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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Topics You Need To Understand For This Page |
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Related Topics and Links |
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If you want a complete lecture on calculus involving parametric equations, we recommend this video.
video by Prof Leonard |
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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 |
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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 |
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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
\(
\begin{array}{rcl}
\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\lim_{\Delta t \to 0}{\frac{[Y(t+\Delta t) - Y(t)](1/\Delta t)}{[X(t+\Delta t) - X(t)](1/\Delta t)}} } \\
& = & \displaystyle{ \frac{\lim_{\Delta t \to 0}{[Y(t+\Delta t) - Y(t)]/\Delta t}}{\lim_{\Delta t \to 0}{[X(t+\Delta t) - X(t)]/\Delta t}} } \\
& = & \displaystyle{ \frac{dy/dt}{dx/dt} ~~~~~ \text{ [qed] } }
\end{array}
\)
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 |
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\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[ \frac{dy}{dx} \right] = \frac{d\left[ dy/dx \right]/dt}{dx/dt} }\) |
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 |
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\(\displaystyle{ \frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] = \frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt} }\) |
Let's calculate some derivatives of parametric equations.
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 \(dy/dx\) 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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Find \(\dfrac{d^2y}{dx^2}\) of \(y=f(x)\) where \(x=t^2+t\), \(y=2t-1\)
Problem Statement
Find the second derivative \(\dfrac{d^2y}{dx^2}\) where \(y=f(x)\) is given by the parametric equations \(x=t^2+t\), \(y=2t-1\) .
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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Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) and determine the concavity at those points.
Problem Statement |
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Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) and determine the concavity at those points.
Final Answer |
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\((1,0)\), concave up
Problem Statement
Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) 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 \(x=\sqrt{t}, y=3t-1\), 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 slope-intercept form.
Problem Statement |
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For the parametric curve \(x=\sqrt{t}, y=3t-1\), 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 slope-intercept form.
Final Answer |
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\(dy/dx=6\sqrt{t}, m=6, y=6x-4\), concave up
Problem Statement
For the parametric curve \(x=\sqrt{t}, y=3t-1\), 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 slope-intercept 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 \((\sqrt{1},3-1)=(1,2)\)
\(y-y_1=m(x-x_1) \to y-2=6(x-1) \to y=6x-4\)
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{6(1/2)t^{-1/2}}{(1/2)t^{-1/2}} = 6 }\)
Since the second derivative is positive at \(t=1\) (and everywhere else for that matter), the curve is concave up.
Final Answer
\(dy/dx=6\sqrt{t}, m=6, y=6x-4\), concave up
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