17Calculus - Derivatives of Parametric Equations
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.
Prof Leonard - Calculus 2 Lecture 10.3: Calculus of Parametric Equations [1hr-34min-18secs]
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.
PatrickJMT - Parametric Differentiation [1min-7secs]
video by PatrickJMT |
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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 \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0 \] | ||
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 |
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Find \(dy/dx\) of the parametric curve \(x=t+5\cos(t)\), \(y=3e^t\).
Solution |
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1373 video
video by PatrickJMT |
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\(x=t\sin(t)\), \(y=t^2+t\)
Problem Statement |
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Find the derivative of the parametric curve \(x=t\sin(t)\), \(y=t^2+t\).
Solution |
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468 video
video by Krista King Math |
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\(x=4t+1\), \(y=t^2+2t\)
Problem Statement |
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Find \(dy/dx\) of the parametric curve \(x=4t+1\), \(y=t^2+2t\).
Solution |
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1372 video
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 |
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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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\(\displaystyle{ \frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] = \frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt} }\) |
Practice
Basic |
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Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t-1\).
Problem Statement |
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Find the second derivative of the parametric equations \(x=t^2+t\), \(y=2t-1\).
Solution |
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56 video
video by Krista King Math |
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Intermediate |
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Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).
Problem Statement |
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Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).
Solution |
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1374 video
video by PatrickJMT |
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Find the second derivative of the parametric curve \( x = t - t^3 \), \( y = 2t + 5 \).
Problem Statement |
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Find the second derivative of the parametric curve \( x = t - t^3 \), \( y = 2t + 5 \).
Solution |
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1375 video
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 |
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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.
Solution |
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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 |
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\((1,0)\), 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 |
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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.
Solution |
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\(dx/dt=(1/2)t^{-1/2}, dy/dt = 3\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3}{(1/2)t^{-1/2}} = 6\sqrt{t} }\)
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 |
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\(dy/dx=6\sqrt{t}, m=6, y=6x-4\), concave up
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Topics You Need To Understand For This Page
Related Topics and Links
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
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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