Find the exact length of the curve. y = (1 − x² ), 0 ≤ x ≤ 1/4.

Solution:

Given,

y = ln(1 - x² ), 0 ≤ x ≤ 1/4

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

\( \int_{a}^{b}\sqrt{1+f'(x)^{2}} .dx \)

y = (1 - x²)

Differentiating w.r.t. x,

dy/dx = (-2x)

Length of the curve,

= \( \int_{0}^{1/4}\sqrt{1+ [-2x]^{2}} .dx \)

= \( \int_{0}^{1/4}\sqrt{1+ 4x^{2}} .dx \)

= \( \frac{\ \frac{2x}{2} \times \sqrt{1 + 4x^{2}}\Biggr|_{0}^{1/4}\ + \frac{1}{2}\times log \left | 2x + \sqrt{1 + 4x^{2}} \right |\Biggr|_{0}^{1/4}}{2} \)

= \( \frac{1}{2}\left \{ \left [ \frac{1}{4}\sqrt{1+\left ( 4\times \frac{1}{16} \right )} \right ] + \frac{1}{2}\left [ log\left ( \frac{1}{2}+\frac{\sqrt{5}}{2} \right )- log\left ( 0+\sqrt{1+0} \right ) \right ]\right \} \)

= \( \left ( \frac{1}{8}\times \frac{\sqrt{5}}{2} \right )+ \left ( \frac{1}{4}log\left ( \frac{\sqrt{5}+1}{2} \right ) \right )\)

= \( \frac{\sqrt{5}}{16}+\left [1/4 log\left ( \frac{\sqrt{5}+1}{2} \right ) \right ] \)

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Find the exact length of the curve. y = (1 − x² ), 0 ≤ x ≤ 1/4.

Summary:

The exact length of the curve, y = (1 - x²), 0 ≤ x ≤ 1/4 is √5/16 + (1/4)log[(√5 +1)/2].