math119: add linear approximations

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eggy 2023-01-16 21:39:40 -05:00
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@ -108,3 +108,60 @@ The order of the variables matter: $f_{xy}$ is the derivative of f wrt. x *and t
!!! warning
In multivariable calculus, **differentiability does not imply continuity**.
### Linear approximations
A **tangent plane** represents all possible partial derivatives at a point of a function.
For two-dimensional functions, the differential could be used to extrapolate points ahead or behind a point on a curve.
$$
\Delta f=f'(a)\Delta d \\
\boxed{y=f(a)+f'(a)(x-a)}
$$
The equations of the two unit direction vectors in $x$ and $y$ can be used to find the normal of the tangent plane:
$$
\vec n=\vec d_1\times\vec d_2 \\
\begin{bmatrix}-f_x(a,b) \\ -f_y(a,b) \\ 1\end{bmatrix} = \begin{bmatrix}1\\0\\f_x(a,b)\end{bmatrix}
\begin{bmatrix}0\\1\\f_y(a,b)\end{bmatrix}
$$
Therefore, the general expression of a plane is equivalent to:
$$
z=C+A(x-a)+B(x-b) \\
\boxed{z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)}
$$
??? tip "Proof"
The general formula for a plane is $c_1(x-a)+c_2(y-b)+c_3(z-c)=0$.
If $y$ is constant such that $y=b$:
$$z=C+A(x-a)$$
which must represent in the x-direction as an equation in the form $y=b+mx$. It follows that $A=f_x(a,b)$. A similar concept exists for $f_y(a,b)$.
If both $x=a$ and $y=b$ are constant:
$$z=C$$
where $C$ must be the $z$-point.
Usually, functions can be approximated via the **tangent at $x=a$.**
$$f(x)\simeq L(x)$$
!!! warning
Approximations are less accurate the stronger the curve and the farther the point is away from $f(a,b)$. A greater $|f''(a)|$ indicates a stronger curve.
!!! example
Given the function $f(x,y)=\ln(\sqrt[3]{x}+\sqrt[4]{y}-1)$, $f(1.03, 0.98)$ can be linearly approximated.
$$
L(x=1.03, y=0.98)=f(1,1)=f_x(1,1)(x-1)+f_y(1,1)(y-1) \\
f(1.03,0.98)\simeq L(1.03,0.98)=0.005
$$