CALCULUS help!?

Question

the length of the arc of the curve y ^2 = x ^3 cut off by the line x=4 is ?

Answer 1

The curve consists of two symmetric parts about x-axis:
y = ±√x^3
At x = 4, y = ±8 (this boundary point values can help us evaluate the final answer.)

By symmetry, you only need to calculate the length of the upper half arc, and then times 2.

Length of the arc = 2∫√(1+y'^2) dx [x: 0...4]

But y' 2= [(3/2)x^(1/2)]^2 = (9/4)x

Length of the arc = 2∫√(1+9x/4) dx [x: 0...4] = 18.1468306

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Scarlet Manuka,

You missed the last step. It should be 10^(3/2) instead of sqrt(10). Check it.

Answer 2

By symmetry we can consider the top half of the curve only and double the answer. (Note that the curve is y = +/- x^(3/2), for x >= 0.)

So the length is L = 2 int(0 to 4) sqrt(1 + y'(x)^2) dx
= 2 int(0 to 4) sqrt (1 + (3/2 x^1/2)^2) dx
= 2 int(0 to 4) sqrt (1 + 9x/4) dx
= 2 (1 + 9x/4)^(3/2) / [(3/2).(9/4)] [0 to 4]
= 2 [(1 + 9)^(3/2) - 1] / (27/8)
= 16/27 (sqrt(10) - 1).

Answer 3

infinity because the curve will only intersect in one place so no arc will be cut off by the line x=4 in this example, rather like cutting the equator in one place only because the equation only exists in real numbers when x>=0

Answer 4

I got about 18.1468 too.

(Starting where Scarlet messed up)

= 2 (1 + 9x/4)^(3/2) / [(3/2).(9/4)]
= 2 (8/27) (9x/4 + 1)^(3/2) [0 to 4]
= 2 (9.3697-.2963)
= 2 (9.073)
= 18.1468...