Find the length of a smooth curve on a closed interval using the arc-length formula.
If \( y = f(x) \) is continuously differentiable on \([a,b]\), the formula above gives the exact length of the graph from \( x = a \) to \( x = b \).
Tip: First find \( f'(x) \), square it, add 1, take the square root, then integrate.
Find the arc length of the graph of \( y = \dfrac{2}{3}x^{3/2} \) on the interval \([0, 4]\).
1. Find the derivative
$$f'(x) = x^{1/2} = \sqrt{x}$$
2. Form the integrand
$$\bigl(f'(x)\bigr)^2 = x$$
$$1 + \bigl(f'(x)\bigr)^2 = 1 + x$$
$$\sqrt{1 + \bigl(f'(x)\bigr)^2} = \sqrt{1 + x}$$
3. Set up and evaluate the integral
$$L = \int_0^4 \sqrt{1 + x}\, dx = \dfrac{2}{3}(1 + x)^{3/2}\Big|_0^4 = \dfrac{2}{3}\bigl(5^{3/2} – 1\bigr) = \dfrac{2}{3}(5\sqrt{5} – 1)$$
Final Answer
\( L = \dfrac{2}{3}(5\sqrt{5} – 1) \)
Straightforward arc-length integrals with clean derivatives
Find the arc length of the graph of \( y = \dfrac{1}{3}x^{3/2} – x^{1/2} \) on the interval \([1, 4]\).
Step 1: Find the derivative
\[ f(x) = \dfrac{1}{3}x^{3/2} – x^{1/2} \]
\[ f'(x) = \dfrac{1}{3}\cdot\dfrac{3}{2}x^{1/2} – \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2}x^{1/2} – \dfrac{1}{2}x^{-1/2} \]
Factor out \(\dfrac{1}{2}\): \[ f'(x) = \dfrac{1}{2}\Bigl(\sqrt{x} – \dfrac{1}{\sqrt{x}}\Bigr) = \dfrac{x – 1}{2\sqrt{x}} \]
Step 2: Square the derivative
\[ \bigl(f'(x)\bigr)^2 = \left(\dfrac{x-1}{2\sqrt{x}}\right)^2 = \dfrac{(x-1)^2}{4x} \]
Step 3: Form \(1 + (f'(x))^2\)
\[ 1 + \bigl(f'(x)\bigr)^2 = 1 + \dfrac{(x-1)^2}{4x} = \dfrac{4x + (x-1)^2}{4x} \]
Expand the numerator: \[ (x-1)^2 = x^2 – 2x + 1 \implies 4x + x^2 – 2x + 1 = x^2 + 2x + 1 = (x+1)^2 \]
Therefore \[ 1 + \bigl(f'(x)\bigr)^2 = \dfrac{(x+1)^2}{4x} \]
Step 4: Take the square root (integrand simplifies nicely)
\[ \sqrt{1 + \bigl(f'(x)\bigr)^2} = \sqrt{\dfrac{(x+1)^2}{4x}} = \dfrac{x+1}{2\sqrt{x}} \]
Step 5: Set up and evaluate the arc-length integral
\[ L = \int_1^4 \dfrac{x+1}{2\sqrt{x}}\,dx = \dfrac{1}{2}\int_1^4 \bigl(x^{1/2} + x^{-1/2}\bigr)\,dx \]
Antiderivative: \[ \dfrac{1}{2}\left[\dfrac{2}{3}x^{3/2} + 2x^{1/2}\right]_1^4 = \left[\dfrac{1}{3}x^{3/2} + x^{1/2}\right]_1^4 \]
Evaluation: \[ \left[\dfrac{1}{3}x^{3/2} + x^{1/2}\right]_1^4 = \Bigl(\dfrac{8}{3}+2\Bigr) – \Bigl(\dfrac{1}{3}+1\Bigr) = \dfrac{14}{3} – \dfrac{4}{3} = \dfrac{10}{3} \]
\( L = \dfrac{10}{3} \)
Find the arc length of the graph of \( y = \ln(\cos x) \) on the interval \(\bigl[0, \dfrac{\pi}{4}\bigr]\).
Derivative
$$f'(x) = -\tan x$$
Integrand
$$\sqrt{1 + \bigl(f'(x)\bigr)^2} = \sqrt{1 + \tan^{2}x} = \sec x$$
Evaluate
$$L = \int_0^{\pi/4} \sec x\, dx = \ln|\sec x + \tan x|\Big|_0^{\pi/4} = \ln(\sqrt{2} + 1) – \ln(1) = \ln(1 + \sqrt{2})$$
\( L = \ln(1 + \sqrt{2}) \)
Find the arc length of the graph of \( y = \dfrac{x^{3}}{6} + \dfrac{1}{2x} \) on the interval \([1, 2]\).
Step 1: Find the derivative
\[ f(x) = \dfrac{x^3}{6} + \dfrac{1}{2x} \]
\[ f'(x) = \dfrac{3x^2}{6} – \dfrac{1}{2x^2} = \dfrac{x^2}{2} – \dfrac{1}{2x^2} \]
Rewrite with a common denominator \(2x^2\): \[ f'(x) = \dfrac{x^2 \cdot x^2}{2 \cdot x^2} – \dfrac{1}{2x^2} = \dfrac{x^4}{2x^2} – \dfrac{1}{2x^2} = \dfrac{x^4 – 1}{2x^2} \]
Step 2: Square the derivative
\[ \bigl(f'(x)\bigr)^2 = \left( \dfrac{x^4 – 1}{2x^2} \right)^2 = \dfrac{(x^4 – 1)^2}{4x^4} \]
Step 3: Form \(1 + (f'(x))^2\)
\[ 1 + \bigl(f'(x)\bigr)^2 = 1 + \dfrac{(x^4 – 1)^2}{4x^4} = \dfrac{4x^4 + (x^4 – 1)^2}{4x^4} \]
Expand the numerator: \[ (x^4 – 1)^2 = x^8 – 2x^4 + 1 \] \[ 4x^4 + x^8 – 2x^4 + 1 = x^8 + 2x^4 + 1 = (x^4 + 1)^2 \]
Therefore \[ 1 + \bigl(f'(x)\bigr)^2 = \dfrac{(x^4 + 1)^2}{4x^4} \]
Step 4: Take the square root (integrand simplifies nicely)
\[ \sqrt{1 + \bigl(f'(x)\bigr)^2} = \sqrt{\dfrac{(x^4 + 1)^2}{4x^4}} = \dfrac{x^4 + 1}{2x^2} = \dfrac{x^2}{2} + \dfrac{1}{2x^2} \]
Step 5: Set up and evaluate the arc-length integral
\[ L = \int_1^2 \left( \dfrac{x^2}{2} + \dfrac{1}{2x^2} \right)\, dx \]
Antiderivative: \[ \int \left( \dfrac{x^2}{2} + \dfrac{1}{2}\,x^{-2} \right)\, dx = \dfrac{x^3}{6} – \dfrac{1}{2x} \]
Evaluate from 1 to 2: \[ \begin{align*} \text{At } x=2:&\quad \dfrac{8}{6} – \dfrac{1}{4} = \dfrac{4}{3} – \dfrac{1}{4} = \dfrac{16}{12} – \dfrac{3}{12} = \dfrac{13}{12}\\[0.8em] \text{At } x=1:&\quad \dfrac{1}{6} – \dfrac{1}{2} = \dfrac{1}{6} – \dfrac{3}{6} = -\dfrac{2}{6} = -\dfrac{1}{3} = -\dfrac{4}{12} \end{align*} \]
\[ L = \dfrac{13}{12} – \left(-\dfrac{4}{12}\right) = \dfrac{13}{12} + \dfrac{4}{12} = \dfrac{17}{12} \]
\( L = \dfrac{17}{12} \)
Requires a careful derivative and a substitution to integrate
Find the arc length of the graph of \( y = \dfrac{3}{2}x^{2/3} + 4 \) on the interval \([1, 27]\).
Step 1: Find the derivative
\[ f(x) = \dfrac{3}{2}x^{2/3} + 4 \]
\[ f'(x) = \dfrac{3}{2} \cdot \dfrac{2}{3} x^{-1/3} = x^{-1/3} = \dfrac{1}{x^{1/3}} \]
Step 2: Form \(1 + (f'(x))^2\)
\[ \bigl(f'(x)\bigr)^2 = \bigl(x^{-1/3}\bigr)^2 = x^{-2/3} \]
\[ 1 + \bigl(f'(x)\bigr)^2 = 1 + x^{-2/3} = \dfrac{x^{2/3} + 1}{x^{2/3}} \]
Step 3: Take the square root (the integrand)
\[ \sqrt{1 + \bigl(f'(x)\bigr)^2} = \dfrac{\sqrt{x^{2/3} + 1}}{x^{1/3}} \]
Step 4: Set up the arc-length integral
\[ L = \int_1^{27} \dfrac{\sqrt{x^{2/3} + 1}}{x^{1/3}}\, dx \]
Step 5: Substitution (keep the original limits)
Let \( u = x^{2/3} + 1 \). Then \[ du = \dfrac{2}{3}x^{-1/3}\, dx \implies x^{-1/3}\, dx = \dfrac{3}{2}\, du \]
The integrand becomes \[ \sqrt{x^{2/3} + 1} \cdot x^{-1/3}\, dx = \sqrt{u} \cdot \dfrac{3}{2}\, du = \dfrac{3}{2}\, u^{1/2}\, du \]
The indefinite integral is \[ \dfrac{3}{2} \int u^{1/2}\, du = \dfrac{3}{2} \cdot \dfrac{2}{3} u^{3/2} = u^{3/2} = \bigl(x^{2/3} + 1\bigr)^{3/2} \]
Step 6: Evaluate from \( x = 1 \) to \( x = 27 \)
\[ L = \Bigl[ \bigl(x^{2/3} + 1\bigr)^{3/2} \Bigr]_1^{27} \]
\[ \begin{align*} \text{At } x = 27:&\quad \bigl(27^{2/3} + 1\bigr)^{3/2} = (9 + 1)^{3/2} = 10^{3/2} = 10\sqrt{10}\\[0.8em] \text{At } x = 1:&\quad \bigl(1^{2/3} + 1\bigr)^{3/2} = (1 + 1)^{3/2} = 2^{3/2} = 2\sqrt{2} \end{align*} \]
\[ L = 10\sqrt{10} – 2\sqrt{2} \]
\( L = 10\sqrt{10} – 2\sqrt{2} \)
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