\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Calculus 2 - Practice Exam 2 (Semester A)

Limits

Using Limits

Limits FAQs

Derivatives

Graphing

Related Rates

Optimization

Other Applications

Integrals

Improper Integrals

Trig Integrals

Length-Area-Volume

Applications - Tools

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Tools

Calculus Tools

Additional Tools

Articles

SV Calculus

MV Calculus

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Tools

Calculus Tools

Additional Tools

Articles

This is the second exam for second semester single variable calculus.

Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

1.5 hours

Questions

11

Total Points

Tools

Calculator

none

Formula Sheet(s)

none

Other Tools

none

Instructions:
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.

[7 points] Evaluate \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Problem Statement

[7 points] Evaluate \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Final Answer

\(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) does not exist

Problem Statement

[7 points] Evaluate \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Solution

Built with GeoGebra

\(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Try direct substitution first.

\(\displaystyle{ \frac{3(-3)^2-10}{(-3)^2+2(-3)-3} = \frac{17}{0} }\)

Since the numerator is non-zero and the denominator is zero, we have three possible answers, \(\pm\infty\) or DNE (does not exist). In order to determine which we have, we need to determine the left and right-sided limits.
It will facilitate our discussion to factor the denominator.
\(\displaystyle{ \frac{3r^2-10}{r^2+2r-3} = \frac{3r^2-10}{(r+3)(r-1)}}\)
First, let's look at the left side, i.e. \(r \lt -3\). When \(r \lt -3\) but very close to -3, the numerator is positive. The term \(r+3\) is negative and \(r-1\) is also negative.
So the signs look like \(\displaystyle{ \frac{+}{(-)(-)} \to + }\), which tells us \(\displaystyle{ \lim_{r \to -3^-}{ \frac{3r^2-10}{r^2+2r-3} } = +\infty }\).

Now let's look at the right side of \(r=-3\). When \(r \gt -3\), the numerator remains positive and \(r-1\) term remains negative. However, since \(r \gt -3\), \(r+3\) is now positive. So the signs look like \(\displaystyle{ \frac{+}{(+)(-)} \to - }\), which tells us \(\displaystyle{ \lim_{r \to -3^+}{ \frac{3r^2-10}{r^2+2r-3} } = -\infty }\).
Since \(\displaystyle{ \lim_{r \to -3^-}{ \frac{3r^2-10}{r^2+2r-3} } \neq \lim_{r \to -3^+}{ \frac{3r^2-10}{r^2+2r-3} } }\), the limit \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) does not exist.
We have included a plot of the function even though the problem statement did not ask for one.

Final Answer

\(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) does not exist

close solution

Log in to rate this practice problem and to see it's current rating.

[7 points] Evaluate \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Problem Statement

[7 points] Evaluate \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Final Answer

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } = -2 }\)

Problem Statement

[7 points] Evaluate \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Solution

Direct substitution yields, \(0/0\) which is indeterminate. So let's try L'Hopitals rule next.

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

\(\displaystyle{ \lim_{x \to 0}{ \frac{(4x)\cos(2x^2)}{(-2x)e^{-x^2}} } = \frac{0}{0} }\)

This is also indeterminate and it looks like we will not get anywhere by repeating L'Hopitals rule. So we need another strategy. We need to try to get rid of either the sine term or the exponential.
With the sine term, if we had something like \(\sin(\theta)/\theta\), the limit as \(\theta\) approaches zero is one. So let's try that.

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Multiply the numerator and denominator by \(2x^2\).

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} \frac{2x^2}{2x^2} } }\)

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{2x^2} \frac{2x^2}{e^{-x^2}-1} } }\)

Since \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{2x^2} } = 1 }\) we are left with

\(\displaystyle{ \lim_{x \to 0}{ \frac{2x^2}{e^{-x^2}-1} } }\)

Now let's try L'Hopitals rule again.

\(\displaystyle{ \lim_{x \to 0}{ \frac{4x}{-2x e^{-x^2}} } }\)

Direct substitution still yields \(0/0\), However, it looks like we are getting somewhere. So use L'Hopitals rule again.

\(\displaystyle{ \lim_{x \to 0}{ \frac{4}{-2x e^{-x^2}(-2x) + e^{-x^2}(-2)} } }\)

Now direct substitution gives us \(4/(-2) = -2\)

Final Answer

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } = -2 }\)

close solution

Log in to rate this practice problem and to see it's current rating.

[7 points] Evaluate \(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } }\) or show that it diverges.

Problem Statement

[7 points] Evaluate \(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } }\) or show that it diverges.

