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Calculus Homework Help

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For you guys taking calculus classes we have a lot to offer (see the topics). We know how challenging the transition from high school to college can be. There's no need to struggle with incomplete study material. That's why we offer our math assistance.

One of the things students always complain is that they don't have enough solved problems to study with. With us, that's not a problem anymore! Submit by e-mail your problems to us for a free quote.

Having a well explained list of solved problems can improve your learning 100 percent. Regrettably, current textbooks come mostly with the solutions to the sample problems, but they don't show detailed solutions.

Make sure to visit our store. There you'll find study guides for sale with hundreds of solved problems on calculus. We have study guides, the most common formulas, etc. Check our new handout, with 200+ solved problems, step-by-step solutions. You can check here an example of our work. Also, follow the link to find a free calculus handout

Calculus Help

This is a list of subjects you may be interested on

  • Functions
    • Polynomials
    • Rational Functions
    • Logarithms
    • Exponentials
    • Trigometric

  • Limits
    • Direct Substitution Property
    • Undetermined Forms
    • Trigometric Limits
    • Factorization
    • Side Limits
    • L'Hopital Rule

  • Continuity
    • Connection with limits
    • Algebraic Properties
    • Intermediate Value Theorem

  • Derivatives
    • Tangent Line
    • Product Rule
    • Quotient Rule
    • Chain Rule
    • Related Rates
    • Mean Value Theorem
    • Maximum and Minimum values

  • Integrals
    • Antiderivatives
    • Fundamental Theorem of Calculus
    • Substitutions and change of variables
    • Partial Fractions
    • Improper Integrals

  • Series
    • Positive Series
    • Convergence criteria
    • Alternating Series

We know how it feels to deal with hard calculus problems. With our help, you'll get that extra edge you need! No need to sweat with poorly explained problems from the books. We can provide solutions that suit your own needs. See some examples:

Problem 1. Find


$\displaystyle \lim_{n\rightarrow \infty} \frac{2n^2+1}{3n^2-5n+2}$

The problem presents the typical case of a quotient of two quantities that go to infinity. In fact, $ 2n^2+1$ and $ 3n^2-5n+2$ both approach to infinity as $ n\rightarrow\infty$ . Which means that the quotient

$\displaystyle \frac{2n^2+1}{3n^2-5n+2}$

is an undetermined form. Somehow we have to cancel the "bad" part, if possible. Notice that since both $ 2n^2+1$ and $ 3n^2-5n+2$ are big, there's a chance for cancelation, so the quotient could be finite.

The trick in this case is to divide by the highest power on the expression (from both numerator and denominator). In this case we will divide both numerator and denominator by $ n^2$ . We get

$\displaystyle \frac{2n^2+1}{3n^2-5n+2}= \frac{2+\frac{1}{n^2}}{ 3-\frac{5}{n}+\frac{2}{n^2} }\rightarrow \frac{2}{3}$

as $ n\rightarrow\infty$ because $ \frac{1}{n^2}\rightarrow 0$ , $ \frac{5}{n}\rightarrow 0$ and $ \frac{2}{n^2} \rightarrow 0$ as $ n\rightarrow\infty$ .

This trick is applied in general for this type of expressions. $ \Box$

Problem 2. Find $ \frac{dy}{dx}$ and $ \frac{d^2 y}{dx^2}$ if

$\displaystyle x^{2/3}+y^{2/3}=a^{2/3}, \qquad a>0 $

Solution. Let's differentiate both sides of the equality. Let's recall that we are assuming that $ y=y(x)$ . That means that the equation gives us a way to solve $ y$ as a function of $ x$ , but we still don't know how to express that function. That assumption requires further justification (meaning, it's not always granted), but we'll take it for granted. So, since $ y=y(x)$ we get using the chain rule:

$\displaystyle \frac{2}{3} x^{-1/2}+\frac{2}{3} y^{-1/3} y'=0 $

where $ y'\equiv \frac{dy}{dx}$ . The 0 on the right hand side comes from differentiating the constant $ a^{2/3}$ . We multiply the equation by $ 3/2$ and we get

$\displaystyle x^{-1/2}+ y^{-1/3} y'=0 $

which means that

$\displaystyle \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} $

Let's get $ y''$ . From the last equation we obtain easily that

$\displaystyle x\left(\frac{dy}{dx}\right)^3=-y $

and now we differentiate this again with respect to $ x$ to get:

$\displaystyle \left(\frac{dy}{dx}\right)^3+3x\left(\frac{dy}{dx}\right)^2 y''=-y'

But $ \left(\frac{dy}{dx}\right)^3=-y/x$ and $ 3x\left(\frac{dy}{dx}\right)^2=3x(y/x)^{2/3}$ . We know put all the elements together:

$\displaystyle y''=\frac{y/x+(y/x)^{1/3}}{3x(y/x)^{2/3}} $

and now we simplify the right hand side of the last equation

$\displaystyle y''=\frac{(y/x)^{2/3}+1}{3x(y/x)^{1/3}}=\frac{y^{2/3}+x^{2/3}}{3x^{4/3}y^{1/3}} $

But by definition: $ x^{2/3}+y^{2/3}=a^{2/3}$ , so we replace this into the previous equation:

$\displaystyle y''=\frac{a^{2/3}}{3x^{4/3}y^{1/3}} $

which gives the final answer.$ \Box$

Check more examples here.

Calculus handout: Here we present you with a free calculus handout that includes calculus and pre-calculus topics. You can download it here

calculus handout

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