## 17Calculus Parametric Equations - Slope & Tangent Lines

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

### 3501 video

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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

### 3502 video

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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

### 3503 video

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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

### 54 video

video by Krista King Math

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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

### 1371 video

video by PatrickJMT

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### parametric calculus 17calculus youtube playlist

You CAN Ace Calculus

 Wikipedia - Parametric Derivative Wikipedia - Arc Length Pauls Online Notes - Arc Length with Parametric Equations Wikipedia - Solid of Revolution

### 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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