Hyperbolic Function: Help Needed

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:

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

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

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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

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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

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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_

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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

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

Dr.Peterson

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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_

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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_

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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?

2020-11-19 (3).png
 

Dr.Peterson

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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

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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.
 

rajputaman04

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Nov 21, 2020
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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?
 
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