Find the exact length of the curve. x = 1/3√y(y - 3), 4 ≤ y ≤ 16.

Solution:

Given, x = 1/3 √y (y - 3), 4 ≤ y ≤ 16

Length of the curve, x = f(y) from y =a to y = b is given by:

$\int_{a}^{b}\sqrt{1+f'(y)^{2}}.dy$

x = 1/3 √y (y - 3)

x = 1/3 (y^3/2- 3y^1/2)

dx/dy = 1/3 [(3/2)y^1/2- (3/2)y^-1/2]

= 1/3 [(3y^1/2/2) - (3/2y^1/2)]

= 1/2 [y^1/2 - 1/y^1/2]

= (y-1)/2√y = f'(y)

Length of the curve

=$\int_{a}^{b}\sqrt{1+f'(y)^{2}} .dy$

=$\int_{4}^{16}\sqrt{1+\left ( \frac{y-1}{2\sqrt{y}} \right )^{2}} .dy$

=$\int_{4}^{16}\sqrt{1 + \frac{y^{2}-2y+1}{4y}}. dy$

=$\int_{4}^{16}\sqrt{ \frac{4y + y^{2}-2y+1}{4y}}. dy$

=$\int_{4}^{16}\frac{\left ( y+1 \right )^{2}}{\left (2\sqrt{y}\right )^{2}}$

=$\int_{4}^{16} \frac{y+1}{2\sqrt{y}}. dy$

=$\int_{4}^{16} \frac{1}{2}\sqrt{y}. dy + \int_{4}^{16}\frac{1}{2\sqrt{y}}. dy$

=$\int_{4}^{16} \frac{1}{2}y^{\frac{1}{2}}. dy + \int_{4}^{16} \frac{1}{2}y^{\frac{-1}{2}}. dy$

=$\frac{1}{2} {\frac{y^{3/2}}{3/2}}\Biggr|_{4}^{16} +\frac{1}{2} {\frac{y^{1/2}}{1/2}}\Biggr|_{4}^{16}$

= [1/3 (64-8)] +[4-2]

= (56/3)+2

= 62/3

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Find the exact length of the curve. x = 1/3√y(y − 3), 4 ≤ y ≤ 16.

Summary:

The exact length of the curve x = 1/3 √y (y - 3), 4 ≤ y ≤ 16, is 62/3.