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17Calculus - Converting Between Rectangular and Polar Coordinates

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On this page, we discuss converting between polar and rectangular (cartesian) coordinates in two main areas, converting points and converting equations.

Equations

As discussed on the main polar coordinates page, the two main equations we use to convert between polar and rectangular (cartesian) coordinates are

\( x = r \cos(\theta)\)

 

\( y = r \sin(\theta)\)

All other equations are derived from these two equations and the equation of a circle with radius \(r\), \(x^2+y^2=r^2\).

This short video clip explains really well how to use trigonometry and the Pythagorean Theorem to get these equations.

PatrickJMT - Converting Between Polar and Rectangular (Cartesian) Coordinates [50secs]

video by PatrickJMT

Rectangular → Polar Points

In this case, we are given a specific point \((x,y)\) and we want to find \(r\) and \(\theta\) that describes this point. This direction is a little more complicated because we need to pay special attention to the signs of \( x\) and y.

The equations we use are \( r = \sqrt{x^2+y^2} \) and \(\theta= \arctan(y/x)\). Let's discuss these equations in more detail. This discussion is critical for you to understand in order to correctly determine the polar coordinates.

The first equation looks easy but there is a hidden assumption that you need to be aware of. Looking at the graph on the right, you know from the Pythagorean Theorem that \( r^2 = x^2 + y^2 \). So far, so good. Now, to solve for \(r\) we take the square root of both sides.

\(\begin{array}{rcl} r^2 & = & x^2 + y^2 \\ \sqrt{r^2} & = & \pm \sqrt{x^2 + y^2} \\ r & = & \pm \sqrt{x^2 + y^2} \end{array}\)

In the second line above, we could have put the \(\pm\) sign on either side, but to be strictly true, we needed it. (Do you see why?)

Now, in the equation \(r=\sqrt{x^2+y^2}\) we dropped the \(\pm\) sign. What does that mean? That means we are assuming that \(r\) is always positive. So, when we convert from rectangular to polar coordinates, we will take \(r\) to be positive. This is a subtle point but you need to keep that in mind. (As a teacher, one of my favorite questions on homework or exams will be to ask what happens when \(r\) is negative. See the practice problems below for examples of this case.)

Okay, now for the equation \(\theta= \arctan(y/x)\). This takes special care. You cannot just divide y by x and plug that value into your calculator. You need to know the sign of both x and y, which will determine the quadrant of your answer and thus the angle \(\theta\). For discussion here, we want the angle to be in the interval \( (-\pi, \pi] \). Notice this is a half-open (or half-closed) interval and includes \(\pi\) but excludes \(-\pi\). Of course, you need to check with your instructor to see what they require.

The best way that we've found to determine this angle, can be found using these equations.

determining \(\theta\) for \(\tan(\theta)=y/x\)

\(\theta = \arctan(y/x)\)

\(x > 0\)

quadrant 1 or 4

\(\theta = \arctan(y/x) \) \(+\pi\)

\(x < 0, y \geq 0\)

quadrant 2

\(\theta = \arctan(y/x)\) \(-\pi\)

\(x < 0, y < 0\)

quadrant 3

\(\theta = +\pi/2\)

\(x = 0, y > 0\)

\(\theta = -\pi/2\)

\(x = 0, y < 0\)

\( \theta = 0 \)

\(x=0, y=0\)

[ source: wikipedia ]

Okay, time for some practice problems.

Practice

given \((x,y)=(0,2)\); calculate \((r,\theta)\)

Problem Statement

given \((x,y)=(0,2)\); calculate \((r,\theta)\)

Final Answer

\( (r,\theta) = (2,\pi/2) \)

Problem Statement

given \((x,y)=(0,2)\); calculate \((r,\theta)\)

Solution

753 video

video by PatrickJMT

Final Answer

\( (r,\theta) = (2,\pi/2) \)

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given \((x,y)=(-3,4)\); calculate \((r,\theta)\)

Problem Statement

given \((x,y)=(-3,4)\); calculate \((r,\theta)\)

Final Answer

\( (r,\theta) = (5,2.215) \)

Problem Statement

given \((x,y)=(-3,4)\); calculate \((r,\theta)\)

Solution

754 video

video by PatrickJMT

Final Answer

\( (r,\theta) = (5,2.215) \)

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given \((x,y)=(1,-5)\); calculate \((r,\theta)\)

Problem Statement

given \((x,y)=(1,-5)\); calculate \((r,\theta)\)

Final Answer

\( (r,\theta) = (\sqrt{26},-1.373) \)

Problem Statement

given \((x,y)=(1,-5)\); calculate \((r,\theta)\)

Solution

755 video

Final Answer

\( (r,\theta) = (\sqrt{26},-1.373) \)

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Rectangular → Polar Equations

In this case, we are given an equation \(f(x,y)=0\) and we need to find an equation \(g(r, \theta)=0\). This is pretty easy. We just substitute for x and y using these equations and simplify.

