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Parametric Equations and Calculus 

On this page we cover the most common calculus problems using parametric equations. The two main topics are differentiation and integration. 
This table lists the topics on this page with links to jump to the topic you are interested in.
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.
PatrickJMT  Parametric Differentiation  
Theorem: Parametric Derivative  

On a smooth curve given by the equations \( x=X(t) \) and \(y = Y(t)\),  
Parametric Derivative ProofParametric 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 1 

Find \(dy/dx\) of the parametric curve \(x=t+5\cos(t)\), \(y=3e^t\). 
solution 
Practice 2 

Find the derivative of the parametric curve \(x=t\sin(t)\), \(y=t^2+t\). 
solution 
Practice 3 

Find \(dy/dx\) of the parametric curve \(x=4t+1\), \(y=t^2+2t\). 
solution 
Differentiation  Slope and Tangent Lines 

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 above theorem.
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 4 

Find the equation of the tangent line to the parametric curve \(x=2t^2+1\), \(y=3t^3+2\) at \(t=1\). 
solution 
Practice 5 

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 
Practice 6 

Find the equation of the tangent line to the parametric curve \(x=t\cos(t)\), \(y=t\sin(t)\) at \(t=\pi\). 
solution 
Practice 7 

Find the equation of the tangent line to the parametric curve \(x=3t^2t\), \(y=\sqrt{t}\) at \(t=4\). 
solution 
Differentiation  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 \( \dot{x} = dx/dt \), 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 

\(\displaystyle{
\frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] =
\frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt}
}\)
Before we go on, let's work some practice problems with what we have learned so far.
Basic Problems 
Practice 8 

Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t1\). 
solution 
Intermediate Problems 
Practice 9 

Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\). 
solution 
Practice 10 

Find the second derivative of the parametric curve \(x=tt^3\), \(y= 2t+5\). 
solution 
Practice 11  

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

The area under a smooth curve defined parametrically as \( x = X(t) \) and \( y = Y(t) \) from \( t=t_0\) to \( t=t_1\) can be calculated using the integral
\(\displaystyle{s = \int_{t_0}^{t_1}{Y(t)X'(t) ~ dt}}\)
where \( X'(t) = dX/dt \).
You may also find this written in a shorthand form as
\(\displaystyle{s = \int_{t_0}^{t_1}{y~dx}}\)
In this notation, \( dx = (dx/dt) dt \).
Practice 13 

Find the area under the curve \(x=1+e^t\), \(y=tt^2\) and above the xaxis. 
solution 
Practice 14  

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t1/t\), \(y=t+1/t\).  
answer 
solution 
Practice 15 

Calculate the area under one arc of the parametric curve \(x=r\thetad\sin\theta\), \(y=rd\cos\theta\). 
solution 
Integration  Arc Length 

When a smooth curve is defined parametrically as \( x=X(t) \) and \( y = Y(t) \), the arc length between the points \( t=t_0 \) and \( t = t_1 \) can be calculated using the integral
\(\displaystyle{ s = \int_{t_0}^{t_1}{\sqrt{[X'(t)]^2 + [Y'(t)]^2} dt}}\)
Notice we use a small s here to represent the arc length. This is the standard symbol you will see in many textbooks. We reserve a capital S to represent surface area.
Here is a quick video clip going over this equation in more detail.
PatrickJMT  Arc Length  
Basic Problems 
Practice 16  

Compute the arc length of the parametric curve \(x(t)=3t^29\), \(y(t)=t^33t\), \(0\leq t\leq 3\).  
answer 
solution 
Practice 17  

Calculate the arc length of the parametric curve \(x(t)=t\sin(t)\) \(y(t)=t\cos(t)\), \(0\leq t\leq1\).  
answer 
solution 
Practice 18  

Calculate the arc length of the parametric curve \(x=1+3t^2\), \(y=4+2t^3\), \(0\leq t\leq1\).  
answer 
solution 
Practice 19  

Calculate the arc length of the parametric curve \(x=\abs{6t}\), \(y=t\), \(0 \leq t \leq 3\).  
answer 
solution 
Intermediate Problems 
Practice 20  

Find the arc length of the parametric curve \(x=2t\), \(y=(2/3)t^{3/2}\) for \(5\leq t\leq12\).  
answer 
solution 
Practice 21  

Calculate the arc length of the parametric curve \(x=t^3\), \(y=t^2\) from \((0,0)\) to \((8,4)\).  
answer 
solution 
Practice 22  

Calculate the arc length of the curve \(x(\theta)=a\cos^3(\theta)\), \(y(\theta)=a\sin^3(\theta)\), \(0\leq\theta\leq2\pi\).  
answer 
solution 
Practice 23  

Calculate the arc length of the parametric curve \(x=e^t+e^{t}\), \(y=52t\), \(0\leq t\leq3\).  
answer 
solution 
Integration  Surface Area 

To calculate the surface area defined by revolving a parametric curve defined as \( x=X(t)\) and \(y=Y(t)\) from \(t=t_0\) to \(t=t_1\) about the xaxis, we use this integral.
\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{ Y(t) \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~ dt}}\)
If you compare this integral to the equation for arc length (in the prevous section) you will see the common factor
\( \sqrt{[X'(t)]^2 + [Y'(t)]^2} \)
Some textbooks write the surface integral differently, taking this into account. You may see it written as
\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{Y(t) ~ ds}}\)
where \( ds = \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~dt \)
When rotating about the yaxis, the integral we use is
\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{X(t) ~ ds}}\)
Notice we use a capital S to represent surface area. This is the standard symbol you will see in many textbooks. We reserve a lowercase s to represent arc length.
Practice 25  

Find the surface area of revolution of the parametric curve \(x=3t^2\), \(y=2t^3\), \(0\leq t\leq5\) rotated about the yaxis.  
answer 
solution 
Integration  Volume 

If we have a parametric curve defined as \( x = X(t) \) and \( y = Y(t) \), we can determine the volume of the solid object defined by revolving this curve about an axis. We will limit our curve from \( t=t_0 \) to \( t=t_1 \).
Revolution About the xaxis  Revolution About the yaxis  

When revolved about the xaxis, the integral is 
Similar to the xaxis integral, we have 