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Calc II - CU Boulder - Fall 2010 - Section005

Webwork 11.5 Hints

Math 2300 Section 005 – Calculus II – Fall 2010

Webwork 11.5 Hints – Wednesday, November 3, 2010

Half-Life Example

  1. I'll modify an example I've used before. Assume that at time t=0 I consume 300mg of caffeine (I believe this is the largest amount legally allowed in a single dose). The amount of caffeine in my system Q is then a function of time that satisfies some proportionality constant k. That is, the amount of caffeine in my system at a time t is a solution to the differential equation


    We don't know what k is, but maybe we know that the half-life of caffeine in my system is 1.5 hours. So, what we want to do is as follows: The differential equation above represents caffeine decay in my system in general. In the specific situation (corresponding to an initial condition) when I digest 300mg at time t=0, there is a specific solution. We need to find that specific solution. Let's do that, and then solve for k.

    \displaystyle\frac{dQ}{dt} \displaystyle=kQ
    \displaystyle\frac{1}{Q}\frac{dQ}{dt} \displaystyle=k
    \displaystyle\int\frac{1}{Q}\frac{dQ}{dt}\, dt \displaystyle=\int k\, dt
    \displaystyle\ln|Q| \displaystyle=kt+C
    \displaystyle Q \displaystyle=Be^{{kt}}

Differential Equations Quiz Redo

Due Date: 
Tuesday, November 2, 2010 - 16:00

This quiz was given last Friday in class. As I mentioned in class today, anyone who missed it or anyone who didn't approach the problems correctly (that is a significant portion of you) can retake it. It is due tomorrow (Tuesday) at 4PM.

Keep in mind, you do not want to start the problems by assuming what it is that you're trying to find.

Book Assignment 11

Due Date: 
Monday, November 8, 2010 - 16:00

Section 11.5 Problem 24
Section 12.1 Problems 30 and 34

Separation of Variables

Wednesday, October 27, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Separation of Variables – October 27, 2010

Definition: We call a differential equation separable if it can be written in the form


for f a function of y and g a function of x.

The Formal Approach to Solving Separable Differntial Equations

If f(y)\neq 0 then we can write


Integrating both sides with respect to the variable x gives

\displaystyle\int\frac{1}{f(y)}\frac{dy}{dx}\, dx \displaystyle=\int g(x)\, dx
\displaystyle\int\frac{1}{f(y)}\, dy \displaystyle=\int g(x)\, dx

What will usually happen is that this introduces a natural log in terms of y. Exponentiating and solving for y then provides a solution in terms of x.

Book Assignment 10

Due Date: 
Monday, November 1, 2010 - 16:00

Section 11.2 Problem 14
Section 11.4 Problems 36 and 42

Slope Fields

Monday, October 25, 2010 - 14:00 - Tuesday, October 26, 2010 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Slope Fields – October 25–26, 2010

Slope Field Examples

  1. Find an infinite collection of solutions to the differential equation


    Solution: It's nearly impossible to ask for even a single solution to this differential equation without knowing more about it. You can graph out the slope field yourself (perhaps not quite as neatly as a computer or calculator, but you can do it). You get a slope field like this.

    Now it appears a bit more obvious that we're dealing with circles centered at the origin. Let's see if that is actually correct by taking a generic circle, centered at the origin, and determining if it satisfies the differential equation. In what follows, we use implicit differentiation to view y as a function of x.

    \displaystyle x^{2}+y^{2} \displaystyle=r^{2}
    \displaystyle\frac{d}{dx}\left[x^{2}+y^{2}\right] \displaystyle=\frac{d}{dx}r^{2}
    \displaystyle 2x+2y\frac{dy}{dx} \displaystyle=0
    \displaystyle 2y\frac{dy}{dx} \displaystyle=-2x
    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{2x}{2y}
    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{x}{y}
    \displaystyle y^{{\prime}} \displaystyle=-\frac{x}{y}.

    So, each circle centered at the origin satisfies the given differential equation.

Taylor Polynomial Error Bounds

Wednesday, October 20, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Taylor Polynomial Error Bounds – October 20, 2010

We'll start by discussing the formal error bound for Taylor polynomials. (I.e., how badly does a Taylor polynomial approximate a function?) Then, we'll see some standard examples. Finally, we'll see a powerful application of the error bound formula.

Lagrange Error Bound for P_{n}(x)

We know that the nth Taylor polynomial is P_{n}(x), and we have spent a lot of time in this chapter calculating Taylor polynomials and Taylor Series. The question is, for a specific value of x, how badly does a Taylor polynomial represent its function? We define the error of the nth Taylor polynomial to be


That is, error is the actual value minus the Taylor polynomial's value. Of course, this could be positive or negative. So, we force it to be positive by taking an absolute value.


The following theorem tells us how to bound this error. That is, it tells us how closely the Taylor polynomial approximates the function. Essentially, the difference between the Taylor polynomial and the original function is at most |E_{n}(x)|. At first, this formula may seem confusing. I'll give the formula, then explain it formally, then do some examples. You may want to simply skip to the examples.

Theorem 10.1 Lagrange Error Bound  Let f be a function such that it and all of its derivatives are continuous. If P_{n}(x) is the nth Taylor polynomial for f(x) centered at x=a, then the error is bounded by


where M is some value satisfying |f^{{(n+1)}}(x)|\le M on the interval between a and x.

Book Assignment 9

Due Date: 
Monday, October 25, 2010 - 16:00

Section 10.2 Problems 28 and 32
Section 10.3 Problems 22 and 24
Section 10.4 Problems 20 and 22
Section 11.1 Problem 22

New Taylor Series From Old

Monday, October 25, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Obtaining New Taylor Series from Old – October 19, 2010

Taylor Series We Currently Know

\displaystyle\sin x \displaystyle=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{{2k+1}}
\displaystyle\cos x \displaystyle=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k)!}x^{{2k}}
\displaystyle e^{x} \displaystyle=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots=\sum _{{k=0}}^{\infty}\frac{x^{k}}{k!}
\displaystyle(1+x)^{p} \displaystyle=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots
\displaystyle\frac{1}{1+x} \displaystyle=(1+x)^{{-1}}=1-x+x^{2}-x^{3}+x^{4}-\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}
\displaystyle\ln(x+1) \displaystyle=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}

There are several ways we could go about obtaining new series from these ones. First, we can substitute inside an existing series. Then, we could differentiate or integrate existing series. We'll consider these with some examples.

Taylor Series Introduction

Monday, October 18, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Taylor Series – October 18, 2010

Definition of a Taylor Series

The Taylor series for f(x) centered at x=0 is

f(x)=f(0)+f^{{\prime}}(0)x+\frac{f^{{\prime\prime}}(0)}{2!}x^{2}+\frac{f^{{(3)}}(0)}{3!}x^{3}+\cdots=\sum _{{k=0}}^{\infty}\frac{f^{{(k)}}(0)}{k!}x^{k}.

Similarly, the Taylor series for f(x) centered at x=a is

f(x)=f(a)+f^{{\prime}}(a)(x-a)+\frac{f^{{\prime\prime}}(a)}{2!}(x-a)^{2}+\cdots=\sum _{{k=0}}^{\infty}\frac{f^{{(k)}}(a)}{k!}(x-a)^{k}.


Since a Taylor series is the same as a Taylor polynomial, but is taken to infinite degree, we already know many Taylor series. From the last section, we know the following.

\displaystyle\sin x \displaystyle=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{{2k+1}}
\displaystyle\cos x \displaystyle=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k)!}x^{{2k}}
\displaystyle e^{x} \displaystyle=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots=\sum _{{k=0}}^{\infty}\frac{x^{k}}{k!}
\displaystyle(1+x)^{p} \displaystyle=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots=\sum _{{k=0}}^{\infty}\binom{p}{k}x^{k}
\displaystyle\ln(x+1) \displaystyle=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}



There is a bit of a technicality here. In section 10.1, we wrote \approx between the functions and their Taylor polynomials. Now, we're writing = instead. (\approx means “approximately equal to,” while = means “equal to.”) The main point to understand here is detailed in the next portion of these notes.

© 2011 Jason B. Hill. All Rights Reserved.