import numpy as np
import matplotlib.pyplot as plt
from IPython.display import display, HTML, IFrame
from ipywidgets import interact,fixed
import pandas as pd
from mpl_toolkits import mplot3d
from mpl_toolkits.mplot3d import axes3d
from matplotlib.patches import FancyArrowPatch
from mpl_toolkits.mplot3d import proj3d
plt.rcParams["figure.figsize"] = [12, 9]
from numpy.linalg import norm
from numpy import cos,sin,tan,arctan,exp,log,pi,sqrt
$\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\bv}[1]{\begin{bmatrix} #1 \end{bmatrix}}$ $\renewcommand{\vec}{\mathbf}$
Announcements¶
- Quiz 3 in recitation this week.
- Curves, tangents
- Motion
- Arc length
- Homework 4 posted, due 2/18
- Midterm 1 - 2/20
- Through HW4 (partial derivatives)
- Some review materials posted to Canvas
- Review on Tues.
One-minute Review¶
A scalar field (e.g., $f(x,y)$) is a function of several variables. Its domain is the subset of input values in $\RR^n$; its image is the set of output values in $\RR$.
Level sets ("curves" for functions of 2 variables) are sets of input points associated to a particular output. For example, contour lines on a topographical map.
Level Curves¶
f = lambda x,y: x*y
g = lambda x,y: x*sin(y)
h = lambda x,y: sqrt(4-x**2-y**2)
k = lambda x,y: log(x**2+y**2)
l = lambda x,y: exp((x-y)*log(2))
@interact
def _(lev=(0.01,1.),prob={"1":[f,r"$z = x y$"],"2":[g,r"$z = x \sin(y)$"],"3":[h,r"$z = \sqrt{4-x^2-y^2}$"],"4":[k,r"$z = \ln (x^2+y^2)$"],"5":[l,r"$z = 2^{x-y}$"]},angle=(-90,120,6),vangle=(0,90,6)):
func,fs = prob
fig = plt.figure(figsize = (12,6))
ax = fig.add_subplot(121,projection='3d')
ax.view_init(vangle,angle)
for c in 'xyz':
# getattr(ax,f"set_{c}lim")([-1,1]);
getattr(ax,f"set_{c}label")(f"${c}$",size=16)
x = y = np.linspace(-1,1,400)
X,Y = np.meshgrid(x,y)
if func == h:
x = np.linspace(0,2*pi,100)
y = np.linspace(0,1.99,100)
x,y = np.meshgrid(x,y)
X = y*cos(x)
Y = y*sin(x)
ax.set_zlim3d([0,4])
else:
X = 5*X
Y = 5*Y
Z = func(X,Y)
# ax.set_autoscale_on(True)
ax.plot_surface(X,Y,Z,alpha=.3,cmap='viridis',rcount=75,ccount=75);
k = np.max(Z)*(lev)+(1-lev)*np.min(Z)
ax.contour(X,Y,Z,offset=k,levels=[k],colors=['red'])
fig.suptitle(fs,fontsize=17)
ax2 = fig.add_subplot(122)
cp = ax2.contour(X,Y,Z,cmap='viridis');
# fig.colorbar(cp); # for colorbar reference
ax2.clabel(cp,fmt='%1.1f'); # inline counour labels.
cp2 = ax2.contour(X,Y,Z,levels=[k],colors=['red'])
ax2.clabel(cp2,fmt='%1.1f'); # inline counour labels.
interactive(children=(FloatSlider(value=0.505, description='lev', max=1.0, min=0.01), Dropdown(description='pr…
Lecture 08¶
Objectives
- Explore limits and continuity of $f(x,y)$.
- Define partial derivatives
- Estimate partial derivatives from contour maps and tables.
Resources
- Content
- Visualization
- Practice
- Mooculus: Continuity Partial Derivatives
- Extras
- CalcBLUE: Partial Derivatives
Limits¶
Consider a function $f:\RR^n \to \RR$ as mapping vectors to scalars. We write $$\lim_{\vec x \to \vec p} f(\vec x) = L$$ if $|f(\vec x) - L|$ can be made arbitrarily small by making $|\vec x - \vec p|$ sufficiently small.
Limits in $\RR^2$¶
Consider a function $f:\RR^2 \to \RR$ as mapping vectors to scalars. We write $$\lim_{(x,y) \to (a,b)} f(x,y) = L$$ if $|f(x,y) - L|$ can be made arbitrarily small by making $\sqrt{(x-a)^2+(y-b)^2}$ sufficiently small.
Examples¶
See This screencast for a few more/more detail.
f = lambda x,y: (x**2 + y**2)
g = lambda x,y: x*y/(x**2 + y**2)
h = lambda x,y: x*y/sqrt(x**2 + y**2)
@interact
def _(lev=(0.01,1.),prob={"1":[f,r"$z = x^2 + y^2$"],"2":[g,r"$z = \frac{xy}{x^2 + y^2}$"],"3":[h,r"$z = \frac{xy}{\sqrt{x^2+ y^2}}$"]},angle=(-90,120,6),vangle=(0,90,6)):
func,fs = prob
fig = plt.figure(figsize = (12,6))
ax = fig.add_subplot(121,projection='3d')
ax.view_init(vangle,angle)
for c in 'xyz':
# getattr(ax,f"set_{c}lim")([-1,1]);
getattr(ax,f"set_{c}label")(f"${c}$",size=16)
x = y = np.linspace(-1,1,400)
X,Y = np.meshgrid(x,y)
Z = func(X,Y)
# ax.set_autoscale_on(True)
ax.plot_surface(X,Y,Z,alpha=.7,cmap='viridis',rcount=75,ccount=75);
k = np.max(Z)*(lev)+(1-lev)*np.min(Z)
ax.contour(X,Y,Z,offset=k,levels=[k],colors=['red'])
ax.set_title(fs)
ax2 = fig.add_subplot(122)
cp = ax2.contour(X,Y,Z,cmap='viridis');
# fig.colorbar(cp); # for colorbar reference
ax2.clabel(cp,fmt='%1.1f'); # inline counour labels.
cp2 = ax2.contour(X,Y,Z,levels=[k],colors=['red'])
ax2.clabel(cp2,fmt='%1.1f'); # inline counour labels.
interactive(children=(FloatSlider(value=0.505, description='lev', max=1.0, min=0.01), Dropdown(description='pr…
Rates of Change¶
Limits went well. Let's try derivatives as a limit of a difference quotient.
$$ \lim_{\langle x,y\rangle \to \langle a,b \rangle} \frac{f(x,y) - f(a,b)}{\langle x,y\rangle - \langle a,b \rangle}$$Blech!¶
One cannot divide by vectors!
A sensible question¶
A group of hikers follows a curving path up a mountain ridge. How steep is their path at the halfway point?
f = lambda x,y: exp(-4*(y-sin(x))**2)*(1-np.abs(x+pi/2)/5)
@interact
def _(func=fixed(f),angle=(-90,120,6),vangle=(0,90,6)):
fig = plt.figure(figsize = (12,6))
ax = fig.add_subplot(111,projection='3d')
ax.view_init(vangle,angle)
for c in 'xyz':
# getattr(ax,f"set_{c}lim")([-1,1]);
getattr(ax,f"set_{c}label")(f"${c}$",size=16)
x = np.linspace(-4,4,601)
y = np.linspace(-2,2,301)
X,Y = np.meshgrid(x,y)
Z = func(X,Y)
ax.plot_surface(X,Y,Z,alpha=.6,cmap='ocean',rcount=100,ccount=100);
t = np.linspace(-pi/2,pi/2,100)
X = t
Y = t/pi - 1/2 - t*(t**2-pi**2/4)/5
Z = func(X,Y)
ax.plot(X,Y,Z,lw=6,color='r',alpha=1)
interactive(children=(IntSlider(value=12, description='angle', max=120, min=-90, step=6), IntSlider(value=42, …
Partial Derivatives¶
We start by considering "one direction at a time".
The partial derivative of a function $f(x,y)$ with respect to $x$ at the point $(a,b)$ is $$f_x(a,b) = \lim_{h\to 0} \frac{f(a+h,b) - f(a,b)}{h}.$$
The partial derivative of a function $f(x,y)$ with respect to $y$ at the point $(a,b)$ is $$f_y(a,b) = \lim_{h\to 0} \frac{f(a,b+h) - f(a,b)}{h}.$$
f = lambda x,y: exp(-4*(y-sin(x))**2)*(1-np.abs(x+pi/2)/5)
@interact
def _(func=fixed(f),angle=(-90,120,6),vangle=(0,90,6),var=['x','y']):
fig = plt.figure(figsize = (12,6))
ax = fig.add_subplot(111,projection='3d')
ax.view_init(vangle,angle)
for c in 'xyz':
# getattr(ax,f"set_{c}lim")([-1,1]);
getattr(ax,f"set_{c}label")(f"${c}$",size=16)
x = np.linspace(-4,4,601)
y = np.linspace(-2,2,301)
X,Y = np.meshgrid(x,y)
Z = func(X,Y)
ax.plot_surface(X,Y,Z,alpha=.6,cmap='ocean',rcount=100,ccount=100);
t = np.linspace(-pi,pi,100)
X = t
Y = np.zeros_like(X)
if var == 'y':
X,Y = Y,X
ax.set_xlim([-4,4])
ax.set_ylim([-2,2])
Z = func(X,Y)
ax.plot(X,Y,Z,lw=6,color='r',alpha=1)
interactive(children=(IntSlider(value=12, description='angle', max=120, min=-90, step=6), IntSlider(value=42, …
Other notation¶
All of these are equivalent.
$$f_x = \frac{\partial f}{\partial x} = \partial_x f = f^{(1,0)}$$and there are many more.
Computing $\frac{\partial f}{\partial x}$¶
In practice, we compute partial derivatives by treating all variables except the variable in question as constant.
- $\displaystyle \frac{\partial}{\partial y} \left( x^2y - \sin(x-2y) \right)$
- $\displaystyle \frac{\partial}{\partial z} \left( \frac{z^2 \tan^{-1}(\sqrt{x^2+1})}{\cosh(xy)} \right)$
Higher Order Derivatives¶
Since the partial derivative of a function is a function, we can iterate the process.
$$f_{xx} = \frac{\partial^2 f}{\partial x^2}$$$$f_{xy} = \frac{\partial^2 f}{\partial y \partial x}$$etc.
Interpretation¶
The case of a second derivative of a single variable easily relates to the one-variable case and the concept of concavity. A function $f$ for which $f_{xx} > 0$ is said to be "concave up in the $x$-direction".
Example: Heat Equation¶
The temperature at time $t$ at position $x$ along a straight bar is given by a function $u(t,x)$. The evolution of the temperature distribution is governed by the heat equation $$u_t = u_{xx}.$$ This is a partial differential equation or PDE, but don't let it intimidate you.
One could interpret this equation as stating, "Where the temperature distribution is concave down, the bar will cool; where it is concave up, the bar will warm."
u = lambda x,t: exp(-x**2/t)/sqrt(2*pi*t)
@interact(t=(.1,7))
def _(t=.1):
x = np.linspace(-3,3,100)
plt.plot(x,u(x,t)+u(x-2,t+1/2),color='k',lw=3)
plt.ylim([0,1.5])
plt.ylabel("temp")
plt.xlabel("position")
y = np.linspace(0,1.5,10)
x,y= np.meshgrid(x,y)
plt.pcolormesh(x,y,u(x,t)+u(x-2,t+1/2),vmin=0,vmax=.6,cmap='rainbow')
interactive(children=(FloatSlider(value=0.1, description='t', max=7.0, min=0.1), Output()), _dom_classes=('wid…
u = lambda x,t: exp(-x**2/t)/sqrt(2*pi*t)
x = np.linspace(-3,3,100)
t = np.linspace(.4,5,150)
x,t = np.meshgrid(x,t)
plt.contourf(x,t,u(x,t));
plt.ylabel("time")
plt.colorbar()
plt.xlabel("position");
Mixed partials¶
A quantity like $\frac{\partial^2 f}{\partial x \partial y}$ is a little harder to wrap ones head around.
Compute all mixed partials of the following funtions:
- $f(x,y) = xy^3 - y \sin x$
- $r(x,t) = \frac{x}{x+t}$
- $u(p,q) = e^{-p\sqrt{q}}$
Clairaut's Theorem¶
If all mixed partials of a function $f$ exist and are continuous in a neighborhood of a point, then $$ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}.$$
Here is a quick illustration justifying Clairaut's Theorem.
Suppose you connect 4 points in space, each at a different height and directly over the corner of a square (side length $\Delta s$).
Exercise¶
Compute the differences of the slopes on opposite sides of the square,
@interact
def _(angle=(-90,120,6),vangle=(0,90,6),b=(.5,4.5)):
fig = plt.figure(figsize = (8,8))
ax = fig.add_subplot(111,projection='3d')
ax.view_init(vangle,angle)
for c in 'xyz':
# getattr(ax,f"set_{c}lim")([-1,1]);
getattr(ax,f"set_{c}label")(f"${c}$",size=16)
ax.plot([0,0,1,1,0],[0,1,1,0,0],[2,1,3,b,2],'r')
ax.set_xlim([-.1,1.1])
ax.set_ylim([-.1,1.1])
ax.set_zlim([0,5])
x = y = np.linspace(-.1,1.1,60)
x,y=np.meshgrid(x,y)
ax.plot_surface(x,y,(1-y)*((1-x)*2+b*x) + y*((1-x)+3*x),alpha=.5,cmap="viridis")
ax.text(1,0,b+.2,"$B$",fontsize=14)
ax.text(0,0,2+.2,"$A$",fontsize=14)
ax.text(0,1,1+.2,"$C$",fontsize=14)
ax.text(1,1,3+.2,"$D$",fontsize=14)
interactive(children=(IntSlider(value=12, description='angle', max=120, min=-90, step=6), IntSlider(value=42, …
Quick exercise¶
Compute $g_{zzxw}$ for $$g(w,x,y,z)= w^2 x^3 y z^2+\sin \left(\frac{x y}{z^2}\right).$$
Estimating Partials¶
Below is a contour plot of a function $f(x,y)$. Estimate the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at each labeled point.
X = Y = np.linspace(-3,3,400)
X,Y = np.meshgrid(X,Y)
Z = (1.5**Y*Y - X) / sqrt(X**2 + Y**2 + 1)
plt.figure(figsize=(10,10))
cs = plt.contour(X,Y,Z)
pts=np.column_stack([[-2,-.8],[-1,2.17],[.57,1]])
plt.scatter(pts[0],pts[1],color='k')
for i,ch in enumerate("ABC"):
plt.text(pts[0,i]-.1,pts[1,i]+.1,"${}$".format(ch),fontsize=14)
plt.grid(True)
plt.clabel(cs,fmt="%1.2f");
Does this make your estimates more or less accurate?
X = Y = np.linspace(-3,3,150)
X,Y = np.meshgrid(X,Y)
Z = (1.5**Y*Y - X) / sqrt(X**2 + Y**2 + 1)
plt.figure(figsize=(10,10))
pts=np.column_stack([[-2,-.8],[-1,2.17],[.57,1]])
plt.grid(True,'both')
for i,ch in enumerate("ABC"):
plt.text(pts[0,i]-.1,pts[1,i]+.1,"${}$".format(ch),fontsize=14)
cs = plt.contour(X,Y,Z,levels=np.arange(-1,3,.2))
plt.scatter(pts[0],pts[1],color='k')
plt.clabel(cs,fmt="%1.2f");
