\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Parametric Equations - Slope & Tangent Lines

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On this page we discuss calculating slope and tangent lines of parametric equations.

To find the equation of a tangent line to a graph given by a set of parametric equations, we need to be able to find the slope by calculating the derivative \( dy/dx \) using the parametric derivative \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \] For horizontal tangent lines, the slope \(dy/dx\) is zero, so we need \( dy/dt = 0 \) and \( dx/dt \neq 0 \). For vertical tangent lines, the slope is undefined, which means that \( dx/dt = 0 \) when \( dy/dt \neq 0 \). In the case where both \( dx/dt = 0\) and \( dy/dt = 0 \) at the same point, we need to handle that case separately, since nothing can be concluded from \( dy/dx = 0/0 \), which is indeterminate.

Once you have found the slope, you can easily find the equation of a tangent line. Go to the Tangent Lines page for more information.

Practice

Unless otherwise instructed, find the equation of the tangent line to these parametric curves at the given points.

\( x=4t \), \( y=3t^2+2 \); \(t=2\).

Problem Statement

Find the equation of the tangent line to the parametric curve \( x=4t \), \( y=3t^2+2 \) at \(t=2\).

Solution

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\( x=t^2-5 \), \( y=t^2-4t \); \( (-4,-3) \)

Problem Statement

Find the equation of the tangent line to the parametric curve \( x=t^2-5 \), \( y=t^2-4t \) at \( (-4,-3) \).

Solution

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\( x=8\cos\theta \), \( y=6\sin\theta \); \( \theta = \pi/3 \)

Problem Statement

Find the equation of the tangent line to the parametric curve \( x=8\cos\theta \), \( y=6\sin\theta \) at \( \theta = \pi/3 \).

Solution

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\(x=2t^2+1\), \(y=3t^3+2\); \(t=1\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=2t^2+1\), \(y=3t^3+2\) at \(t=1\).

Solution

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\(x=3(t-\sin(t))\), \(y=3(1-\cos(t))\); \(t=\pi/2\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=3(t-\sin(t))\), \(y=3(1-\cos(t))\) at \(t=\pi/2\).

Solution

55 video

video by Krista King Math

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\(x=t\cos(t)\), \(y=t\sin(t)\); \(t=\pi\)

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=t\cos(t)\), \(y=t\sin(t)\) at \(t=\pi\).

Solution

467 video

video by Krista King Math

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\(x=3t^2-t\), \(y=\sqrt{t}\); \(t=4\).

Problem Statement

Find the equation of the tangent line to the parametric curve \(x=3t^2-t\), \(y=\sqrt{t}\) at \(t=4\).

Solution

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, find the equation of the tangent line to these parametric curves at the given points.

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