Final Answer

\(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } = \frac{\ln 3}{2} }\)

Problem Statement

[7 points] Evaluate \(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } }\) or show that it diverges.

Solution

This is an improper integral with at least one issue that we need to deal with, an infinite upper limit. The lower limit looks okay since at \(x=2\), the integrand is defined. The other thing we need to check for is a problem within the limits of integration.
For this integrand, the problem points would be where the denominator is zero. So \(x^2-1=0 \to x=\pm 1\). However, both of those points are outside the limits of integration, so we do not need to consider them.
So, it looks like the only issue is the upper limit. Let's write the limit.
\(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } = \lim_{b \to \infty}{ \int_2^b{ \frac{1}{x^2-1} dx } } }\)
Now we will evaluate the integral. Since we want to save some writing, we will drop the limits of integration for now.

\(\displaystyle{ \int{ \frac{1}{x^2-1} dx } }\)

Since the denominator will factor, we can use partial fractions.

\(\displaystyle{ \int{ \frac{-1/2}{x+1} + \frac{1/2}{x-1} dx } }\)

\( (-1/2)\ln|x+1| + (1/2)\ln|x-1| \)

Now evaluate the limit.

\(\displaystyle{ (1/2)\lim_{b \to \infty}{ [ \ln|x-1| - \ln|x+1| ]_2^b } }\)

\(\displaystyle{ (1/2)\lim_{b \to \infty}{ [ \ln|b-1| - \ln|b+1| - \ln|1| + \ln|3| ] } }\)

The last two terms are constants, but we need to look closer at the first two. Direct substitution yields \(\infty - \infty\) which is indeterminate. So, to use L'Hopitals Rule, we need to a fraction.

\(\displaystyle{ \lim_{b \to \infty}{ [ \ln|b-1| - \ln|b+1| ] } }\)

\(\displaystyle{ \lim_{b \to \infty}{ \ln \left| \frac{b-1}{b+1} \right| } }\)

Apply L'Hopitals Rule

\(\displaystyle{ \lim_{b \to \infty}{ \ln \left| \frac{1}{1} \right| } = 0 }\)

So now we have \( (1/2)( 0 - 0 + \ln 3 ) = (1/2)\ln 3 \)

Final Answer

\(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } = \frac{\ln 3}{2} }\)

close solution

Log in to rate this practice problem and to see it's current rating.

[7 points] Evaluate \(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) or show that it diverges.

Problem Statement

[7 points] Evaluate \(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) or show that it diverges.

Final Answer

\(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) does not converge

Problem Statement

[7 points] Evaluate \(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) or show that it diverges.

Solution

\(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\)

\(\displaystyle{ \lim_{b \to \infty}{ \int_{\pi}^{b}{ \sin(2x) dx } } }\)

\(\displaystyle{ \lim_{b \to \infty}{ [\cos(2x)/2]_{\pi}^b } }\)

\(\displaystyle{ (1/2)\lim_{b \to \infty}{ [ \cos(2b) - \cos(2\pi) ] } }\)

\( (1/2)\displaystyle{ \lim_{b \to \infty}{ \cos(2b) } } - (1/2) \)

The limit in the last equation does not converge.

Final Answer

\(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) does not converge

close solution

Log in to rate this practice problem and to see it's current rating.

[14 points] Find the convergence set for the power series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(x+1)^k } }\)

Problem Statement

[14 points] Find the convergence set for the power series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(x+1)^k } }\)

Hint

The convergence set is what 17calculus calls the interval of convergence, which includes information about the endpoints.

Problem Statement

[14 points] Find the convergence set for the power series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(x+1)^k } }\)

Final Answer

The convergence set is \((-4,2)\).

Problem Statement

[14 points] Find the convergence set for the power series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(x+1)^k } }\)

Hint

The convergence set is what 17calculus calls the interval of convergence, which includes information about the endpoints.

Solution

First we will use the Ratio Test to determine the radius of convergence and then check the endpoints to get the interval of convergence.

\(\displaystyle{ \lim_{k \to \infty}{ \left| \frac{a_{k+1}}{a_k} \right| } < 1 }\)

\(\displaystyle{ a_k = \frac{k^2}{3^k}(x+1)^k }\)

\(\displaystyle{ a_k = \frac{(k+1)^2}{3^{k+1}}(x+1)^{k+1} }\)

\(\displaystyle{ \lim_{k \to \infty}{ \left| \left[ \frac{(k+1)^2(x+1)^{k+1} }{3^{k+1}} \right] \left[ \frac{3^k}{k^2(x+1)^k} \right] \right| } < 1 }\)

\(\displaystyle{ \lim_{k \to \infty}{ \left| \frac{(k+1)^2}{k^2} \frac{3^k}{3^{k+1}} \frac{(x+1)^{k+1}}{(x+1)^k} \right| } < 1 }\)

\(\displaystyle{ \lim_{k \to \infty}{ \left| \left[ \frac{k+1}{k} \right]^2 \frac{1}{3} (x+1) \right| } < 1 }\)

\(\displaystyle{ \left| \frac{x+1}{3} \right| < 1 }\)

\(\displaystyle{ \left| x+1 \right| < 3 }\)

\( -3 \lt x+1 \lt 3 \)

radius of convergence \( -4 \lt x \lt 2 \)

Now we need to check the endpoints.

For \(x=-4\), we have \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(-3)^k } }\)

\(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k k^2 } }\)

Since \(\displaystyle{ \lim_{k \to \infty}{(-1)^k k^2} \neq 0 }\) the series diverges by the divergence test.

For \(x=2\), we have \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(3)^k } }\)

\(\displaystyle{ \sum_{k=1}^{\infty}{ k^2 } }\)

Since \(\displaystyle{ \lim_{k \to \infty}{k^2} \neq 0 }\) the series diverges by the divergence test.

Final Answer

The convergence set is \((-4,2)\).

close solution

Log in to rate this practice problem and to see it's current rating.

[10 points] Given the power series \(\displaystyle{ \frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots }\) for \(-1 \lt t \lt 1 \),
find a power series for \(\displaystyle{ \frac{x^2}{x^2+3} }\).

Problem Statement

[10 points] Given the power series \(\displaystyle{ \frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots }\) for \(-1 \lt t \lt 1 \),
find a power series for \(\displaystyle{ \frac{x^2}{x^2+3} }\).

Final Answer

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n-1} x^{2n}}{3^n} } }\)

Problem Statement

[10 points] Given the power series \(\displaystyle{ \frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots }\) for \(-1 \lt t \lt 1 \),
find a power series for \(\displaystyle{ \frac{x^2}{x^2+3} }\).

Solution

We need to get our power series \(\displaystyle{ \frac{x^2}{x^2+3} }\) into the form \(\displaystyle{ \frac{1}{1-t} }\).

\(\displaystyle{ \frac{x^2}{x^2+3} }\)

\(\displaystyle{ \frac{x^2}{3+x^2} }\)

\(\displaystyle{ \frac{-x^2}{-3-x^2} \frac{-1/3}{-1/3} }\)

\(\displaystyle{ \frac{x^2/3}{1+x^2/3} }\)

\(\displaystyle{ \frac{x^2/3}{1-(-x^2/3)} }\)

Comparing the last equation with \(\displaystyle{ \frac{1}{1-t} }\), we can see that \(t=-x^2/3\). Now use that in the series equation to get

\(\displaystyle{ (x^2/3)\frac{1}{1-(-x^2/3)} }\)

\(\displaystyle{ \frac{x^2}{3} \left[ 1 + \frac{-x^2}{3} + \left( \frac{-x^2}{3} \right)^2 + \left( \frac{-x^2}{3} \right)^3 + \ldots \right] }\)

\(\displaystyle{ \frac{x^2}{3} - \frac{x^4}{3^2} + \frac{x^6}{3^3} - \frac{x^8}{3^4} + \ldots }\)

Although this answer is probably sufficient, we can write it in a more compact form.
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n-1} x^{2n}}{3^n} } }\)

Final Answer

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n-1} x^{2n}}{3^n} } }\)

close solution

Log in to rate this practice problem and to see it's current rating.

[4 points] Write a geometric series that converges to 5.

Problem Statement

[4 points] Write a geometric series that converges to 5.

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{4}{5} \right]^n } }\)

Problem Statement

[4 points] Write a geometric series that converges to 5.

Solution

A geometric series that converges is \(\displaystyle{ \sum_{n=0}^{\infty}{ r^n } = \frac{1}{1-r} }\) where \( 0 < |r| < 1 \).
If we want this equal to 5, we need solve \(\displaystyle{ \frac{1}{1-r} = 5 }\) for r.
\(\begin{array}{rcl} \displaystyle{ \frac{1}{1-r} } & = & 5 \\ \displaystyle{ \frac{1}{5} } & = & 1-r \\ r & = & 1 - \displaystyle{ \frac{1}{5} } \\ r & = & \displaystyle{ \frac{4}{5} } \end{array}\)
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{4}{5} \right]^n } }\) is the series and, since \( 0 < |r=4/5| < 1 \) holds, the series converges.

