Drew Youngren dcy2@columbia.edu
The partial derivative of a function $f(x_1,\ldots,x_n)$ with respect to $x_i$ is \[\frac{\partial f}{\partial x_i} = \lim_{h\to 0} \frac{f(x_1,\ldots,x_i+h,\ldots,x_n) - f(x_1,\ldots,x_n)}{h}.\]
A linear function on $\RR^n$ has the form \[L(x_1, \ldots, x_n) = a_0 + \sum_{i=1}^n a_i x_i\] where the $a_i$'s are constants.
*really affine linear
In one variable, a linear function $L(x)$ is one whose graph is a line\[y = ax + b.\]
In two variables, a linear function $L(x,y)$ is one whose graph is a plane \[z = ax + by + c. \]
In one dimension, a function $f(x)$ is differentiable at $a$ if $f'(a)$ exists.
Graphically, this means the graph resembles a line at small scales.
In 2-D, a function $f(x,y)$ is differentiable at a point if its graph looks like a plane close-up.
A scalar field $f(x_1, \ldots, x_n)$ is differentiable at $(p_1, \ldots, p_n)$ provided there is a linear function $L(x_1, \ldots, x_n)$ such that \[f(x_1, \ldots, x_n) - L(x_1, \ldots, x_n) = \sum_{i=1}^n \epsilon_i (x_i - p_i)\] where each $\epsilon_i \to 0$ as $\vec x \to \vec p$.
Show $f(x) = x^3$ is differentiable at $x=1$.
Solution. Let $L(x) = 3x - 2$. Then, \[ f(x) - L(x) = x^3 - 3x + 2 \] \[ = (x^2 + x - 2)(x - 1) \]
We start with the graph of a smooth 2-D function \[ z = f(x,y) = \ln(x + 2y + 1) \] and calculate the tangent plane at $(1,2, \ln 6)$.
Use linearization to estimate values of a function nearby known points.
\[f(x,y,z) = e^{-x^2 + 3y + 3z}\] Estimate $f(3.01, 2.02, 0.99)$.
Estimate $\sqrt{2/7}$.
Estimate $\sqrt{2/7}$.
Estimate $\sqrt{2/7}$.
An alternative, compact approach to the same concept is to estimate changes in functions using differentials.
We write $\Delta f = f(x,y) - f(a,b) \approx $ \[df = f_x(a,b)dx + f_y(a,b)dy\]
A cylindrical aluminum can is is 5cm high with a diameter of 6cm and is 0.4mm thick. Estimate the volume of the aluminum.
Solution. Let $V = \pi r^2 h$. The volume of the aluminum is approximately \[ dV = 2\pi r h\, dr + \pi r^2 \,dh = \pi(30\times0.04 + 9\times 0.08) \]