How do you find the length of the curve $y=sqrtx-1/3xsqrtx$ from x=0 to x=1?

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Answer 1 · 2017-05-24T23:28:50

$4/3$

The length of a curve can be calculated with this integral
$int_a^b=sqrt(1+(dy/dx)^2)dx$

Well, we better get deriving! First, take the derivative of $y$ .

$dy/dx=d/dx[x^(1/2)]-d/dx[1/3x^(3/2)]$

$dy/dx=1/2(x^(-1/2)-x^(1/2))=(1-x)/(2sqrt(x))$

$(dy/dx)^2=(1-2x+x^2)/(4x)$

So

$sqrt(1+(dy/dx)^2)=sqrt(1+(1-2x+x^2)/(4x))=sqrt((1+2x+x^2)/(4x))=(x+1)/(2sqrt(x))$

Finally

$int_0^1(x+1)/(2sqrt(x))dx=int_0^1x/(2sqrtx)+1/(2sqrtx)dx$

Evaluate indefinite

$intx/(2sqrtx)+1/(2sqrtx)dx$
$=1/2intx/sqrtx+1/sqrtxdx$

Solve and ignore constants for now

$intx/sqrtx=2/3x^(3/2)$

$int1/sqrtx=2sqrt(x)$

Piece them together and ignore constants for now

$1/2intx/sqrtx+1/sqrtxdx$
$=1/2(2sqrt(x)+2/3x^(3/2))$
$=(sqrt(x)(x+3))/3$

Finally, solve the definite integral:
$int_0^1(x+1)/(2sqrt(x))dx=(sqrt(x)(x+3))/3|_0^1$
$=4/3$

$:.$ The length of the curve is $4/3$

How do you find the length of the curve y=sqrtx-1/3xsqrtxy=sqrtx-1/3xsqrtx from x=0 to x=1?