42 tangent and arc length in polar coordinates
Transcript of 42 tangent and arc length in polar coordinates
Tangent and Arc Length in Polar Equations
Tangent and Arc Length in Polar Equations
Given the polar function r = f(), it may be realized as
the parametric equations:
x() = f()cos()
y() = f()sin().
Tangent and Arc Length in Polar Equations
Given the parametric equations
x = x()
y = y()
the dervative . dy
dx =dy/d
dx/d
Given the polar function r = f(), it may be realized as
the parametric equations:
x() = f()cos()
y() = f()sin().
Tangent and Arc Length in Polar Equations
Given the parametric equations
x = x()
y = y()
the dervative .
Hence for r = f(), the slope at the point (r=f(), ) isdy
dx =dy/d
dx/d=
d(f()cos())/d
d(f()sin())/d
dy
dx =dy/d
dx/d
Given the polar function r = f(), it may be realized as
the parametric equations:
x() = f()cos()
y() = f()sin().
Tangent and Arc Length in Polar Equations
Given the parametric equations
x = x()
y = y()
the dervative .
Given the polar function r = f(), it may be realized as
the parametric equations:
x() = f()cos()
y() = f()sin().
Hence for r = f(), the slope at the point (r=f(), ) isdy
dx =dy/d
dx/d=
d(f()cos())/d
d(f()sin())/d=
f '()sin() + f()cos()
f '()cos() – f()sin()
dy
dx =dy/d
dx/d
Slopes in Polar Equations
Example: a. Find the slope at r = cos(2) at = /6.
Example: a. Find the slope at r = cos(2) at = /6.
(1/2,/6)
Slopes in Polar Equations
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(),
(1/2,/6)
Slopes in Polar Equations
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(), hence
dy
dx =dy/d
dx/d=
(1/2,/6)
Slopes in Polar Equations
-2sin(2)sin() + cos(2)cos()
(1/2,/6)
Slopes in Polar Equations
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(), hence
dy
dx =dy/d
dx/d=
=-2sin(2)cos() – cos(2)sin()
(1/2,/6)
Slopes in Polar Equations
-2sin(2)sin() + cos(2)cos()
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(), hence
dy
dx =dy/d
dx/d=
at = /6,dy
dx=
-2sin(/3)sin(/6) + cos(/3)cos(/6)
-2sin(/3)cos(/6) – cos(/3)sin(/6)
(1/2,/6)
Slopes in Polar Equations
=-2sin(2)cos() – cos(2)sin()
-2sin(2)sin() + cos(2)cos()
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(), hence
dy
dx =dy/d
dx/d=
=-3 * ½ + ½ * 3/2
-3* 3/2 – ½ * ½ (1/2,/6)
Slopes in Polar Equations
at = /6,dy
dx=
-2sin(/3)sin(/6) + cos(/3)cos(/6)
-2sin(/3)cos(/6) – cos(/3)sin(/6)
=-2sin(2)cos() – cos(2)sin()
-2sin(2)sin() + cos(2)cos()
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(), hence
dy
dx =dy/d
dx/d=
=-3 * ½ + ½ * 3/2
-3* 3/2 – ½ * ½
=-23 + 3
-6 – 1
=3 7
(1/2,/6)
Slopes in Polar Equations
at = /6,dy
dx=
-2sin(/3)sin(/6) + cos(/3)cos(/6)
-2sin(/3)cos(/6) – cos(/3)sin(/6)
=-2sin(2)cos() – cos(2)sin()
-2sin(2)sin() + cos(2)cos()
Example: a. Find the slope at r = cos(2) at = /6.
We have x() = cos(2)cos() and y() = cos(2)sin(), hence
dy
dx =dy/d
dx/d=
Example: b. Find where the slope is 0 for r = 1+ cos().
Slopes in Polar Equations
Example: b. Find where the slope is 0 for r = 1+ cos().
Slopes in Polar Equations
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
Slopes in Polar Equations
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Slopes in Polar Equations
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
Slopes in Polar Equations
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
-S2 + C + C2 = 0
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
-S2 + C + C2 = 0
-(1 – C2) + C + C2 = 0
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
d-S2 + C + C2 = 0
-(1 – C2) + C + C2 = 0
2C2 + C – 1 = 0
(2C – 1)(C + 1) = 0
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
-S2 + C + C2 = 0
-(1 – C2) + C + C2 = 0
2C2 + C – 1 = 0
(2C – 1)(C + 1) = 0
cos() = ½ = ±/3
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
-S2 + C + C2 = 0
-(1 – C2) + C + C2 = 0
2C2 + C – 1 = 0
(2C – 1)(C + 1) = 0
cos() = ½ = ±/3
cos() = -1 = -π
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
d-S2 + C + C2 = 0
-(1 – C2) + C + C2 = 0
2C2 + C – 1 = 0
(2C – 1)(C + 1) = 0
cos() = ½ = ±/3
cos() = -1 = 3/2
However, 1+cos() is not
differentiable at -π since
dx/d = 0. Hence = ±/3.
