Calculus Help Samples
SAMPLE CALCULUS PROBLEMS:
Problem 1. Find
The problem presents the typical case of a quotient of two quantities that go to infinity. In fact, and both approach to infinity as . Which means that the quotient
is an undetermined form. Somehow we have to cancel the "bad" part, if possible. Notice that since both and 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 . We get
as because , and as .
This trick is applied in general for this type of expressions.
Problem 2. Find and if
Solution. Let's differentiate both sides of the equality. Let's recall that we are assuming that . That means that the equation gives us a way to solve as a function of , 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 we get using the chain rule:
where . The 0 on the right hand side comes from differentiating the constant . We multiply the equation by and we get
which means that
Let's get . From the last equation we obtain easily that
and now we differentiate this again with respect to to get:
But and . We know put all the elements together:
and now we simplify the right hand side of the last equation
But by definition: , so we replace this into the previous equation:
which gives the final answer.
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We provide a quality problem solving service on the following calculus topics:
- Rational Functions
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- Trigometric Limits
- Side Limits
- L'Hopital Rule
- Connection with limits
- Algebraic Properties
- Intermediate Value Theorem
- Tangent Line
- Product Rule
- Quotient Rule
- Chain Rule
- Related Rates
- Mean Value Theorem
- Maximum and Minimum values
- Fundamental Theorem of Calculus
- Substitutions and change of variables
- Partial Fractions
- Improper Integrals
- Positive Series
- Convergence criteria
- Alternating Series
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