Calculus Small Changes, Big Ideas

Section 6.4 Arc length

• Differential of arc length:

  \begin{equation*} \dd s=\sqrt{(\dd{x})^{2}+(\dd{y})^{2}} \end{equation*}

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• Arc length:

  \begin{equation*} s=\int_{a}^{b}\dd{s}=\int_{a}^{b}\sqrt{\qty(\dv{x}{t})^{2}+\qty(\dv{y}{t})^{2}}\dd{t} \end{equation*}

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• If the curve is a function \(y=f(x)\text{,}\) the formula simplifies:

  \begin{equation*} s=\int_{a}^{b} \sqrt{1+f'(x)^{2}}\dd{x} \end{equation*}

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• Easy to find arc length of a circle, but not doable for an ellipse! More on that later.

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• If \((x,y)\) is considered as the position of a particle, then the arc length is the distance traveled.

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• Polar arc length:

  \begin{equation*} s=\int_{\alpha}^{\beta} \sqrt{r^{2}+\qty(\dv{r}{\theta})^{2}}\dd{\theta} \end{equation*}

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• Why is the inverse of the sine function called the ARCsine function? Because it gives the angle whose sine is a given number, and the “arc” refers to the arc length of the unit circle corresponding to that angle.

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• Fix \(r=1\text{,}\) so \(x=\cos\theta\text{,}\) \(y=\sin\theta\text{,}\) and \(m=\tan\theta\text{.}\) Then we can express the inverse trigonometric functions in terms of these variables:

  \begin{equation*} \theta = \arcsin y = \arccos x = \arctan m \end{equation*}

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• The derivatives of the inverse trigonometric functions can be found using implicit differentiation. For example, if \(\theta=\arcsin y\text{,}\) then we have \(\sin \theta = y\text{.}\) Differentiating both sides with respect to \(x\) gives:

  \begin{equation*} \cos \theta \cdot \dv{\theta}{y} = 1 \end{equation*}

  Solving for \(\DS\dv{\theta}{y}\) yields:

  \begin{equation*} \dv{\theta}{y} = \frac{1}{\cos \theta} = \frac{1}{\sqrt{1-y^2}} \end{equation*}

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• Similarly, for \(\theta=\arccos x\text{,}\) we have \(\cos \theta = x\text{.}\) Differentiating both sides gives:

  \begin{equation*} -\sin \theta \cdot \dv{\theta}{x} = 1 \end{equation*}

  Solving for \(\DS\dv{\theta}{x}\) yields:

  \begin{equation*} \dv{\theta}{x} = -\frac{1}{\sin \theta} = -\frac{1}{\sqrt{1-x^2}} \end{equation*}

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• For \(\theta=\arctan m\text{,}\) we have \(\tan \theta = m\text{.}\) Differentiating both sides gives:

  \begin{equation*} \sec^2 \theta \cdot \dv{\theta}{m} = 1 \end{equation*}

  Solving for \(\DS\dv{\theta}{m}\) yields:

  \begin{equation*} \dv{\theta}{m} = \frac{1}{\sec^2 \theta} = \frac{1}{1+m^2} \end{equation*}

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• Put another way, the derivatives of \(\arcsin\text{,}\) \(\arccos\text{,}\) and \(\arctan\) can be thought of as \(\DS\dv{\theta}{y}\text{,}\) \(\DS\dv{\theta}{x}\text{,}\) and \(\DS\dv{\theta}{m}\) respectively, where \(\theta\) is the angle corresponding to the given trigonometric value.

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• Using \(x\) as a generic independent variable, we can express the derivatives of the inverse trigonometric functions in terms of \(x\text{:}\)

  \begin{align*} \dv{x}\arcsin x \amp = \frac{1}{\sqrt{1-x^2}}\\ \dv{x}\arccos x \amp = -\frac{1}{\sqrt{1-x^2}}\\ \dv{x}\arctan x \amp = \frac{1}{1+x^2} \end{align*}

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