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
d-S2 + C + C2 = 0
-(1 – C2) + C + C2 = 0
2C2 + C – 1 = 0
(2C – 1)(C + 1) = 0
cos() = ½ = ±/3
cos() = -1 = 3/2
However, 1+cos() is not
differentiable at -π since
dx/d = 0. Hence = ±/3.
Slopes in Polar Equations
-sin()sin() +(1+ cos())cos() = 0dy
d=Set
dy
dx=
dy/d
dx/dWe want = 0, or
dy
d = 0.
Example: b. Find where the slope is 0 for r = 1+ cos().
We have x() = (1+cos())cos() and y() = (1+cos())sin().
(1+3/2, /3)
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
drrdddyddx
b
a
b
a
2222 ))('())/()/(
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
drrdddyddx
b
a
b
a
2222 ))('())/()/(
Example: Let r = 2sin(). Find the arc-length for = 0 to =2π.
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
drrdddyddx
b
a
b
a
2222 ))('())/()/(
Example: Let r = 2sin(). Find the arc-length for = 0 to =2π.
r =1
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
drrdddyddx
b
a
b
a
2222 ))('())/()/(
Example: Let r = 2sin(). Find the arc-length for = 0 to =2π.
r2 + (r')2 = (2sin())2 + (2cos())2
= 4sin2()+4cos()2 = 4. Hence the
arc-length is: r =1
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
drrdddyddx
b
a
b
a
2222 ))('())/()/(
Example: Let r = 2sin(). Find the arc-length for = 0 to =2π.
424))('(
2
0
2
0
2
0
22
dddrr
r =1
r2 + (r')2 = (2sin())2 + (2cos())2
= 4sin2()+4cos()2 = 4. Hence the
arc-length is:
Arc Length for Polar CurvesGiven the polar equation r = f() in parametric form
x = r*cos(), y = r*sin(). We may easily verify that
(dx/d)2+(dy/d)2 = r2 + (dr/d)2.
Hence the arc-length from =a to =b is:
drrdddyddx
b
a
b
a
2222 ))('())/()/(
Example: Let r = 2sin(). Find the arc-length for = 0 to =2π.
424))('(
2
0
2
0
2
0
22
dddrr
r =1
Note that we get the circumference of two circles since the
Graphs traced out two cicles as goes from 0 to 2π.
r2 + (r')2 = (2sin())2 + (2cos())2
= 4sin2()+4cos()2 = 4. Hence the
arc-length is:
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r = 1 – cos()
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
r = 1 – cos()
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
r = 1 – cos()
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c.
r = 1 – cos()
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
r = 1 – cos()
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
Use the double formula to remove the root:
r = 1 – cos()1 – cos(2A)
2sin2( ) =A or 1 – cos()2sin2 ( ) =
2
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
Use the double formula to remove the root:
r = 1 – cos()1 – cos(2A)
2sin2( ) =A or 1 – cos()2sin2 ( ) =
2
Use symmetry, we double the integral from 0 to π:
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
Use the double formula to remove the root:
ddrr
b
a
0
22 ))cos(1(22))('(
r = 1 – cos()1 – cos(2A)
2sin2( ) =A or 1 – cos()2sin2 ( ) =
2
Use symmetry, we double the integral from 0 to π:
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
Use the double formula to remove the root:
dddrr
b
a
0
2
2
0
22 )(sin2*22))cos(1(22))('(
r = 1 – cos()1 – cos(2A)
2sin2( ) =A or 1 – cos()2sin2 ( ) =
2
Use symmetry, we double the integral from 0 to π:
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
Use the double formula to remove the root:
d
dddrr
b
a
0
2
0
2
2
0
22
)sin(4
)(sin2*22))cos(1(22))('(
r = 1 – cos()1 – cos(2A)
2sin2( ) =A or 1 – cos()2sin2 ( ) =
2
Use symmetry, we double the integral from 0 to π:
Arc Length for Polar CurvesExample: Find the arc-length of one round of r = 1 – cos().
The graph is a cardiod.
r2 + (r')2 = (1 – cos())2 + (sin())2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate 2 – 2c as is.
Use the double formula to remove the root:
8)8(0|)cos(8)sin(4
)(sin2*22))cos(1(22))('(
02
0
2
0
2
2
0
22
d
dddrr
b
a
r = 1 – cos()1 – cos(2A)
2sin2( ) =A or 1 – cos()2sin2 ( ) =
2
Use symmetry, we double the integral from 0 to π: