PEPCVF - Derivatives and Arc Length of Parametric Curves

Derivatives and Arc Length of Parametric Curves

Slopes and Tangent Lines

Recall that for a curve y = f(x), f '(x) gives the slope of the curve at (x, y). Just as not all functions have derivatives, not all parametric curves have defined slopes. For this reason, the parametric curves addressed in this lesson are restricted to smooth parametric curves. A smooth curve is defined as a curve with parametric representation x = f(t) and y = g(t) on an interval I where f and g have continuous first derivatives on I and are not simultaneously 0 (which prevents the curve from having any corners or cusps), except possibly at the endpoints of I.

A parametrized curve C defined by x = f(t), y = g(t), and a < t < b is differentiable at t if f and g are differentiable at t. At a point on a differentiable parametrized curve where y is also a differentiable function of x, the derivatives dy/dt, dx/dt, and dy/dx are related by the Chain Rule and written equation image indicator $LaTeX: \frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$ $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$ . If dx/dt ≠ 0, then the formula for finding dy/dx as a function of t from dy/dt and dx/dt is $LaTeX: \frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ $\frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ . Once a derivative is found, the slope for a given value of t can be calculated and an equation of a tangent line to the curve can be determined in much the same way equations of tangent lines were found for non-parametrically defined curves. The presentation below shows how to find the derivative and equation of a tangent line to a curve written in parametric form.

To explore parametric derivatives using an interactive applet, CLICK HERE. Links to an external site.

image of a cycloid where k = 4 If the parametric equations for a curve define y as a twice-differentiable function of x, then the second derivative may be calculated as follows: equation image indicator $LaTeX: \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ $\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ . Just as second derivatives of functions such as y = f(x) are used to determine where a curve is concave up or concave down, second derivatives of parametrically defined curves provide information on the concavity of curves such as the given astroid.

The next two presentations illustrate how the second derivative of parametric equations is used to predict and verify a curve's concavity.

Arc Length

If a curve C is described by the parametric equations x = f(t), y = g(t), a < t < b, where f ' and g' are continuous on [a, b] and C is traversed exactly once as t increases from a to b, then the length of C is equation image indicator $LaTeX: L=\int_a^b\sqrt[]{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$ $L=\int_a^b\sqrt[]{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$ . The video illustrates the use of this formula with two examples.

Returning to the cycloid problem from the previous lesson, the length of one arch of the cycloid is found as follows.

image of arch of a cycloid

Arc Length of a Cycloid

Find the length of one arch of the cycloid x = a(t-sin(t)), y = a(1-cos(t)).

Solution: One arch is traced by the parameter interval equation image indicator $LaTeX: 0\le t\le2\pi$ $0\le t\le2\pi$ .

equation image indicator $LaTeX: \frac{dx}{dt}=a\left(1-\cos t\right);\frac{dy}{dt}=a\cos t\\ L=\int_0^{2p}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt=\int_0^{2p}\sqrt{a^2(1-\cos t)^2+a^2 \sin ^2t}\textbf{d}t\\ \int_0^{2p}\sqrt{a^2(1-2\cos t + \cos ^2t +\sin ^2t)^2+a^2 \sin ^2t}\textbf{d}t=a\int_0^{2p}\sqrt{2(1-\cos t}\textbf{d}t$ $\frac{dx}{dt}=a\left(1-\cos t\right);\frac{dy}{dt}=a\cos t\\ L=\int_0^{2p}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt=\int_0^{2p}\sqrt{a^2(1-\cos t)^2+a^2 \sin ^2t}\textbf{d}t\\ \int_0^{2p}\sqrt{a^2(1-2\cos t + \cos ^2t +\sin ^2t)^2+a^2 \sin ^2t}\textbf{d}t=a\int_0^{2p}\sqrt{2(1-\cos t}\textbf{d}t$

Using the identity equation image indicator with ,

equation image indicator $LaTeX: \sqrt[]{2\left(1-\cos x\right)}=\sqrt[]{4\sin^2\left(\frac{x}{2}\right)}=2\left|\sin\left(\frac{x}{2}\right)\right|=2\sin\left(\frac{x}{2}\right)\\ L=2a\int_0^{2p}\sin\frac{x}{2}\textbf{d}x=2a[-2\cos\frac{x}{2}]_0^{2p}=2a[2+2]=8a$ $\sqrt[]{2\left(1-\cos x\right)}=\sqrt[]{4\sin^2\left(\frac{x}{2}\right)}=2\left|\sin\left(\frac{x}{2}\right)\right|=2\sin\left(\frac{x}{2}\right)\\ L=2a\int_0^{2p}\sin\frac{x}{2}\textbf{d}x=2a[-2\cos\frac{x}{2}]_0^{2p}=2a[2+2]=8a$

To use the interactive parametric arc length applet, click on the down arrow to change from Standard to Parametric and CLICK HERE Links to an external site..

Derivatives and Arc Length of Parametric Curves Practice

#5 is enrichment only. You are not expected to be able to do this problem using the trapezoidal rule but you should be able to find the length of the curve using the formula for arc length.

Derivatives and Arc Length of Parametric Curves: Even More Problems!

Please review all content and feedback and contact your instructor if additional clarification is needed. Once you feel confident with the material, go to the Quizzes widget and complete Parametric Equations, Derivatives, and Arc Length Quiz.

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