I sincerely hope you can provide me with some help. I am currently studying for my advanced Calculus and I got stuck on this question.
The question in regard is: Calculate the linearizations of φ and ψ around 2.
With the functions being given by, $φ(x)^2$ + $sin(φ(x)ψ(x))$ = 2 − x, and $φ(x) + ψ(x)$ = $x^2$
What I tried Well, what I have tried to do was to create both functions where the right-hand side would be zero and then differentiate them both with respect to x from this you get a system of equations and then just solve them for both.
I see that I have to use the chain rule ( and product rule for sin(phi) to get them but when I differentiate $φ(x)^2$ I get$ 2φ(x)_x$ whereas the answer gives me 2φ(x) I have not tried differentiating $sin(φ(x)ψ(x))$ yet but I expect it to become something like $φ_x(x)ψ(x)cos(φ(x)ψ(x))$+$φ(x)ψ_x(x)cos(φ(x)ψ(x))$
If somebody could help me it would be greatly appreciated. Picture attached with answers from the professor.enter image description here
You want to write both functions as
$$\phi(x) \approx \phi(2) + \phi'(2) \, (x-2)$$
$$\psi(x) \approx \psi(2) + \psi'(2) \, (x-2)$$
So you need to calculate $\phi(2)$, $\psi(2)$, $\phi'(2)$ and $\psi'(2)$.
$$$$
(a) $\phi(2)$ and $\psi(2)$
$$\begin{matrix} \phi^2(x) + \sin(\,\phi(x)\,\psi(x)\,) = 2 − x & \rightarrow & \phi^2(2) + \sin(\,\phi(2)\,\psi(2)\,) = 0\\ \phi(x) + \psi(x) = x^2 & \rightarrow & \phi(2) + \psi(2) = 4 \end{matrix}$$
This system is transcendental and may have more than one solution. But $\phi(2) = 0$ and $\psi(2) = 4$ is clearly one of them.
$$$$
(b) $\phi'(2)$ and $\psi'(2)$
Differentiate each side of both equations to get:
$$\begin{matrix} 2\,\phi(x)\,\phi'(x) + \cos(\,\phi(x)\,\psi(x)\,) \, [\,\phi'(x)\,\psi(x)+\phi(x)\,\psi'(x)\,] = -1 & \rightarrow & 2\,\phi(2)\,\phi'(2) + \cos(\,\phi(2)\,\psi(2)\,) \, [\,\phi'(2)\,\psi(2)+\phi(2)\,\psi'(2)\,] = -1\\ \phi'(x) + \psi'(x) = 2x & \rightarrow & \phi'(2) + \psi'(2) = 4 \end{matrix}$$
$$ \left[\begin{matrix} 2\,\phi(2) + \cos(\,\phi(2)\,\psi(2)\,) \, \psi(2) & \cos(\,\phi(2)\,\psi(2)\,) \, \phi(2)\\ 1 & 1 \end{matrix}\right] \left[\begin{matrix} \phi'(2)\\ \psi'(2) \end{matrix}\right] = \left[\begin{matrix} -1\\ 4 \end{matrix}\right] $$
For $\phi(2) = 0$ and $\psi(2) = 4$, we get:
$$ \left[\begin{matrix} 4 & 0\\ 1 & 1 \end{matrix}\right] \left[\begin{matrix} \phi'(2)\\ \psi'(2) \end{matrix}\right] = \left[\begin{matrix} -1\\ 4 \end{matrix}\right] $$
And finally, $\phi'(2) = -1/4$ and $\psi'(2) = 17/4$.