Find the exact length of the curve x = 1/3 √y (y - 3), 16 ≤ y ≤ 25.
Solution:
Given, x = 1/3 √y (y - 3), 16 ≤ y ≤ 25
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 (y3/2- 3y1/2)
dx/dy = 1/3 [(3/2)y1/2- (3/2)y-1/2]
= 1/3 [(3y1/2/2) - (3/2y1/2)]
= 1/2 [y1/2 - 1/y1/2]
= (y-1)/2√y = f'(y)
Length of the curve
=\( \int_{a}^{b}\sqrt{1+f'(y)^{2}} .dy \)
=\( \int_{16}^{25}\sqrt{1+\left ( \frac{y-1}{2\sqrt{y}} \right )^{2}} .dy \)
=\( \int_{16}^{25}\sqrt{1 + \frac{y^{2}-2y+1}{4y}}. dy \)
=\( \int_{16}^{25}\sqrt{ \frac{4y + y^{2}-2y+1}{4y}}. dy \)
=\( [\int_{16}^{25}\sqrt{\frac{\left ( y+1 \right )^{2}}{\left (2\sqrt{y}\right )^{2}}}] \)
=\( \int_{16}^{25} \frac{y+1}{2\sqrt{y}}. dy \)
=\( \int_{16}^{25} \frac{1}{2}\sqrt{y}. dy + \int_{16}^{25}\frac{1}{2\sqrt{y}}. dy \)
=\( \int_{16}^{25} \frac{1}{2}y^{\frac{1}{2}}. dy + \int_{16}^{25} \frac{1}{2}y^{\frac{-1}{2}}. dy \)
=\( \frac{1}{2} {\frac{y^{3/2}}{3/2}}\Biggr|_{16}^{25} +\frac{1}{2} {\frac{y^{1/2}}{1/2}}\Biggr|_{16}^{25}\)
= [1/3 (125-64)] +[5-4]
= (61/3)+1
= 64/3
Find the exact length of the curve x = 1/3 √y (y - 3), 16 ≤ y ≤ 25.
Summary:
The exact length of the curve x = 1/3 √y (y - 3), 16 ≤ y ≤ 25, is 64/3.
visual curriculum