quizzes

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Review Quiz 2

Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Monday, November 15, 2010

We will be having a review in class every day until the next midterm exam on Wednesday of this week. For this quiz, you are given 10 minutes to decide which problems you think you can solve and which ones you want to see solved during our review session today.

  1. Use the Lagrange Error Bound Formula for P_{n}(x) to find a reasonable bound for the error in approximating the quantity e^{{0.40}} with a third-degree Taylor polynomial for the function f(x)=e^{x} about x=0. Choose the best error estimate.

    Solution: We use the formula

    |E_{3}(x)|\le\frac{M}{4!}|x|^{4}

    where M\ge|f^{{(4)}}(x)| on the interval [0,0.4] (i.e., the interval between where our Taylor polynomial is centered and where we're approximating the value). Since f^{{(4)}}(x)=e^{x} has a maximum value of e^{{0.4}} on the interval [0,0.4], we obtain

    |E_{3}(0.4)|\le\frac{e^{{0.4}}}{4!}0.4^{4}.
  2. Use the Lagrange Error Bound Formula for P_{n}(x) to find a reasonable bound for the error in approximating the quantity 17/\sqrt{3} with a third-degree Taylor polynomial for the function

    g(x)=\frac{17}{\sqrt{4-x}}

    about x=0. Choose the best error estimate.

Review Quiz 1

Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Friday, November 12, 2010

We will be having a review in class every day until the next midterm exam on Wednesday of next week. For this quiz, you are given 10 minutes to decide which problems you think you can solve and which ones you want to see solved during our review session today.

  1. If f(2)=g(2)=h(2)=0, f^{{\prime}}(2)=h^{{\prime}}(2)=0, g^{{\prime}}(2)=22, f^{{\prime\prime}}(2)=3, g^{{\prime\prime}}(2)=5 and h^{{\prime\prime}}(2)=7, calculate

    \lim _{{x\rightarrow 2}}\frac{f(x)}{h(x)}.

    Solution: Using L'Hopital's rule we have

    \lim _{{x\rightarrow 2}}\frac{f(x)}{h(x)}=\lim _{{x\rightarrow 2}}\frac{f^{{\prime}}(x)}{h^{{\prime}}(x)}=\lim _{{x\rightarrow 2}}\frac{f^{{\prime\prime}}(x)}{h^{{\prime\prime}}(x)}=\frac{3}{7}.
  2. Find the first four nonzero terms of the Taylor series about 0 for the function

    \frac{t}{1+t}.

    Solution: Notice that

    \frac{d}{dt}\frac{t}{1+t}=\frac{1}{(1+t)^{2}}=(1+t)^{{-2}}.

    Since this is a binomial series with p=-2 we will expand this series and then integrate it term by term to obtain the series that we need. We have

    \displaystyle(1+t)^{{-2}} \displaystyle=1-2t+\frac{(-2)(-3)}{2!}t^{2}+\frac{(-2)(-3)(-4)}{3!}t^{3}+\cdots
    \displaystyle=1-2t+\frac{6}{2!}t^{2}-\frac{24}{3!}t^{3}+\cdots
    \displaystyle=1-2t+3t^{2}-4t^{3}+\cdots

    Now integrate term by term and find

    \displaystyle\int(1+t)^{{-2}}\, dt \displaystyle=\int 1-2t+3t^{2}-4t^{3}+\cdots\, dt
    \displaystyle\frac{t}{1+t} \displaystyle=t-t^{2}+t^{3}-t^{4}+\cdots

    Since we're only looking for the first four terms, we're done.

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.

Midterm 2 Review Quiz

Due Date: 
Tuesday, October 12, 2010 - 16:00

Name:

Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Due Tuesday, October 12, 2010

This quiz is intended to be a review, so it contains more problems than usual. I won't be grading each problem individually, but the quiz will be graded as a whole. Solutions will be posted on Tuesday.

  1. Find the sum

    \sum _{{n=3}}^{{12}}\frac{3^{n}+6}{4^{n}}

    Hint: Write this as two geometric series.

Sequence and Series Quiz

Due Date: 
Friday, October 1, 2010 - 16:00 - Monday, October 4, 2010 - 16:00

This quiz is due on Monday at 4PM. Obviously, only questions 1 and 2 count for credit. Due to my current lack of creativity, there is no replacement question on this quiz.

