Question
Find the length of the curve. You may use your calculator.
f(x)=x^(1/3)+x^(2/3) [0,2]
I understand that the function needs to be written as x in terms of y because there's a vertical tangent at x=0, but I don't understand how to go about the problem other than that.
f(x)=x^(1/3)+x^(2/3) [0,2]
I understand that the function needs to be written as x in terms of y because there's a vertical tangent at x=0, but I don't understand how to go about the problem other than that.
Answers
The vertical slope does not matter, once you go through the hoops. The answer is shown here:
http://www.wolframalpha.com/input/?i=plot+x%5E(1%2F3)%2Bx%5E(2%2F3)+for+x%3D0..2
The problem I have is evaluating that integral! I tried something like
u = 1+2∛x
du = 2/3 x^(-2/3)
x = (u-1)^3/8
and the integral then becomes
∫1/2 √(9(u-1)^4/8 + u^2) du
which is no easier to handle. Looks like this is really a calculator problem.
http://www.wolframalpha.com/input/?i=plot+x%5E(1%2F3)%2Bx%5E(2%2F3)+for+x%3D0..2
The problem I have is evaluating that integral! I tried something like
u = 1+2∛x
du = 2/3 x^(-2/3)
x = (u-1)^3/8
and the integral then becomes
∫1/2 √(9(u-1)^4/8 + u^2) du
which is no easier to handle. Looks like this is really a calculator problem.
bobpursley
analyze how Wolfram does it.
https://www.wolframalpha.com/input/?i=arc+length+of+%3Dx%5E(1%2F3)%2Bx%5E(2%2F3)++from+x%3D0+to+x%3D2
https://www.wolframalpha.com/input/?i=arc+length+of+%3Dx%5E(1%2F3)%2Bx%5E(2%2F3)++from+x%3D0+to+x%3D2
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