Wednesday, December 8, 2010

`x = 1/3(y^2 + 2)^(3/2) , 0

Arc length (L) of the function x=h(y) on the interval [c,d] is given by the formula,


 `L=int_c^dsqrt(1+(dx/dy)^2)dy` , if x=h(y) and c `<=`  y `<=`  d,


`x=1/3(y^2+2)^(3/2)`


`dx/dy=1/3(3/2)(y^2+2)^(3/2-1)(2y)`


`dx/dy=y(y^2+2)^(1/2)`


Plug in the above derivative in the arc length formula,


`L=int_0^4sqrt(1+(y(y^2+2)^(1/2))^2)dy`


`L=int_0^4sqrt(1+y^2(y^2+2))dy`


`L=int_0^4sqrt(1+y^4+2y^2)dy`


`L=int_0^4sqrt((y^2+1)^2)dy`


`L=int_0^4(y^2+1)dy`


`L=[y^3/3+y]_0^4`


`L=[4^3/3+4]-[0^3/3+0]`


`L=[64/3+4]`


`L=[(64+12)/3]`


`L=76/3`


Arc length of the function over the given interval is `76/3`

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