\( x = r \cos(\theta)\)

 

\( y = r \sin(\theta)\)

Practice

Basic

given \(x^2+y^2=1\); calculate \(g(r,\theta)=0\)

Problem Statement

given \(x^2+y^2=1\); calculate \(g(r,\theta)=0\)

Final Answer

\( r-1 = 0 \)

Problem Statement

given \(x^2+y^2=1\); calculate \(g(r,\theta)=0\)

Solution

762 video

Final Answer

\( r-1 = 0 \)

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given \(y=2x+1\); calculate \(g(r,\theta)=0\)

Problem Statement

given \(y=2x+1\); calculate \(g(r,\theta)=0\)

Final Answer

\(\displaystyle{ r + \frac{1}{2\cos(\theta)-\sin(\theta)} = 0 }\)

Problem Statement

given \(y=2x+1\); calculate \(g(r,\theta)=0\)

Solution

763 video

Final Answer

\(\displaystyle{ r + \frac{1}{2\cos(\theta)-\sin(\theta)} = 0 }\)

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given \(y=3/x\); calculate \(g(r,\theta)=0\)

Problem Statement

given \(y=3/x\); calculate \(g(r,\theta)=0\)

Final Answer

\(\displaystyle{ r^2 - \frac{3}{\cos(\theta)\sin(\theta)} = 0 }\)

Problem Statement

given \(y=3/x\); calculate \(g(r,\theta)=0\)

Solution

764 video

Final Answer

\(\displaystyle{ r^2 - \frac{3}{\cos(\theta)\sin(\theta)} = 0 }\)

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given \(y=x^2\); calculate \(g(r,\theta)=0\)

Problem Statement

given \(y=x^2\); calculate \(g(r,\theta)=0\)

Final Answer

\( r - \tan(\theta)\sec(\theta) = 0 \)

Problem Statement

given \(y=x^2\); calculate \(g(r,\theta)=0\)

Solution

771 video

Final Answer

\( r - \tan(\theta)\sec(\theta) = 0 \)

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given \(xy=1\); calculate \(g(r,\theta)=0\)

Problem Statement

given \(xy=1\); calculate \(g(r,\theta)=0\)

Final Answer

\( r^2 - \sec(\theta)\csc(\theta) = 0 \)

Problem Statement

given \(xy=1\); calculate \(g(r,\theta)=0\)

Solution

772 video

Final Answer

\( r^2 - \sec(\theta)\csc(\theta) = 0 \)

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Intermediate

For the two 3-dim curves, \(2z=x^2-y^2+2x\) and \(3z=4x^2+y^2-2x\), determine the projection of the intersection of these curves in the xy-plane and describe the projection in polar coordinates.

Problem Statement

For the two 3-dim curves, \(2z=x^2-y^2+2x\) and \(3z=4x^2+y^2-2x\), determine the projection of the intersection of these curves in the xy-plane and describe the projection in polar coordinates.

Hint

In the xy-plane, the z coordinate is zero.

Problem Statement

For the two 3-dim curves, \(2z=x^2-y^2+2x\) and \(3z=4x^2+y^2-2x\), determine the projection of the intersection of these curves in the xy-plane and describe the projection in polar coordinates.

Hint

In the xy-plane, the z coordinate is zero.

Solution

1764 video

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Polar → Rectangular Points

In this case, we are given specific \(r\) and \(\theta\) values and we want to find the point \((x,y)\). This is the easiest direction because we already have the equations in the form we need, i.e. \( x = r \cos(\theta)\) and \( y = r \sin(\theta)\).

Practice

Basic

calculate \((x,y)\); given \((r,\theta)=(4,\pi/3)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(4,\pi/3)\)

Final Answer

\( (x,y) = (2,2\sqrt{3}) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(4,\pi/3)\)

Solution

756 video

Final Answer

\( (x,y) = (2,2\sqrt{3}) \)

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calculate \((x,y)\); given \((r,\theta)=(2,\pi)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(2,\pi)\)

Final Answer

\( (x,y) = (-2,0) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(2,\pi)\)

Solution

757 video

Final Answer

\( (x,y) = (-2,0) \)

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calculate \((x,y)\); given \((r,\theta)=(1,\pi/4)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(1,\pi/4)\)

Final Answer

\( (\sqrt{2}/2, \sqrt{2}/2) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(1,\pi/4)\)

Solution

765 video

Final Answer

\( (\sqrt{2}/2, \sqrt{2}/2) \)

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calculate \((x,y)\); given \((r,\theta)=(-2,2\pi/3)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(-2,2\pi/3)\)

Final Answer

\( (1, -\sqrt{3}) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(-2,2\pi/3)\)

Solution

766 video

Final Answer

\( (1, -\sqrt{3}) \)

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calculate \((x,y)\); given \((r,\theta)=(1,-\pi/3)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(1,-\pi/3)\)

Final Answer

\( (1/2, -\sqrt{3}/2) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(1,-\pi/3)\)

Solution

767 video

Final Answer

\( (1/2, -\sqrt{3}/2) \)

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calculate \((x,y)\); given \((r,\theta)=(3,3\pi/2)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(3,3\pi/2)\)

Final Answer

\( (0, -3) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(3,3\pi/2)\)

Solution

768 video

Final Answer

\( (0, -3) \)

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calculate \((x,y)\); given \((r,\theta)=(2,-\pi/4)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(2,-\pi/4)\)

Final Answer

\( (\sqrt{2}, -\sqrt{2}) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(2,-\pi/4)\)

Solution

769 video

Final Answer

\( (\sqrt{2}, -\sqrt{2}) \)

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Intermediate

calculate \((x,y)\); given \((r,\theta)=(-7,-2\pi/3)\)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(-7,-2\pi/3)\)

Final Answer

\( (x,y) = (7/2, 7\sqrt{3}/2) \)

Problem Statement

calculate \((x,y)\); given \((r,\theta)=(-7,-2\pi/3)\)

Solution

758 video

Final Answer

\( (x,y) = (7/2, 7\sqrt{3}/2) \)

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Polar → Rectangular Equations

Equations - - Here, we are given a function \(g(r,\theta)=0\) and we need to find an equation \(f(x,y)=0\). One of the keys to solving these problems is to use the identity \( \cos^2(\theta) + \sin^2(\theta) = 1\). In terms of the above equations, we have
\(\begin{array}{rcl} x^2 + y^2 & = & [ r \cos(\theta) ]^2 + [ r \sin(\theta) ]^2 \\ & = & r^2[ \cos^2(\theta) + r \sin^2(\theta)] \\ & = & r^2 \end{array}\)

Practice

Basic

calculate \(f(x,y)=0\); given \(r=3\cos(\theta)\)

Problem Statement

calculate \(f(x,y)=0\); given \(r=3\cos(\theta)\)

Final Answer

\( x^2 - 3x + y^2 = 0 \)

Problem Statement

calculate \(f(x,y)=0\); given \(r=3\cos(\theta)\)

Solution

759 video

Final Answer

\( x^2 - 3x + y^2 = 0 \)

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calculate \(f(x,y)=0\); given \(r=-5\cos(\theta)\)

Problem Statement

calculate \(f(x,y)=0\); given \(r=-5\cos(\theta)\)

Final Answer

\( x^2 + 5x + y^2 = 0 \)

Problem Statement

calculate \(f(x,y)=0\); given \(r=-5\cos(\theta)\)

Solution

770 video

Final Answer

\( x^2 + 5x + y^2 = 0 \)

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Intermediate

calculate \(f(x,y)=0\); given \(r^2=-3\sec(\theta)\)

Problem Statement

calculate \(f(x,y)=0\); given \(r^2=-3\sec(\theta)\)

Final Answer

\( x^4 + x^2y^2 - 9 = 0 \)

Problem Statement

calculate \(f(x,y)=0\); given \(r^2=-3\sec(\theta)\)

Solution

760 video

Final Answer

\( x^4 + x^2y^2 - 9 = 0 \)

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calculate \(f(x,y)=0\); given \(r=3\sin(\theta)\tan(\theta)\)

Problem Statement

calculate \(f(x,y)=0\); given \(r=3\sin(\theta)\tan(\theta)\)

Final Answer

\( x^3 + xy^2 - 3y^2 = 0 \)

Problem Statement

calculate \(f(x,y)=0\); given \(r=3\sin(\theta)\tan(\theta)\)

Solution

761 video

Final Answer

\( x^3 + xy^2 - 3y^2 = 0 \)

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

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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