On this page we get started with applying calculus to parametric equations by covering derivatives of parametric equations and higher order derivatives.
If you want a complete lecture on calculus involving parametric equations, we recommend this video.
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Let's start out with a quick video clip giving us an introduction to finding the derivative \(dy/dx\) if the function \(y=f(x)\) is given in parametric equations \(x(t)\) and \(y(t)\). The theorem is given below.
video by PatrickJMT 

Theorem: Parametric Derivative  

On a smooth curve given by the equations \( x=X(t) \) and \(y = Y(t)\), the slope of the curve at the point \((x,y)\) is \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \]  
Parametric Derivative Proof
We present two proofs here, a short informal version and a longer, more formal version. 
Before we go on, let's practice using this theorem.
Practice
Unless otherwise instructed, determine \(dy/dx\) of the given parametric curves. Give your answers in exact, simplified form.
\(x=t+5\cos(t)\), \(y=3e^t\)
Problem Statement 

Find \(dy/dx\) of the parametric curve \(x=t+5\cos(t)\), \(y=3e^t\).
Solution 

video by PatrickJMT 

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\(x=t\sin(t)\), \(y=t^2+t\)
Problem Statement 

Find the derivative of the parametric curve \(x=t\sin(t)\), \(y=t^2+t\).
Solution 

video by Krista King Math 

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\(x=4t+1\), \(y=t^2+2t\)
Problem Statement 

Find \(dy/dx\) of the parametric curve \(x=4t+1\), \(y=t^2+2t\).
Solution 

video by PatrickJMT 

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Higher Order Derivatives
In order to determine concavity of a graph and other information, you will need higher order derivatives. We list two of them below from which you can extract a pattern.
Second Derivative 

\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[ \frac{dy}{dx} \right] = \frac{d\left[ dy/dx \right]/dt}{dx/dt} }\) 
You may also find the notation \( \dot{x} = dx/dt \) and \( \dot{y} = dy/dt \), in which case the above derivative can be written \(\displaystyle{ \frac{d^2y}{dx^2} = \frac{\dot{x} \ddot{y}  \dot{y} \ddot{x}}{\dot{x}^3} }\)
Third Derivative 

\(\displaystyle{ \frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] = \frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt} }\) 
Practice
Basic 

Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t1\).
Problem Statement 

Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t1\).
Solution 

video by Krista King Math 

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Intermediate 

Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).
Problem Statement 

Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).
Solution 

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Find the second derivative of the parametric curve \( x = t  t^3 \), \( y = 2t + 5 \).
Problem Statement 

Find the second derivative of the parametric curve \( x = t  t^3 \), \( y = 2t + 5 \).
Solution 

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Find the horizontal tangent points to the curve \(x=1t, y=t^2\) and determine the concavity at those points.
Problem Statement 

Find the horizontal tangent points to the curve \(x=1t, y=t^2\) and determine the concavity at those points.
Final Answer 

\((1,0)\), concave up
Problem Statement 

Find the horizontal tangent points to the curve \(x=1t, y=t^2\) and determine the concavity at those points.
Solution 

How do you know when you have a horizontal tangent? When the slope is zero.
\(dx/dt=1, dy/dt=2t\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{1} }\)
\(dy/dx = 0\) when \(2t=0 \to t=0\)
When \(t=0\), \((x,y) = (1,0)\).
To determine concavity at that point, we need to calculate the second derivative.
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{2}{1} = 2 }\)
Since the second derivative is positive everywhere, the curve is concave up.
Final Answer 

\((1,0)\), concave up
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For the parametric curve \(x=\sqrt{t}, y=3t1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slopeintercept form.
Problem Statement 

For the parametric curve \(x=\sqrt{t}, y=3t1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slopeintercept form.
Final Answer 

\(dy/dx=6\sqrt{t}, m=6, y=6x4\), concave up
Problem Statement 

For the parametric curve \(x=\sqrt{t}, y=3t1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slopeintercept form.
Solution 

\(dx/dt=(1/2)t^{1/2}, dy/dt = 3\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3}{(1/2)t^{1/2}} = 6\sqrt{t} }\)
slope at \(t=1, m=6\sqrt{1} = 6\)
for the equation of tangent line, we need the \((x,y)\) point at \(t=1\), so \((\sqrt{1},31)=(1,2)\)
\(yy_1=m(xx_1) \to y2=6(x1) \to y=6x4\)
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{6(1/2)t^{1/2}}{(1/2)t^{1/2}} = 6 }\)
Since the second derivative is positive at \(t=1\) (and everywhere else for that matter), the curve is concave up.
Final Answer 

\(dy/dx=6\sqrt{t}, m=6, y=6x4\), concave up
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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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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