__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

__back into__

*ct**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....