Find the exact length of the curve, y = ln1 – x2, 0≤ x ≤ 16.
The exact length of a curve is a geometrical concept that addressed by integral calculus. It is a method for calculating the exact lengths of line segments.
Answer: The exact length of the curve y = ln1 – x2, 0≤ x ≤ 16 is 257.093 units
Let’s solve it step by step.
Explanation:
For the given curve, We will use the formula for the length of the arc(L) it.
Given function ⇒ y = ln 1 – x2
Now, differentiate the given function (y = ln 1 – x2) with respect to “x”
dy/dx = d( ln 1 – x2)/dx
dy/dx = 0 - 2x ------------- (1)
For finding arc length (L), we will use the following formula,
L = \(\int_{a}^{b}\) √[1 + (dy/dx)2] dx.
Substituting the value of dy/dx in length of curve formula from (1)
L = √[1 + (-2x)2] dx.
\(\int_{a}^{b} \sqrt{1+4 x^{2}} d x\)
Evaluate the indefinite integral
\(\int \sqrt{1+4 x^{2}} d x\)
Using the formula for integration by parts, we have
\(x\sqrt{1+4 x^{2}}-\int \dfrac{4x^2}{\sqrt{1+4x^2}}d x\)
= \(x\sqrt{1+4 x^{2}}-\int \dfrac{1-1+4x^2}{\sqrt{1+4x^2}}d x\)
= \(x\sqrt{1+4 x^{2}}+\int \dfrac{1}{\sqrt{1+4x^2}}d x - \int\sqrt{1+4x^2} dx\)
= \(2\int\sqrt{1+4x^2} dx=x\sqrt{1+4 x^{2}}+\frac{1}{2}\ln|x+\sqrt{1+4x^2}|\)
= \(\int\sqrt{1+4x^2} dx=\frac{1}{2}x\sqrt{1+4 x^{2}}+\frac{1}{4}\ln|x+\sqrt{1+4x^2}| + C\)
Now, determine the definite integral
\(\int_0^{16} \sqrt{1+4 x^{2}} d x \)
= \(\frac{1}{2}x\sqrt{1+4 x^{2}}+\frac{1}{4}\ln|x+\sqrt{1+4x^2}| |_0^{16}\)
= \(8\sqrt{1+4\times16^{2}}+\frac{1}{4}\ln|16+\sqrt{1+4\times 16^2}| | - 0\)
= 257.093 units
L ≈ 257.093 units