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(a) Sketch the curve $ y^3 = x^2 $.(b) Use Formulas 3 and 4 to set up two integrals for the arclength from $ (0, 0) $ to $ (1, 1) $.Observe that one of these is an improper integral and evaluate both of them.(c) Find the length of the arc of this curve from $ (-1, 1) $ to $ (8, 4) $.

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a) Graph of the curve $y^{3}=x^{2}$ is shown below $:$b) $\frac{13 \sqrt{13}-8}{27}$c) $L=\frac{1}{27}[13 \sqrt{13}+80 \sqrt{10}-16]$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 1

Arc Length

Applications of Integration

The B.

October 4, 2020

Where does she get the 13 in the bounds?

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Lectures

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(a) Sketch the curve $y^{3…

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$\begin{array}{l}{\text { …

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(a) Sketch the curve y3

01:49

a. Write and simplify the …

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05:05

Find the arc length of the…

00:43

Consider the curve $y=x^{2…

it was clear. So when you read here, so for part A, we're gonna Graff Why Cube is equal to the square and it looks something like this for part B. We're gonna use formula three and four. So we have why cube is equal to X square. This gives this why is equal to X to the 2/3 power. Differentiating and respect to X gives us d Y over. The X is equal to 2/3 next to the 1/2 next to the negative 1/3 excuse. So we're gonna find the length of the curve between zero comma zero and one comma one. So we're gonna integrate from 0 to 1 square root of one plus 2/3 X to the negative 1/3 square DX, then continue here. This gives us from 01 square root, but nine x to the 2/3 power plus four over three x to the 1/3 power E X. This is an improper integral because that integrity is not defined. Uh, X is equal zero so improper we could integrate this by using limits so limit as T approaches zero positive on degree from t one square root of nine x to the 2/3 plus four over three x to the 1/3 d X because substitute you is equal to nine x to the 2/3 plus four. So an x six x to the negative 1/3 the ex is equal to do you. This changes our limits of integration. So it goes from T 21 and from to 92 3rd plus four 13 we have square work of you over 18. You have to put our limit most t approaches zero positive, do you? And this gives us 13 square root of 13 minus eight over 27 using formula for we're going to take square both sides for why cute is equal to X square. So we get why 2 3/2 power equal to plus or minus X. We're gonna differentiate in respect to what? So we get three house alive to the 1/2 power is equal to the X over de Juan. You have to find the length the curve between zero comma zero and one comma one. So he set up this we took roll in respect to why 1/2 square, do you? Why, this gives us from 0 to 1 square of four plus nine. Why over to D Why we substitute you is equal to four plus nine. Why that makes 90. Why equal to do you in our limits of integration? Change from 0124 13 and then it goes Square it of you over 18 do you? And this also gives us the same answer which is 13 and square roots. Well 13 minus eight over 27. Her part. See? So we're going to use our flank formula. But before we do that well, right X is equal to T cube. Why is eat equal to t square and T is between negative one and two included. So if we put this into our blank formula yeah, l is equal to from negative 1 to 2 square root of three t square square plus two tea square d t. This gives us negative. We could have 10 t square root. I'm 90 square plus four e t. And then we integrate from 0 to 2 tea square root of nine t squared plus floor DT. We stuffed it to you for 90 square plus four. That means 18 T D t is equal to do you. This gives us l is equal to negative won over 18 into grow from 13 four square root of U. D T plus one over 18 for 40 square root of you, do you? And her final answer is one over 27 then 13 square root of 13 less eating square root of 10 minus 16.

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