How do you find the length of the curve $y^2 = 16(x+1)^3$ where x is between [0,3] and $y>0$ ?

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Answer 1 · 2015-02-24T18:59:51

The answer is: $L=1/54(sqrt145-37sqrt37)$ .

The curve can be written:

$y=+-4sqrt((x+1)^3)$ , but we need only the branch with $y>0$ , so:

$y=4sqrt((x+1)^3)$ .

To find the lenght of a curve written as a function and in cartesian coordinates we have to remember this formula:

$L=int_a^bsqrt(1+[f'(x)]^2)dx$ .

So, we have first of all calculate the derivative of:

$y=4(x+1)^(3/2)rArry'=4*3/2(x+1)^(1/2)=6sqrt(x+1)$

$L=int_0^3sqrt(1+36(x+1))dx=int_0^3sqrt(36x+37)dx=$

$=1/36int_0^3 36sqrt(36x+37)dx=1/36[(36x+37)^(1/2+1)/(1/2+1)]_0^3=$

$=1/36*2/3[sqrt((36x+37)^3)]_0^3=1/54(sqrt145-sqrt(37^3))=1/54(sqrt145-37sqrt37)$ .

How do you find the length of the curve y^2 = 16(x+1)^3y^2 = 16(x+1)^3 where x is between [0,3] and y>0y>0?