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Example Questions
Example Question #91 :Parametric Form
Find the arc length of the curve (Round to three significant digits):
Finding the length of the curve requires simply applying the formula:
Where:
Since we are also givenand, we can easily compute the derivatives of each:
Applying these into the above formula results in:
Looking at this integral, it seems pretty intimidating at first, but we can do some algebraic manipulation to make it look like a simpleu-substitutionproblem. First we notice that there is aterm that is common within the two separate functions. If we factor it out then the integral becomes:
Now we can apply the rules ofu-substitution:
Therefore:
Now we must deal with the new bounds of our integral:
Our new integral becomes:
Plugging this result into a calculator results in:
Because the problem statement requires us to round this tothreesignificant figures, the final result is:
This is one of the answer choices.
Example Question #92 :Parametric Form
Find the arc length of the curve:
Finding the length of the curve requires simply applying the formula:
Where:
Since we are also givenand, we can easily compute the derivatives of each:
Applying these into the above formula results in:
We can factor out the common, and pull it outside of the square-root, and we will notice one of the most common trigonometric identities:
The term inside the square-root symbol can be simplified to.
This is one of the answer choices.
Example Question #93 :Parametric Form
Find the arc length of the curve:
Finding the length of the curve requires simply applying the formula:
Where:
Since we are also givenand, we can easily compute the derivatives of each, using theProduct Rule:
Applying these into the above formula results in:
Simplifying the above will require these two formulas:
It may also be useful to know this formula:
We can factor out the commonto make the above expression easier to look at:
We can take theoutside of the square-root by cancelling out therepresenting the "square". Then we can apply formulas&to the trigonometric expressions:
We can now simplify the terms inside the square-root to get:
If we factor out the common "2" above, we are left with the trigonometric identity, which simplifies to, since:
Therefore the integral now becomes:
This is a simple integral which can be solved usingu-substitution. But first, we can factor out the constant term, outside of the integral:
We will make our substitutions:
We also need to change the bounds of the new integral:
Our new integral becomes:
This is one of the answer choices.
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