Take Home Quiz - Improper Integrals

Due Date: 
Monday, September 13, 2010 - 16:00

MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Due Monday, September 13, 2010

  1. Evaluate the integral

    \int _{1}^{4}\frac{dx}{(x-2)^{{2/3}}}\, dx
  2. Evaluate the integral

    \int _{{-\infty}}^{0}\frac{e^{x}\, dx}{3-2e^{x}}
  3. Replacement Question: While trying to escape a wildfire, a family of four prairie dogs comes to a bridge. They realize that the bridge is old and only two of them may cross at a single time. Making matters worse, it is night and they only have one headlamp. So, two will cross and then one will return with the headlamp, then two more will cross and one will return, etc. When two prairie dogs cross the bridge, they will always travel at the pace of the slower prairie dog. Daddy Prarie Dog can cross the bridge in 1 minute, Mommy Prairie Dog can cross in 2 minutes, Junior Prairie Dog crosses in 5 minutes and Baby Prairie Dog crosses in 10 minutes. If the fire will arrive at their location and burn the bridge, taking with it any possibility of escape, in 17 minutes and 30 seconds, is there a chance for the family to survive?

Partial Fractions Quiz

MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Friday, September 3, 2010

  1. Explain how you know that an integration problem might involve integration by parts. That is, what clues you in to the idea that integration by parts might be a valid approach to solving an integral?

    Solution: The main thing I am looking for here is something along the lines of the following: The integrand appears to be a product of two functions u and v^{{\prime}}, where we can find u^{{\prime}} and v and the integral of u^{{\prime}}v is in some way easier than the original integral. But, there are other instances when integration by parts comes up that don't look like this, such as when integrating \ln x.

  2. Solve the integral, showing all calculations that lead to your result. (You may use a calculator, but your calculations need to be capable of being followed as if the integral is computed by hand.)

    \int\frac{3}{(x-2)(x-5)}\, dx

    Solution: This is an obvious partial fractions problem. First, write

    \int\frac{3}{(x-2)(x-5)}\, dx=3\int\frac{1}{(x-2)(x-5)}\, dx.

    and then compute the partial fractions.

    \frac{1}{(x-2)(x-5)}=\frac{A}{x-2}+\frac{B}{x-5}

    Multiplying on both sides by (x-2)(x-5) and then putting the polynomial on the right in standard form gives

    \displaystyle 1 \displaystyle=A(x-5)+B(x-2)
    \displaystyle 1 \displaystyle=(A+B)x-5A-2B

    This results in the system of equations:

    \displaystyle A+B \displaystyle=0
    \displaystyle-5A-2B \displaystyle=1

    which has the solution A=-\frac{1}{3} and B=\frac{1}{3}. So, we have

    \displaystyle 3\int\frac{1}{(x-2)(x-5)}\, dx \displaystyle=3\int\frac{-1/3}{x-2}+\frac{1/3}{x-5}\, dx
    \displaystyle=-\int\frac{1}{x-2}\, dx+\int\frac{1}{x-5}\, dx
    \displaystyle=-\ln|x-2|+\ln|x-5|+C.

Friday Quiz 1

MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Friday, August 27, 2009

  1. Explain how you know that an integration problem might involve substitution. That is, what clues you in to the idea that substitution might be a valid approach to solving an integral?

    Solution: I will accept answers that make sense here. What I would write is as follows. The biggest hint that an integral might involve substitution is when the integral contains an inside function and that inside function's derivative (up to some constant) is somewhere in the outside of the integrand.

  2. Solve the integral, showing all calculations that lead to your result. (You may use a calculator, but your calculations need to be capable of being followed as if the integral is computed by hand.)

    \int _{{-\pi}}^{\pi}\pi\sin(\pi^{2}x)\, dx

    Solution: There are several ways to solve this. We know what the curve \sin x looks like, and integrating it from -n to n for any real number n will give us zero. If we scale that curve by \pi vertically and compress horizontally by a factor of \pi^{2}, we expect the same to be true. So, right away, we know that we'll end with zero. Using the substitution w=\pi^{2}x we find the following.

    \displaystyle\int _{{-\pi}}^{\pi}\pi\sin(\pi^{2}x)\, dx \displaystyle=\frac{1}{2}\int _{{-\pi^{3}}}^{{\pi^{3}}}\sin(w)\, dw
    \displaystyle=\frac{1}{2}\left[-\cos w\right]_{{-\pi^{3}}}^{{\pi^{3}}}
    \displaystyle=\frac{1}{2}\left[-\cos(\pi^{3})+\cos(-\pi^{3})\right]
    \displaystyle=\frac{1}{2}\left[-\cos(\pi^{3})+\cos(\pi^{3})\right]
    \displaystyle=0.

    The second to last step is true because \cos(\theta)=\cos(-\theta) for all \theta.

Office Quiz

Due Date: 
Monday, August 23, 2010 - 14:00 - Friday, August 27, 2010 - 16:00

The purpose of this quiz is for you to be able to find my office. It also lets me know when office hours would be best for you. Fill out the attached pdf and deliver it to my office before Friday, August 27 at 3 PM. If I am not at my office, then either leave the quiz on my desk (assuming someone else is at the office) or slide it under the door (assuming nobody is at the office and the door is closed).

My office is in Math 340, in the Mathematics Department near the intersection of Colorado and Folsom. (See the link to the left.)

© 2011 Jason B. Hill. All Rights Reserved.