How do you find the length of the curve for $y = x^{\frac{3}{2}}$ $y=x^(3/2)$ for (0,6)?

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How do you find the length of the curve for $y = x^{\frac{3}{2}}$ $y=x^(3/2)$ for (0,6)?

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Answer 1 · 2018-03-10T23:21:21

See below.

Explanation:

We need $\int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $int_a^b sqrt(1+(dy/dx)^2) dx$ .

$\int_{0}^{6} \sqrt{1 + \frac{9 x}{4}} d x = \frac{4}{9} \int_{0}^{6} \sqrt{1 + \frac{9 x}{4}} \frac{9}{4} d x$ $int_0^6 sqrt(1+(9x)/4) dx = 4/9 int_0^6 sqrt(1+(9x)/4) 9/4dx$

$= \frac{4}{9} {[\frac{2}{3} {(1 + \frac{9 x}{4})}^{\frac{3}{2}}]}_{0}^{\frac{3}{2}}$ $= 4/9 [2/3(1+(9x)/4)^(3/2)]_0^6$

$= \frac{8}{27} {⎡ ⎣ \frac{{(4 + 9 x)}^{\frac{3}{2}}}{8} ⎤ ⎦}_{0}^{6}$ $= 8/27[(4+9x)^(3/2)/8]_0^6$

$= \frac{1}{27} (58 \sqrt{58} - 8)$ $= 1/27(58sqrt58 - 8)$

Answer 2 · 2018-03-10T23:57:17

16.064 (to 3 places of decimals)

Explanation:

Denoting some short segment of the curve by $d s$ $ds$ , and abusing terminology a little, note that it is possible to approximate $d s$ $ds$ using the Pythagorean relationship

$d s^{2} = {d x}^{2} + {d y}^{2}$ $ds^2 = dx^2 + dy^2$

Extracting ${d x}^{2}$ $dx^2$ as a factor from the right hand side

$d s^{2} = (1 + \frac{{d y}^{2}}{{d x}^{2}}) {d x}^{2}$ $ds^2 = (1 + dy^2/dx^2) dx^2$

Taking square roots of both sides

$d s = \sqrt{(1 + \frac{{d y}^{2}}{{d x}^{2}})} d x$ $ds = sqrt((1 + dy^2/dx^2)) dx$

from which

$s = \int \sqrt{(1 + {(\frac{d y}{d x})}^{2})} d x$ $s = int sqrt((1 + (dy/dx)^2)) dx$

Now noting

$y = x^{\frac{3}{2}}$ $y = x^(3/2)$

implies

$\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}}$ $dy/dx = 3/2 x^(1/2)$

so

$s = \int \sqrt{(1 + {(\frac{3}{2} x^{\frac{1}{2}})}^{2})} d x$ $s = int sqrt((1 + (3/2 x^(1/2))^2)) dx$

that is

$s = \int (\sqrt{1 + \frac{9}{4} x}) d x$ $s = int (sqrt(1 + 9/4 x)) dx$

so the required solution is

$\int_{0}^{6} (\sqrt{1 + \frac{9}{4} x}) d x$ $int_0^6 (sqrt(1 + 9/4 x)) dx$

$= \frac{1}{2} \int_{0}^{6} (\sqrt{4 + 9 x}) d x$ $= 1/2 int_0^6 (sqrt(4 + 9x)) dx$ (after rearrangement)

Gritting teeth and going for integration by substitution with

$u = 4 + 9 x$ $u = 4+9x$

so that

$\frac{d u}{d x} = 9$ $(du)/dx= 9$

from which

$\frac{1}{9} \int d u = \int d x$ $1/9 int du = int dx$

and noting required limits under the substitution

$u (0) = 4 and u (6) = 58$ $u(0) = 4 and u(6) = 58$

so

$s = (\frac{1}{2}) (\frac{1}{9}) \int_{4}^{58} (\sqrt{u}) d u$ $s = (1/2)(1/9) int_4^58 (sqrt(u)) du$

that is

$s = (\frac{1}{2}) (\frac{1}{9}) (\frac{2}{3}) {[u^{\frac{3}{2}}]}_{4}^{\frac{3}{2}}$ $s = (1/2)(1/9)(2/3) [u^(3/2)]_4^58$

$= (\frac{1}{27}) (58^{\frac{3}{2}} - 4^{\frac{3}{2}})$ $= (1/27)(58^(3/2) - 4^(3/2))$

$= 16.064$ $= 16.064$ (3 places of decimals)

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