Problem

(a) Sketch the curve $ y^3 = x^2 $. (b) Use Form…

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Problem 33 Hard Difficulty

Sketch the curve with equation $ x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1 $ and use symmetry to find its length.

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Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 1

Arc Length

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Video Transcript

Hey, it's clear. So when you read here So the other equation acts to 2/3 power plus one to the 2/3 power is equal to one. So here we're gonna graph are equation no good of one on good of one. You know, look, something like this. So, using the symmetry of the graph, we see that the length of the complete curve is equal to the length of the curve in the first quadrant right here and then we just multiply by four. So using our or Klink formula, we have to first find the derivative. So we got 2/3 X. It's a negative one. Third was 2/3. Why'd the negative 1/3 times D y over? The X is equal to zero. We got d Y over DX is equal to negative X to the negative 1/3 over, allied to the negative 1/3 and it's also equal to negative. Why to the 1/3 over X to the 1/3. So when we plug this into our Klink formula, yeah Oh, from 0 to 1 square root of one plus x, It's a negative 1/3 power over. Why'd the negative 1/3 power Negative square, T X. You continue here? This is equal to from 0 to 1 square root off X is that 2/3 power Plus why the 2/3 power over X to the two periods Power defense. We're gonna plug in X to the 2/3 power plus y two. The 2/3 power equals one. So this equals from 0 to 1 square root of one over X to the 2/3 the axe. This is equal to from 0 to 1 next to the negative 1/3 X. This gives us three hounds and then we have to multiply this by four. It gives us six. So the complete length of the curve it's six.

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Problem

(a) Sketch the curve $ y^3 = x^2 $. (b) Use Form…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 8

Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Samuel Hannah

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals