Asked by Chelsea
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
Answered by
Steve
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.
Answered by
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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