We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$ at x=3 with the initial condition y(1)=1 and step size h=0.5 .

Possible Answers:

(3,3)

(3,1)

(3,2.75)

(3,2)

Correct answer:

(3,2)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x)=\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$

h=0.5

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.5=1.5$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.5*ln(1/1)=1$

(x_1,y_1)=(1.5,1)

$x_{2}=x_1+h=1.5+0.5=2$

$y_{2}=y_1+h*f(x_1,y_1)=1+0.5*ln(1/1.5)=1.333$

(x_2,y_2)=(2,1.333)

We continue using Euler's method until x=3 . The results of Euler's method are in the table below.

Problem 5

Report an Error

Example Question #12 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$ at x=3 with the initial condition y(1)=1 and step size h=0.2 .

Possible Answers:

(3,2.5)

(3,2)

(3,1.25)

(3,3)

Correct answer:

(3,2.5)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x)=\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$

h=0.2

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.2=1.2$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.2*ln(1/1)=1$

(x_1,y_1)=(1.2,1)

$x_{2}=x_1+h=1.2+0.2=1.4$

$y_{2}=y_1+h*f(x_1,y_1)=1+0.2*ln(1/1.2)=1.167$

(x_2,y_2)=(1.4,1.167)

We continue using Euler's method until x=3 . The results of Euler's method are in the table below.

Problem 6

Report an Error

Example Question #13 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=e^{x}$ at x=5 with the initial condition y(0)=1 and step size h= 1 .

Possible Answers:

(5,32.19)

(5,86.79)

(5,65.49)

(5,94.77)

Correct answer:

(5,86.79)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,1)

$x_{1}=x_0+h=0+1=1$

$y_{1}=y_0+h*f(x_0,y_0)=1+1*e^{0}=2$

(x_1,y_1)=(1,2)

$x_{2}=x_1+h=1+1=2$

$y_{2}=y_1+h*f(x_1,y_1)=2+1*e^1=4.718$

(x_2,y_2)=(2,4.718)

We continue using Euler's method until x=5 . The results of Euler's method are in the table below.

Problem 9

Note: Using a step size of gives an answer that is not close to the correct answer, y(5)=148.413 . This is because the step size is too small.

Report an Error

Example Question #14 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=e^{x}$ at x=5 with the initial condition y(0)=1 and step size h= 0.5 .

Possible Answers:

(5,129.26)

(5,114.62)

(5,132.77)

(5,69.61)

Correct answer:

(5,114.62)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,1)

$x_{1}=x_0+h=0+0.5=0.5$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.5*e^{0}=1.5$

(x_1,y_1)=(0.5,1.5)

$x_{2}=x_1+h=0.5+0.5=1$

$y_{2}=y_1+h*f(x_1,y_1)=1.5+0.5*e^{0.5}=2.324$

(x_2,y_2)=(1,2.324)

We continue using Euler's method until x=5 . The results of Euler's method are in the table below.

Problem 10

Note: Using a larger step size got us closer the correct answer, y(5)=148.413 .

Report an Error

Example Question #1651 : Calculus Ii

Euler's explicit method is often used to find a numerical solution to a differential equation.Given the step size and initial condition, it utilizes the tangent line equation to find the consecutive value of the solution function.

Find an approximation to a given differential equation using Euler's explicit method:

$\frac{dy}{dx}=y-x$

Use step size $\Delta x=1$ , and range of from 0 to 3.

Initial condition is .

In the answer there should be three consecutive values of solution function for given range and step size of x.

Possible Answers:

$\newline y(1)=2\newline y(2)=7\newline y(3)=12$

$\newline y(1)=4\newline y(2)=9\newline y(3)=17$

$\newline y(1)=2\newline y(2)=5\newline y(3)=7$

$\newline y(1)=1\newline y(2)=3\newline y(3)=5$

$\newline y(1)=2\newline y(2)=5\newline y(3)=12$

Correct answer:

$\newline y(1)=2\newline y(2)=5\newline y(3)=12$

Explanation:

Recall formula of the tangent line to a two-dimensional function from Calculus 1:

$y(x)=y(x_0)+y^{'}(x_0)(x-x_0);$

For the Euler's method, this equation can be rewritten as:

$y(i+1)=y(i)+y'(i)\Delta x;$

Where is the next value of the solution function, is the derivative of the previous step.

We are given a differential equation $\frac{dy}{dx}=y-x$ . Therefore, if we know the value of and , we can find the . Consecutively, we can plug it into the Euler's method equation and find the .

Here is how this process is executed for the y(1):

1) Identify the initial condition and step size. This is the only information that Euler's method uses to calculate the approximation of solution function, other than the equation itself. In this case, we can say, that for the first step,

$\newline y(i)=y(0)=1;\newline x(i)=x(0)=0;$

Note, that for the whole process, step size remains the same, $\Delta x=1$

2) Calculate the derivative of function at point using given differential equation. In this case,

$\frac{dy}{dx}(i)=y(i)+x(i)=y(0)+0=1$

3) Note that in step 2 we found . Now we have all components to plug in to the Euler's method formula. Again, for our first step:

$y(i+1)=y(1)=y(i)+y'(i)\Delta x=1+1*1=2$

The same algorithm is used for y(2) and y(3). Note, that if we know step size and range of x, we will always know . For each consecutive step, is the value of the solution function, found in the previous step.

Report an Error

Example Question #1 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1}$

Possible Answers:

$\frac{1}{2}$

The limit does not exist.

$\infty$

Correct answer:

Explanation:

Let's examine the limit

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1}$

first.

$\lim_{x\rightarrow \infty }x= \lim_{x\rightarrow \infty } \left ( 2^{x} - 1 \right )= \infty$

$\frac{\mathrm{d} }{\mathrm{d} x}x = 1$

and

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{ x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x}e^{ x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$=\ln 2 \cdot e^{ x \ln 2} - 0 = \ln 2 \cdot 2^{x}$ ,

so by L'Hospital's Rule,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}}$

Since $\lim_{x\rightarrow \infty } \ln 2 \cdot 2^{x} = \infty$ ,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}} = 0$

Now, for each x > 0 , $0 \leq \sin^{2} x \leq 1$ ; therefore,

$0 \leq \frac{x \sin^{2} x}{2^{x} - 1} \leq \frac{x }{2^{x} - 1}$

By the Squeeze Theorem,

$0 \leq \lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} \leq \lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = 0$

and

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} = 0$

Report an Error

Example Question #2 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$

Possible Answers:

The limit does not exist.

$\ln \frac{5}{3}$

$\frac{\ln 5}{\ln 3}$

$\infty$

$\frac{5}{3}$

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\left ( 5^{x} - 1 \right ) = \lim_{x\rightarrow \infty }\left ( 3^{x} - 1 \right ) = \infty$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 5^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 5} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 5 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 5\cdot e^{x \ln 5 } - 0 = \ln 5\cdot 5^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 3^{x} - 1 \right ) = \ln 3 \cdot 3 ^{x}$

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1} = \lim_{x\rightarrow \infty } \frac{\ln 5 \cdot 5^{x} }{\ln 3 \cdot 3^{x} }$

$=\frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty } \frac{ 5^{x} }{ 3^{x} }= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x}$

But $\lim_{x\rightarrow \infty }a^{x} = \infty$ for any a > 1 , so

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x} = \infty$

Report an Error

Example Question #3 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$

Possible Answers:

$\log_{2}7$

$\infty$

$\ln \frac{7}{2}$

$\frac{1}{2}\ln 7$

The limit does not exist.

Correct answer:

$\log_{2}7$

Explanation:

$\lim_{x\rightarrow 0 }\left ( 7^{x} - 1 \right ) = 7^{0} - 1 = 1- 1 = 0$

and

$\lim_{x\rightarrow 0 }\left ( 2^{x} - 1 \right ) = 2^{0} - 1 = 1- 1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 7^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 7} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 7 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 7\cdot e^{x \ln 7 } - 0 = \ln 7\cdot 7^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \ln 2 \cdot 2 ^{x}$

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1} = \lim_{x\rightarrow 0 } \frac{ \ln 7\cdot 7^{x }}{ \ln 2\cdot 2^{x }}$

$= \frac{ \ln 7 \cdot 7^{0 }}{ \ln 2\cdot 2^{0 }} = \frac{ \ln 7\cdot 1}{ \ln 2\cdot 1} = \frac{ \ln 7}{ \ln 2 }= \log_{2}7$

Report an Error

Example Question #4 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$

Possible Answers:

