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 257.16475 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 substitution \(x = \dfrac{1}{2}×tan(t) \) to transform the expression
\(\int \sqrt{1+4 \times\left(\frac{1}{2} \times \tan (t)\right)^{2}} \times \frac{1}{2} \times \sec (t)^{2} d t\)
Use the properties of integrals
\(\frac{1}{2} \times \int \sqrt{1+4 \times\left(\frac{1}{2} \times \tan (t)\right)^{2}} \sec (t)^{2} d t\)
Use the properties of exponents
\(\frac{1}{2} \times \int \sqrt{1+4 \times \frac{1}{4} \times \tan (t)^{2}} \sec (t)^{2} d t\)
Reduce the number
\(\frac{1}{2} \times \int \sqrt{1+\tan (t)^{2}} \sec (t)^{2} d t\)
Simplify the expression
\(\frac{1}{2} \times \int \sqrt{\sec (t)^{2}} \sec (t)^{2} d t\)
Simplify the root
\(\frac{1}{2} \times \int \sec (t) \sec (t)^{2} d t\)
Write in exponential form
\(\frac{1}{2} \times \int \sec (t)^{3} d t\)
Evaluale the integral
\(\frac{1}{2} \times\left(\frac{1}{2} \times \sec (t) \tan (t)+\frac{1}{2} \times \int \sec (t) d t\right)\)
Evaluate the integral
\(\left.\frac{1}{2} \times\left(\frac{1}{2} \times \sec (t) \tan (t)+\frac{1}{2} \times ln(|\sec (t)+\tan (t)|\right)\right)\)
Substitute back
\(\frac{1}{2} \times\left(\frac{1}{2} \times \sec \left(\arctan \left(\frac{x}{\frac{1}{2}}\right)\right) \tan \left(\arctan \left(\frac{x}{\frac{1}{2}}\right)\right)+\frac{1}{2} \times \ln \left(\left|\sec \left(\arctan \left(\frac{x}{\frac{1}{2}}\right)\right)+\tan \left(\arctan \left(\frac{x}{\frac{1}{2}}\right)\right)\right|\right)\right.\)
Simplify
\(\frac{x \sqrt{1+4 x^{2}}}{2}+\frac{1}{4} \times \ln \left(\left|\sqrt{1+4 x^{2}}+2 x\right|\right)\)
Return the limits and substitute, a = 0 and b = 16.
\(\left.\left(\frac{x \sqrt{1+4 x^{2}}}{2}+\frac{1}{4} \times \ln \left(\left|\sqrt{1+4 x^{2}}+2 x\right|\right)\right)\right|_{0} ^{16}\)
Calculate the expression.
\(\begin{array}{l} \frac{16 \sqrt{1+4 \times 16^{2}}}{2}+\frac{1}{4} \times \ln \left(\left|\sqrt{1+4 \times 16^{2}}+2 \times 17\right|\right)-\left(\frac{0 \sqrt{1+4 \times 0^{2}}}{2}+\frac{1}{4} \times \ln \left(\left|\sqrt{1+4 \times 0^{2}}+2 \times 0\right|\right)\right)\\ \end{array}\)
Simplifying further, we get
\(\frac{40 \sqrt{41}}{2}+\frac{1}{4} \times \ln (5\sqrt{41}+32)\)
L ≈ 257.16475 units