! :)
 
Yes, there is a value for v! It is b tanh(ct). You were told that right at the start.
 
chlxowns_ said:
There is a value for [MATH]v[/MATH]? And where do I substitute it? I am so sorry, this is my first time encountering this type of question... Please bear with me...
Click to expand...
Yes, [MATH]v[/MATH] is the function that you want to be a solution of the differential equation, [MATH]m\frac{dv}{dt}=mg−kv^2[/MATH].

You were given the function to try, [MATH]v=b\tanh(ct)[/MATH], and you replaced [MATH]\frac{dv}{dt}[/MATH] with its expression, but didn't replace [MATH]v[/MATH] with its.
 
Dr.Peterson said:
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.
Click to expand...

So...

[MATH]mbc(1-tanh^2(ct)= -(mgkb tanh^2(ct))/(1-tanh^2(ct)[/MATH]
[MATH]c= -( gk tanh^2(ct))/(1-tanh^2(ct))[/MATH]
How do i reach the objective of the question from here?

Thank you...
 
Dr.Peterson said:
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.
Click to expand...

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

2020-11-19 (3).png
 
chlxowns_ said:
So...

[MATH]mbc(1-tanh^2(ct)= -(mgkb tanh^2(ct))/(1-tanh^2(ct)[/MATH]
[MATH]c= -( gk tanh^2(ct))/(1-tanh^2(ct))[/MATH]
How do i reach the objective of the question from here?

Thank you...
Click to expand...
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.
 
chlxowns_ said:
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
Click to expand...
Now distribute the LHS, and, since you want it to be true for any value of t, equate terms:

[MATH]mbc = mg[/MATH]​
[MATH]-mbc\tanh^2(ct) = -kb^2\tanh^2(ct)[/MATH]​

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