$\ln 2$

$\frac{1}{\ln2}$

Correct answer:

Explanation:

$\lim_{x\rightarrow 0^{+} } x \sin x= 0 \cdot \sin 0 = 0$

and

$\lim_{x\rightarrow 0^{+} } \left ( 2 ^{x} - 1 \right )= 2^{0} -1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}x \sin x = \frac{\mathrm{d} }{\mathrm{d} x} x \cdot \sin x + \frac{\mathrm{d} }{\mathrm{d} x}\sin x \cdot x$

$= 1 \cdot \sin x + \cos x \cdot x = \sin x + x \cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$= \ln 2\cdot e^{x \ln 2} - 0 = \ln 2\cdot 2^{x }$

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1} = \lim_{x\rightarrow 0^{+} } \frac{\sin x + x \cos x}{ \ln 2\cdot 2^{x }}$

$= \frac{\sin 0 + 0 \cos 0}{ \ln 2\cdot 2^{0 }} = \frac{ 0 + 0 }{ \ln 2\cdot 1} = 0$

Report an Error

Example Question #1 : L'hospital's Rule

Solve:

${}\lim_{x \to \infty} x^\frac{5}{x}}$

Possible Answers:

e^5

ln(5)

Correct answer:

Explanation:

Substitution is invalid. In order to solve ${}{ \lim_{x \to \infty} x^\frac{5}{x}}$ , rewrite this as an equation.

${} y= { \lim_{x \to \infty} x^\frac{5}{x}}$

Take the natural log of both sides to bring down the exponent.

${} \lim_{x \to \infty} ln(y)= \lim_{x \to \infty} \frac{5}{x} \cdot ln(x)$

${} \lim_{x \to \infty} ln(y)= 5\lim_{x \to \infty} \frac{ln(x)}{x}$

Since ${} \lim_{x \to \infty} \frac{ln(x)}{x}$ is in indeterminate form, ${}\frac{\infty}{\infty}$ , use the L'Hopital Rule.

L'Hopital Rule is as follows:

$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty}\frac{f'(x)}{g'(x)}$

${} \lim_{x \to \infty} \frac{ln(x)}{x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = 0$

This indicates that the right hand side of the equation is zero.

${} \lim_{x \to \infty} ln(y)= 0$

Use the term to eliminate the natural log.

${} \lim_{x \to \infty} e^l^n^(^y^)= e^0$

$\lim_{x \to \infty} y=e^0=1$

Report an Error

View Calculus 2 Tutors

Mohammad
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Mathematics. Rutgers University-New Brunswick, Master of Science, Stat...

View Calculus 2 Tutors

Jason
Certified Tutor

Brigham Young University, Bachelor, Neuroscience.

View Calculus 2 Tutors

Reza
Certified Tutor

University of Sherbrooke, Doctor of Philosophy, Mathematics. University of Manitoba, Master of Science, Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in San Diego, Biology Tutors in Chicago, Math Tutors in Seattle, LSAT Tutors in Washington DC, Algebra Tutors in Houston, GRE Tutors in Phoenix, Computer Science Tutors in Phoenix, SAT Tutors in Phoenix, Reading Tutors in Boston, Calculus Tutors in Dallas Fort Worth

Popular Courses & Classes

Spanish Courses & Classes in Los Angeles, MCAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Washington DC, ACT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Houston, Spanish Courses & Classes in Chicago, SAT Courses & Classes in New York City, SAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Boston, ISEE Courses & Classes in San Francisco-Bay Area

Popular Test Prep

ACT Test Prep in Denver, GMAT Test Prep in Houston, LSAT Test Prep in Miami, SAT Test Prep in Miami, SSAT Test Prep in Phoenix, ACT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Boston, SAT Test Prep in Seattle, ISEE Test Prep in Boston, LSAT Test Prep in Chicago

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Euler's Method

Example Question #12 : Euler's Method

Example Question #13 : Euler's Method

Example Question #14 : Euler's Method

Example Question #1651 : Calculus Ii

Example Question #1 : Euler's Method And L'hopital's Rule

Example Question #2 : Euler's Method And L'hopital's Rule

Example Question #3 : Euler's Method And L'hopital's Rule

Example Question #4 : Euler's Method And L'hopital's Rule

Example Question #1 : L'hospital's Rule

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: