Length of a Curve

Solution to Example 1:

We first calculate the derivative:

\[ \dfrac{dy}{dx} = 0.2x \]

Use the formula for the arc length:

\[ L = \int_{-15}^{10} \sqrt{1 + (0.2x)^2} \, dx \]

Use trigonometric substitution \( \tan u = 0.2x \). Taking the derivative of both sides gives:

\[ \sec^2 u \, du = 0.2 \, dx \implies dx = 5\sec^2 u \, du \]

Solving \( \tan u = 0.2x \) for \( u \) gives \( u = \arctan(0.2x) \). The limits of integration after substitution are:

• \( u_1 = \arctan(0.2(-15)) = \arctan(-3) \approx -1.24904 \) (when \( x = -15 \))
• \( u_2 = \arctan(0.2(10)) = \arctan(2) \approx 1.10714 \) (when \( x = 10 \))

Substituting into the integral:

\[ L = 5 \int_{-1.24904}^{1.10714} \sqrt{1 + \tan^2 u} \sec^2 u \, du \]

Using the identity \( \sqrt{1 + \tan^2 u} = |\sec u| \):

\[ L = 5 \int_{-1.24904}^{1.10714} |\sec u| \sec^2 u \, du \]

Since \( \sec u > 0 \) on the interval \( [-1.24904, 1.10714] \), \( |\sec u| = \sec u \), yielding:

\[ L = 5 \int_{-1.24904}^{1.10714} \sec^3 u \, du \]

Using the integral of \( \sec^3 u \):

\[ L = \left[ \dfrac{5}{2} \left( \sec u \tan u + \ln|\sec u + \tan u| \right) \right]_{-1.24904}^{1.10714} \approx 43.05 \]

Solution to Example 2:

Calculate the derivative:

\[ \dfrac{df}{dx} = \dfrac{d}{dx}[\ln(\sin x)] = \dfrac{\cos x}{\sin x} = \cot x \]

Applying the formula for arc length:

\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 + (\cot x)^2} \, dx \]

Use the trigonometric identity \( 1 + \cot^2 x = \csc^2 x \):

\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{\csc^2 x} \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} |\csc x| \, dx \]

Since \( \csc x > 0 \) on the closed interval \( \left[\dfrac{\pi}{4}, \dfrac{\pi}{2}\right] \), \( |\csc x| = \csc x \):

\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc x \, dx \]

Using the standard integral formula \( \displaystyle \int \csc x \, dx = \ln\left|\tan\left(\dfrac{x}{2}\right)\right| + C \):

\[ L = \left[ \ln\left|\tan\left(\dfrac{x}{2}\right)\right| \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \ln(1) - \ln\left(\tan\left(\dfrac{\pi}{8}\right)\right) \approx 0 - (-0.88137) \approx 0.88137 \]

Solution to Example 3:

Calculate the derivative using the chain rule:

\[ \dfrac{df}{dx} = \dfrac{-2(x+2)}{2\sqrt{9 - (x+2)^2}} = -\dfrac{x+2}{\sqrt{9 - (x+2)^2}} \]

Applying the arc length formula:

\[ L = \int_{-4}^{0} \sqrt{1 + \left( -\dfrac{x+2}{\sqrt{9 - (x+2)^2}} \right)^2} \, dx \]

Simplify the integrand:

\[ 1 + \dfrac{(x+2)^2}{9 - (x+2)^2} = \dfrac{9 - (x+2)^2 + (x+2)^2}{9 - (x+2)^2} = \dfrac{9}{9 - (x+2)^2} \]

Taking the square root:

\[ L = \int_{-4}^{0} \dfrac{3}{\sqrt{9 - (x+2)^2}} \, dx \]

Using substitution \( u = x+2 \), \( du = dx \). When \( x = -4 \), \( u = -2 \); when \( x = 0 \), \( u = 2 \):

\[ L = \int_{-2}^{2} \dfrac{3}{\sqrt{9 - u^2}} \, du = 3 \left[ \arcsin\left(\dfrac{u}{3}\right) \right]_{-2}^{2} \] \[ = 3 \left( \arcsin\left(\dfrac{2}{3}\right) - \arcsin\left(-\dfrac{2}{3}\right) \right) = 6 \arcsin\left(\dfrac{2}{3}\right) \approx 4.38 \]