We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Other Writing Equations

Find the equation tangent to $y=x\sin x$ at $x=\pi$

Possible Answers:

Equation of tangent line is undefined

$y=-\pi x+\pi^2$

$y=\pi x$

$y=\pi x+\pi^2$

$y=-\pi x+\pi$

Correct answer:

$y=-\pi x+\pi^2$

Explanation:

First, find the y coordinate of the function at $x=\pi$

$y=\pi \sin \pi = 0$

Thus, we have the point $(\pi,0)$ .

To find the slope of the tangent line, take the derivative and then plug in $x=\pi$

${y}'=\sin x +x\cos x$

${y}'(\pi)=\sin \pi +\pi\cos \pi=-\pi$

Thus, our slope is $-\pi$ .

Knowing that the general formula of a straight line is $y=mx+b$ where is the slope and that slope is $-\pi$ . Thus, $y=-\pi x+b$ . To find , plug in the coordinate $(\pi,0)$ and then solve for .

$0=-\pi (\pi)+b$

$b=\pi^2$

So now the equation of the tangent line becomes $y=-\pi x +\pi^2$

Report an Error

Example Question #3 : Other Writing Equations

Take the indefinite integral of

$\int x(x^6+7)^2dx$

Possible Answers:

$(x^2+7)(x^6+7)^2$

$x^2(x^6+7)^3+C$

Indefinite integral does not exist

$\frac{1}{14}x^{14}+\frac{7}{4}x^8+\frac{49}{2}x^2+C$

$\frac{1}{13}x^{13}+2x^7+49x+C$

Correct answer:

$\frac{1}{14}x^{14}+\frac{7}{4}x^8+\frac{49}{2}x^2+C$

Explanation:

First, foil the integral so it is easier to manage.

$\int (x^{13}+14x^7+49x) dx$

then perform the indefinite integral the normal way you would do

$\frac{1}{14}x^{14}+\frac{7}{4}x^8+\frac{49}{2}x^2+C$

Report an Error

Example Question #4 : Other Writing Equations

Find the limit of $lim_{x \to 0}\frac{sin(2x)}{x}$ .

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

In this limit, we have $\frac{0}{0}$ as $x\rightarrow 0$ .

Because of this, we can use L'Hospital's rule. Differentiating the top and the bottom of the function, we get

$\frac{2cos(2x)}{1}$ .

If we evaluate at x=0 , we don't have $\frac{0}{0}$ anymore and we get

$\frac{2cos(0)}{1}=\frac{2*1}{1}=2$ .

So the answer is 2.

Report an Error

Example Question #5 : Other Writing Equations

Find $lim_{x \to 0}\frac{3^x-2^x}{x}.$

Possible Answers:

ln(6)

$ln\frac{2}{3}$

$ln\frac{3}{2}$

-ln(6)

Correct answer:

$ln\frac{3}{2}$

Explanation:

Because we have a $\frac{0}{0}$ in our limit, we can use L'Hospital's Rule. Applying this, we get

$lim _{x \to 0} \frac{3^x-2^x}{x}=lim_{x \to 0}\frac{ln(3)*3^x-ln(2)*2^x}{1}$ .

Evaluating this as ${x \to 0}$ we get

$\frac{ln(3)*3^0-ln(2)*2^0}{1}=ln(3)-ln(2)=ln(3/2).$

Report an Error

Example Question #191 : Equations

What is the equation of the tangent line at x = 15 for f(x) = x⁴ + 5x² + 44x – 3?

Possible Answers:

y = 55x + 13382

y = 13694x – 153003

y = 153103x – 13694

y = 2848x + 34

None of the other answers

Correct answer:

y = 13694x – 153003

Explanation:

First we must solve for the general derivative of f(x) = f(x) = x⁴ + 5x² + 44x – 3.

f'(x) = 4x³ + 10x + 44

Now, the slope of the tangent line for f(15) is equal to f'(15):

f'(15) = 4(15)³ + 10 *15 + 44 = 13694.

To find the tangent line, we need at least one point on the line. To find this, we can use f(15) to get the y value of the point of tangency, which will suffice for our use:

f(15) = 15⁴ + 5(15)² + 44 * 15 – 3 = 50625 + 1125 + 660 – 3 = 52407

Now, using the point-slope form of the line, we get:

y - 52407 = 13694 * (x – 15)

Simplify:

y – 52407 = 13694x – 205410

y = 13694x – 153003

Report an Error

Example Question #196 : Writing Equations

For the following function, use implicit differentiation to find the equation of the tangent line at x=0 :

$y^2e^{2x}=3y+x^2$

Possible Answers:

$y=-\frac{7}{2}x-\frac{1}{2}$

y=2x+1

$y=-\frac{7}{2}x+\frac{1}{2}$

y=-6x-3

y=-6x+3

Correct answer:

y=-6x+3

Explanation:

Our first step is to apply implicit differentiation to the function to find y’. This derivative tells us the slope of our function and will therefore give us the slope of our tangent line at the given point:

$2yy'e^{2x}+2y^2e^{2x}=3y'+2x$

Rearranging the equation, factoring out y’, and dividing, we find y’ is:

$y'=\frac{2x-2y^2e^{2x}}{2ye^{2x}-3}$

Before we can plug in the values of our given point to find the slope of the tangent line, we must first find its y value by plugging x=0 into the original function, which gives us:

y^2e^0=3y+0

y^2-3y=0

y(y-3)=0

y=3

So now that we know our given point is (0,3), we plug these values into our equation for y’, which gives us the slope of the tangent line at that point.

$m=y'(0,3)=\frac{2(0)-2(3)^2e^0}{2(3)e^0-3}=-\frac{18}{3}=-6$

Now that we have our slope (m), an x coordinate (x1), and a y coordinate (y1), we can simply plug these values into the point-slope formula for a line:

y-y_1=m(x-x_1)

y-3=-6(x-0)

y=-6x+3

Report an Error

Example Question #4 : Other Writing Equations

Find two positive numbers whose sum is 450 and whose product is a maximum.

Possible Answers:

x=325,y=125

x=100,y=350

x=210,y=240

x=400,y=50

x=225,y=225

Correct answer:

x=225,y=225

Explanation:

This problem deals with the concept of optimization. We start by simply writing out the equations described in words by the problem:

x+y=450

xy=C

The problem asks that we maximize the product of the two unknown numbers, C, but we must first solve our first equation for y. Then we substitute it into our second equation to ensure that it is in terms of x alone. We then take the derivative of the equation with respect to x, and set it equal to 0 to solve for the x value that corresponds to the critical point of the function:

$x+y=450\rightarrow y=450-x$

C(x)=xy=x(450-x)=450x-x^2

C'(x)=450-2x

$450-2x=0\rightarrow x=225$

Because we only know that this x value corresponds to a critical point, and not necessarily whether that point is a minimum or a maximum, we must check the value of the second derivative at this point to see whether the function is concave up or concave down:

C''(x)=-2

Because the value of the second derivative at this point is negative, we know that the function is concave down, so our one and only critical point in this case occurs at a global maximum, verifying that the product of the two numbers is maximized when x=225. Now that we know the value of x, we substitute it into the equation for y to find our second unknown number:

y=450-x=450-225=225

Report an Error

Example Question #4 : How To Write Equations

What is the equation of a line with slope of and -intercept of ?

Possible Answers:

y=4x+4

y=4x+2

y=2x+4

y=4x

y=2x+2

Correct answer:

y=4x+2

Explanation:

To write this equation we need to read the question carefully. Since the slope is we know the slope is represented by .

Also since the y-intercept is , when is then y=2 .

Therefore plugging our values into the slope intercept form:

y=mx+b we get,

y=4x+2 .

Report an Error

Example Question #11 : How To Write Equations

My friend has 100 dollars at time t = 0 days. She earns dollars every day, and spends dollars every day (so at day , she has 103-2=101 dollars). Write an equation for how much money she has at time .

Possible Answers:

y = 100 + 1

y = 100 + 3t

y = 100 + t

y = 100 - t

y = t

Correct answer:

y = 100 + t

Explanation:

My friend starts off with 100 dollars, and every day, she makes a net dollar. This is because she earns three dollars and spends one dollar.

Net=Earn-Spent

1=3-2

So, we need an equation that reflects that she has 100 +1 *(number of days), which is given by

y = 100 + t .

Report an Error

Example Question #11 : How To Write Equations

Identify the inner and the outer functions of the following equation (let f(x) be the outer equation and g(x) be the inner equation:

$y=\sqrt[3]{1-4x}$

Possible Answers:

f(x)=x

$g(x)=\sqrt[3]{1-4x}$

$f(x)=\sqrt[3]{x}$

g(x)=1-4x

$g(x)=\sqrt[3]{x}$

f(x)=1-4x

$f(x)=\sqrt[3]{1-4x}$

g(x)=x

Correct answer:

$f(x)=\sqrt[3]{x}$

g(x)=1-4x

Explanation:

The first section of the equation to be resolved (in this case 1-4x) is the innermost function and the second section to solve is the outer function. So:

$f(x)=\sqrt[3]{x}$

g(x)=1-4x

becuase g(x) must be performed first before plugging it into f(x).

Report an Error

View Calculus Tutors

Aayush
Certified Tutor

University of Illinois at Urbana-Champaign, Bachelor of Science, Pre-engineering.

View Calculus Tutors

Chernyeh
Certified Tutor

Purdue University-Main Campus, Bachelor of Science, Business Administration and Management.

View Calculus Tutors

Nana Kena
Certified Tutor

Kwame Nkrumah University of Science and Technology, Bachelor of Science, Mathematics. Kwame Nkrumah University of Science and...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Chemistry Tutors in Washington DC, Computer Science Tutors in Phoenix, SAT Tutors in Boston, LSAT Tutors in San Diego, Calculus Tutors in San Francisco-Bay Area, English Tutors in Washington DC, English Tutors in Los Angeles, Spanish Tutors in Houston, ACT Tutors in Miami, Statistics Tutors in New York City

Popular Courses & Classes

ISEE Courses & Classes in Philadelphia, ACT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Washington DC, SAT Courses & Classes in Denver, Spanish Courses & Classes in Boston, SSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Boston, Spanish Courses & Classes in Atlanta, ISEE Courses & Classes in Denver

Popular Test Prep

SSAT Test Prep in Houston, LSAT Test Prep in San Diego, SSAT Test Prep in Seattle, SAT Test Prep in Phoenix, ISEE Test Prep in New York City, SSAT Test Prep in Los Angeles, SSAT Test Prep in Washington DC, SAT Test Prep in Miami, LSAT Test Prep in Phoenix, ACT Test Prep in Washington DC

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 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2 : Other Writing Equations

Example Question #3 : Other Writing Equations

Example Question #4 : Other Writing Equations

Example Question #5 : Other Writing Equations

Example Question #191 : Equations

Example Question #196 : Writing Equations

Example Question #4 : Other Writing Equations

Example Question #4 : How To Write Equations

Example Question #11 : How To Write Equations

Example Question #11 : How To Write Equations

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: