: Find the length of the are of y2 = x3 from x = of y2 = x3 from x = 0

1.	: Find the length of the are of y2 = x3 from x = of y2 = x3 from x = 0 to x = 5. Given equation of the curve
Answer» $\large\underline{\sf{Solution-}}$ Given CURVE is $\red{\rm :\longmapsto\: {y}^{2} = {x}^{3}}$ can be rewritten as $\rm :\longmapsto\:y = \sqrt{ {x}^{3} }$ So, curve is to be considered in first quadrant to find the length of arc from x = 0 to x = 5. Now, On differentiating both SIDES w. r. t. x, we get $\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{3} } }\dfrac{d}{dx} {x}^{3}$ $\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{3} } } {3x}^{2}$ $\bf\implies \:\dfrac{dy}{dx} = \dfrac{3}{2} \sqrt{x}$ Now, we KNOW Length of arc y = f(x) from x = a to x = B is given by $\red{\rm :\longmapsto\:Length \: of \: arc \: = \displaystyle\int ^b_a \: \sqrt{1 + {\bigg(\dfrac{dy}{dx} \bigg) }^{2} } dx}$ So, on substituting the values, we get $\red{\rm :\longmapsto\:Length \: of \: arc \: = \displaystyle\int ^5_0 \: \sqrt{1 + {\bigg(\dfrac{9x}{4} \bigg) } } dx}$ $\red{\rm :\longmapsto\:Length \: of \: arc \: = \displaystyle\int ^5_0 \: \sqrt{\dfrac{4 + 9x}{4}} dx}$ $\red{\rm :\longmapsto\:Length \: of \: arc \: = \dfrac{1}{2} \displaystyle\int ^5_0 \: \sqrt{4 + 9x} dx}$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{2} \times \dfrac{ {\bigg(4 + 9x \bigg) }^{\dfrac{3}{2} } }{\dfrac{3}{2} \times 9 }\bigg \|^5_0}$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \times {\bigg(4 + 9x \bigg) }^{\dfrac{3}{2} }\bigg \|^5_0}$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg \|{\bigg(4 + 45 \bigg) }^{\dfrac{3}{2} } - {\bigg(4 + 0 \bigg) }^{\dfrac{3}{2} }}\bigg \|$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg \|{\bigg(49 \bigg) }^{\dfrac{3}{2} } - {\bigg(4 \bigg) }^{\dfrac{3}{2} }}\bigg \|$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg \|{\bigg( {7}^{2} \bigg) }^{\dfrac{3}{2} } - {\bigg( {2}^{2} \bigg) }^{\dfrac{3}{2} }}\bigg \|$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg \|343 - 8\bigg \|}$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg \|335\bigg \|}$ $\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{335}{27} \: units}$

: Find the length of the are of y2 = x3 from x = of y2 = x3 from x = 0 to x = 5. Given equation of the curve

Answer»

$\large\underline{\sf{Solution-}}$

Given CURVE is

$\red{\rm :\longmapsto\: {y}^{2} = {x}^{3}}$

can be rewritten as

$\rm :\longmapsto\:y = \sqrt{ {x}^{3} }$

So, curve is to be considered in first quadrant to find the length of arc from x = 0 to x = 5.

Now,

On differentiating both SIDES w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{3} } }\dfrac{d}{dx} {x}^{3}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{ {x}^{3} } } {3x}^{2}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{3}{2} \sqrt{x}$

Now, we KNOW Length of arc y = f(x) from x = a to x = B is given by

$\red{\rm :\longmapsto\:Length \: of \: arc \: = \displaystyle\int ^b_a \: \sqrt{1 + {\bigg(\dfrac{dy}{dx} \bigg) }^{2} } dx}$

So, on substituting the values, we get

$\red{\rm :\longmapsto\:Length \: of \: arc \: = \displaystyle\int ^5_0 \: \sqrt{1 + {\bigg(\dfrac{9x}{4} \bigg) } } dx}$

$\red{\rm :\longmapsto\:Length \: of \: arc \: = \displaystyle\int ^5_0 \: \sqrt{\dfrac{4 + 9x}{4}} dx}$

$\red{\rm :\longmapsto\:Length \: of \: arc \: = \dfrac{1}{2} \displaystyle\int ^5_0 \: \sqrt{4 + 9x} dx}$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{2} \times \dfrac{ {\bigg(4 + 9x \bigg) }^{\dfrac{3}{2} } }{\dfrac{3}{2} \times 9 }\bigg |^5_0}$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \times {\bigg(4 + 9x \bigg) }^{\dfrac{3}{2} }\bigg |^5_0}$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg |{\bigg(4 + 45 \bigg) }^{\dfrac{3}{2} } - {\bigg(4 + 0 \bigg) }^{\dfrac{3}{2} }}\bigg |$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg |{\bigg(49 \bigg) }^{\dfrac{3}{2} } - {\bigg(4 \bigg) }^{\dfrac{3}{2} }}\bigg |$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg |{\bigg( {7}^{2} \bigg) }^{\dfrac{3}{2} } - {\bigg( {2}^{2} \bigg) }^{\dfrac{3}{2} }}\bigg |$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg |343 - 8\bigg |}$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{1}{27} \bigg |335\bigg |}$

$\red{\rm :\longmapsto\:Length \: of \: arc = \dfrac{335}{27} \: units}$

: Find the length of the are of y2 = x3 from x = of y2 = x3 from x = 0 to x = 5. Given equation of the curve

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: Find the length of the are of y2 = x3 from x = of y2 = x3 from x = 0 to x = 5. Given equation of the curve​

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: Find the length of the are of y2 = x3 from x = of y2 = x3 from x = 0 to x = 5. Given equation of the curve