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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. 
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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.
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 