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{4}{5} \right]^n } }\)

close solution

Log in to rate this practice problem and to see it's current rating.

[7 points] Determine whether the series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k \frac{2k}{k^2+1} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Problem Statement

[7 points] Determine whether the series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k \frac{2k}{k^2+1} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Final Answer

The series converges by the Alternating Series Test.

Problem Statement

[7 points] Determine whether the series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k \frac{2k}{k^2+1} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Solution

Since we obviously have an alternating series, we will start with the alternating series test.
Condition 1 - \(\displaystyle{ \lim_{k \to \infty}{ a_k } = 0 }\) must hold
For this problem, \(\displaystyle{ a_k = \frac{2k}{k^2+1} }\). The limit \(\displaystyle{ \lim_{k \to \infty}{ \frac{2k}{k^2+1} } = 0 }\) holds. So condition 1 holds.
Condition 2 - \( 0 \lt a_{k+1} \leq a_k \) must hold
\(\displaystyle{ a_{k+1} = \frac{2(k+1)}{(k+1)^2+1} }\)
Since \(k \gt 0 \), then the first part of the inequality holds, i.e. \( 0 \lt a_{k+1} \). Now we to see if the second part holds.

\(a_{k+1} \leq a_k \)

\(\displaystyle{ \frac{2(k+1)}{(k+1)^2+1} \leq \frac{2k}{k^2+1} }\)

\( (k+1)(k^2+1) \leq k[ (k+1)^2 + 1 ] \)

\( k^3 + k + k^2 +1 \leq k^3 + 2k^2 + 2k \)

\( 1 \leq k^2+k \)

The last inequality holds for \(k \geq 1 \). If there any question in your mind, you can take the derivative \(2k+1\) and notice that it is always positive. So the values \(k^2+k\) are increasing. And since at \(k=1\) we have \(1^2+1 =2\), the values of \(k^2+k\) are always greater than 1.
Therefore, condition 2 holds as well. Since both conditions hold, the series converges by the alternating series test.

Final Answer

The series converges by the Alternating Series Test.

close solution

Log in to rate this practice problem and to see it's current rating.

[7 points] Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^2} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Problem Statement

[7 points] Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^2} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Final Answer

The series converges by the direct comparison test.

Problem Statement

[7 points] Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^2} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Solution

This problem is solved in practice problem 162 on the direct comparision test page.

Final Answer

The series converges by the direct comparison test.

close solution

Log in to rate this practice problem and to see it's current rating.

[16 points] (a) Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).
(b) Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Problem Statement

[16 points] (a) Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).
(b) Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Hint

You may use the formula \(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} }\) for some number c between a and x.

Problem Statement

[16 points] (a) Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).
(b) Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Final Answer

(a) \( P_2(x) = (x-1) - (1/2)(x-1)^2 \)
(b) \(\displaystyle{ |R_2(x)| = \left| \frac{-1}{3c^3} (x-1)^{3} \right| }\) or \(\displaystyle{ |R_2(x)| = \frac{8}{3} \left|(x-1)^{3} \right| }\)

Problem Statement

[16 points] (a) Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).
(b) Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Hint

You may use the formula \(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} }\) for some number c between a and x.

Solution

(a) The second order Taylor polynomial is given by
\(\displaystyle{ P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 }\)
Let's build a table with the values that we need and then plug them into this equation for \(a=1\).

\( f(x) = \ln x \)

\( f(1) = \ln 1 = 0 \)

\( f'(x) = 1/x \)

\(f'(1) = 1/1 = 1 \)

\( f''(x) = (-1)x^{-2} \)

\( f''(1) = -1 \)

So now our Taylor polynomial is \( P_2(x) = 0 + 1(x-1) + (-1/2)(x-1)^2 \)
Simplifying this a bit gives us \( P_2(x) = (x-1) - (1/2)(x-1)^2 \)

Final Answer - Part (a)

\( P_2(x) = (x-1) - (1/2)(x-1)^2 \)

(b) For this part of the solution, we use the hint \(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} }\). The bound is given by taking the absolute value of this expression, i.e. \(\displaystyle{ |R_n(x)| = \left| \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \right| }\). In our case \(n=2\), so we have \(\displaystyle{ |R_2(x)| = \left| \frac{f^{(3)}(c)}{3!} (x-1)^{3} \right| }\).
The derivative we need is \(f^{(3)}(x) = -2x^{-3} \), so we have
\(\displaystyle{ |R_2(x)| = \left| \frac{-2(c^{-3})}{6} (x-1)^{3} \right| }\)
Simplifying a bit gives us our final answer.

