On this page we get started with applying calculus to parametric equations by covering derivatives of parametric equations and higher order derivatives.
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If you want a complete lecture on calculus involving parametric equations, we recommend this video.
video by Prof Leonard 

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 \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \]
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 \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \] 
We present two proofs here, a short informal version and a longer, more formal version.
Short Informal Proof
Given \( y(t) \) and \( x(t)\), we can write \( y = y(x(t)) \). We want to find \( dy/dx \).
Using the chain rule on \( y(x(t)) \), we have \( dy/dt = dy/dx \cdot dx/dt\).
Solving for \( dy/dx \) we have \( \displaystyle{ dy/dx = \frac{dy/dt}{dx/dt} }\) [qed]
Longer More Formal Proof
Given the two points \((x_1,y_1) = (X(t), Y(t))\) and \((x_2,y_2) = (X(t+\Delta t), Y(t+\Delta t)) \) and \(\Delta t > 0 \) on a smooth curve, let \( \Delta x = x_2  x_1 = X(t+\Delta t)  X(t) \) and \( \Delta y = y_2  y_1 = Y(t+\Delta t)  Y(t) \).
As \( \Delta t \to 0 \), we know that \(\Delta x \to 0 \) and we can write
\(\displaystyle{
\frac{dy}{dx} = \lim_{\Delta x \to 0}{\frac{\Delta y}{\Delta x}} =
\lim_{\Delta t \to 0}{\frac{Y(t+\Delta t)  Y(t)}{X(t+\Delta t)  X(t)}}
}\)
Now we can multiply the numerator and denominator by \(1/\Delta t\) and use limit laws to give us
\(
\begin{array}{rcl}
\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\lim_{\Delta t \to 0}{\frac{[Y(t+\Delta t)  Y(t)](1/\Delta t)}{[X(t+\Delta t)  X(t)](1/\Delta t)}} } \\
& = & \displaystyle{ \frac{\lim_{\Delta t \to 0}{[Y(t+\Delta t)  Y(t)]/\Delta t}}{\lim_{\Delta t \to 0}{[X(t+\Delta t)  X(t)]/\Delta t}} } \\
& = & \displaystyle{ \frac{dy/dt}{dx/dt} ~~~~~ \text{ [qed] } }
\end{array}
\)
Before going on, let's work some practice problems.
Practice  First Derivative
Unless otherwise instructed, calculate the first derivative of these parametric curves.
\(x=t+5\cos(t)\), \(y=3e^t\)
Problem Statement
Find \(dy/dx\) where \(y=f(x)\) is given by the parametric equations \(x=t+5\cos(t)\), \(y=3e^t\) .
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, we 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 

\(\displaystyle{ \frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] = \frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt} }\) 
Let's calculate some higher order derivatives of parametric equations.
Practice  Higher Order Derivatives
Unless otherwise instructed, calculate the first and second derivatives of these parametric equations.
Find \(\dfrac{d^2y}{dx^2}\) of \(y=f(x)\) where \(x=t^2+t\), \(y=2t1\)
Problem Statement
Find the second derivative \(\dfrac{d^2y}{dx^2}\) where \(y=f(x)\) is given by the parametric equations \(x=t^2+t\), \(y=2t1\) .
Solution
video by Krista King Math 

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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
video by PatrickJMT 

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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
video by PatrickJMT 

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\( x = 4t^3  t^2 + 7t \), \( y = t^4  6 \)
Problem Statement
Find \(dy/dx\) and \(d^2y/dx^2\) where \(y=f(x)\) is given by the parametric equations \( x = 4t^3  t^2 + 7t \), \( y = t^4  6 \) .
Solution
The solution can be found on this page.
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\( x = e^{7t} + 2 \), \( y = 6e^{2t} + e^{3t}  4t \)
Problem Statement
Find \(dy/dx\) and \(d^2y/dx^2\) where \(y=f(x)\) is given by the parametric equations \( x = e^{7t} + 2 \), \( y = 6e^{2t} + e^{3t}  4t \) .
Solution
The solution can be found on this page.
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