Do the following for the curve y4x2 on 02 a set up an integr

Do the following for the curve y=4x^2 on [0,2]
a. set up an integral for the length of the curve
b. use your grapher\'s or computers integral evaluator to find the curves length numerically.

please explain

Solution

arc length = integral of (1+y\'^2)dx....limit from 0 to 2
y = 4x^2
y\' = 8x
arc length = integral of (1+64x^2)dx
arc length = 1/8 integral of (1/64 + x^2)dx
integral of (a^2 + x^2)dx = x/2 (a^2+x^2) + a^2/2 ln|x + (a^2+x^2)
applying formula
arc length = 1/8 [x/2 (1/64 + x^2) + 1/128 ln|x + (1/64+x^2)
apply limits
arc length = 1/8[(1/64 + 4) + 1/128 ln|2+(1/64 + 4)] - 1/128 ln|(1/64)
arc length = 1/8[257/256 + 1/128 ln|2+257/256| - 1/128 ln|1/8|

Do the following for the curve y=4x^2 on [0,2] a. set up an integral for the length of the curve b. use your grapher\'s or computers integral evaluator to find

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