Note: It is not clear from the problem statement if you are asked to stop here or to determine the upper bound on the error by finding \(max{|1/(3c^3)|}\). So it would be good to ask your instructor. If they want you to find the maximum on the given interval, it occurs when \(c=1/2\), so your expression would be \(\displaystyle{ |R_2(x)| = \frac{8}{3} \left|(x-1)^{3} \right| }\).

Final Answer - Part (b)

\(\displaystyle{ |R_2(x)| = \left| \frac{-1}{3c^3} (x-1)^{3} \right| }\) or \(\displaystyle{ |R_2(x)| = \frac{8}{3} \left|(x-1)^{3} \right| }\)

Final Answer

(a) \( P_2(x) = (x-1) - (1/2)(x-1)^2 \)
(b) \(\displaystyle{ |R_2(x)| = \left| \frac{-1}{3c^3} (x-1)^{3} \right| }\) or \(\displaystyle{ |R_2(x)| = \frac{8}{3} \left|(x-1)^{3} \right| }\)

close solution

Log in to rate this practice problem and to see it's current rating.

[14 points] (a) Name the curve with polar equation \(r=2\sin\theta\) and sketch the graph.
(b) Set up and evaluate an integral in polar coordinates for the area inside the graph of part (a).

Problem Statement

[14 points] (a) Name the curve with polar equation \(r=2\sin\theta\) and sketch the graph.
(b) Set up and evaluate an integral in polar coordinates for the area inside the graph of part (a).

Final Answer

(a) circle of radius 1 centered at \((0,1)\)
(b) \(A = \pi\)

Problem Statement

[14 points] (a) Name the curve with polar equation \(r=2\sin\theta\) and sketch the graph.
(b) Set up and evaluate an integral in polar coordinates for the area inside the graph of part (a).

Solution

Built with GeoGebra

(a) If you don't automatically know what this graph is, you can convert to rectangular coordinates and it is easy to see what the plot should look like from the resulting equation.

\( r = 2\sin \theta \)

Multiply both sides by r.

\( r^2 = 2r\sin\theta \)

On the left we use \( r^2 = x^2+y^2 \). On the right we use \(y=r\sin\theta\).

\(x^2+y^2 = 2y \)

\( x^2 + y^2 - 2y = 0 \)

Complete the square on the y-term.

\( x^2 + y^2 - 2y + 1 - 1 = 0 \)

\( x^2 + (y-1)^2 = 1 \)

From the last equation, it is easy to see that we have a circle centered at \((0,1)\) with radius 1.

(b) The area of a circle is \(A = \pi r^2\). In this case, \(A = \pi\). This helps us check our answer.

\(\displaystyle{ A = \frac{1}{2}\int_{\theta_1}^{\theta_2}{ r^2 ~ d\theta } }\)

For this problem, we will find the area of the right half of the circle and multiply by 2.

\(\displaystyle{ A = (2)\frac{1}{2}\int_{0}^{\pi/2}{ (2\sin\theta)^2 ~ d\theta } }\)

\(\displaystyle{ A = \int_0^{\pi/2}{ 4\sin^2\theta ~ d\theta } }\)

\(\displaystyle{ A = 4 \int_0^{\pi/2}{ \frac{1-\cos(2\theta)}{2} ~ d\theta } }\)

\(\displaystyle{ A = 2 \int_0^{\pi/2}{ 1-\cos(2\theta) ~ d\theta } }\)

\(\displaystyle{ A = 2\left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\pi/2} }\)

\(\displaystyle{ A = 2 \left[ \frac{\pi}{2} - \frac{\sin \pi}{2} \right] = \pi }\)

Final Answer

(a) circle of radius 1 centered at \((0,1)\)
(b) \(A = \pi\)

close solution

Log in to rate this practice problem and to see it's current rating.

You CAN Ace Calculus

Topics You Need To Understand For This Page

Limits

L'Hopitals Rule

Trig Limits

Improper Integrals

Partial Fractions

Infinite Series

Polar Coordinates

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

To bookmark this page and practice problems, log in to your account or set up a free account.

Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

effective study techniques

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

Save 20% on Under Armour Plus Free Shipping Over $49!

Shop Amazon - New Textbooks - Save up to 40%

The Humongous Book of Calculus Problems

Save 20% on Under Armour Plus Free Shipping Over $49!

Prime Student 6-month Trial

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.