10.1 review of parametric equations - math.tamu.edubaurispa/math152/152ch10.pdf · math 152, c...

Math 152, c©Benjamin Aurispa

10.1 Review of Parametric Equations

Recall that often, instead of representing a curve using just x and y (called a Cartesian equation), it is moreconvenient to define x and y using parametric equations using a parameter such as t. This means thatx and y are both separately defined as functions of this parameter, x = f(t) and y = g(t).

If you are given parametric equations, you can sometimes create a Cartesian equation by eliminating theparameter. How?

• If possible, solve one of the parametric equations for t and use substitution.

• If the parametric equations involve trig functions, use a trig identity.

Sometimes, there may be a restriction on the values of t or the values of x and y may have bounds you needto watch out for.

Example: Eliminate the parameter to find a Cartesian equation for the following curves, sketch a graph,and denote the direction that the curve is traced out as t increases.

• x = t+ 1, y = t3

• x = t2, y = t− 2, −2 ≤ t ≤ 4

• x =√t, y = 2− t

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• x = 2 cos θ, y = 2 sin θ

• x = 3 + sin t, y = 1 + cos t

• x = −4 cos θ, y = 5 sin θ, 0 ≤ θ ≤ π

• x = cos t, y = sec t, 0 ≤ t ≤ π3

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10.2 Arc Length and Surface Area with Parametric Curves

Given a parametric curve defined by x = f(t), y = g(t), the length of the curve from t = a to t = b is:

L =

∫ b

a

√[f ′(t)]2 + [g′(t)]2 dt or

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

dt

Examples:

• Find the length of the curve x = t2 + 4, y = t3 + 1, from the point (4, 1) to the point (8, 9).

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• Find the length of the curve x = 4et − 4t, y = 16et/2, 0 ≤ t ≤ 1.

Surface Area:The surface area of a solid of revolution where a curve x = x(t), y = y(t) with arc length ds is rotated aboutan axis is

S =

∫2πr ds =

∫2πr

√(dx

dt

)2

+

(dy

dt

)2

dt

If revolving about the x-axis, r = y and if revolving about the y-axis, r = x.

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Examples:

• Find the surface area of the solid obtained by rotating the curve x = 3t3 − t, y = 3t2, 0 ≤ t ≤ 1, aboutthe x-axis.

• Find the surface area of the solid formed by rotating the curve x = e3t + e−3t, y = 10 − 6t, 0 ≤ t ≤ 2,about the y-axis.

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Section 10.3 Polar Coordinates

The polar coordinate system is a new way of thinking about graphing curves. Instead of the “rectangular”coordinate system (x, y), polar coordinates are of the form (r, θ) where θ is the angle made with the positivex-axis and r is the distance from the origin.

Examples: Plot the following points given in polar coordinates (r, θ).

A. (2, π/3)

B. (3,−5π/6)

C. (−4, 3π/4)

To convert from polar to rectangular coordinates, we use the following:

x = r cos θ

y = r sin θ

Find the Cartesian coordinates of the polar points plotted above.

(a) (2, π/3)

(b) (3,−5π/6)

(c) (−4, 3π/4)

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Going from Cartesian to polar coordinates, we use the following facts:

r2 = x2 + y2

tan θ =y

x

However, the way we write a point in polar coordinates is NOT unique. Convert the following points withCartesian coordinates into polar coordinates in three different ways.

(i) r > 0, 0 ≤ θ ≤ 2π (ii) r < 0, 0 ≤ θ ≤ 2π (iii) r > 0, −2π ≤ θ ≤ 0

• (−1,√

3)

• (−3,−3)

Find a polar equation for the curve represented by the following Cartesian equations.

• x2 + y2 = 25

• x2 + y2 = 6y

• y = x2

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Find a Cartesian equation for the following polar curves.

• r = 2 cos θ

• r = 4 csc θ

Polar equations are usually written where r is a function of θ, i.e. as r = f(θ). Sketch graphs of the followingpolar equations.

• r = 3 • θ = π/6

Sketch the region 2 ≤ r < 5, 2π/3 ≤ θ < 7π/6

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• r = 3 cos θ

• r = −4 sin θ

Summary of Circles:

r = a r = a cos θ, a > 0 r = a cos θ, a < 0

r = a sin θ, a > 0 r = a sin θ, a < 0

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Cardioids are polar graphs of the form r = a± a cos θ or r = a± a sin θ.

• r = 3 + 3 sin θ

• r = 2− 2 cos θ

Summary of Cardioids

r = a+ a cos θ, a > 0 r = a− a cos θ, a > 0

r = a+ a sin θ, a > 0 r = a− a sin θ, a > 0

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Roses are polar graphs of the form r = a sin(kθ) or r = a cos(kθ).

• r = sin(2θ)

• r = cos(3θ)

Summary of Roses:

r = cos(kθ), k odd, k petals r = cos(kθ), k even, 2k petals

r = sin(kθ), k odd, k petals r = sin(kθ), k even, 2k petals

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10.4 Areas and Lengths in Polar Coordinates

Given a polar equation r = f(θ), the area bounded by the curve between θ = a and θ = b is given by

A =

∫ b

a

1

2r2 dθ =

∫ b

a

1

2[f(θ)]2 dθ

Find the area of the region bounded by r = e2θ within the sector 0 ≤ θ ≤ π/2.

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Find the area of the region bounded by the circle r = 3 cos θ within the sector π/6 ≤ θ ≤ π/3.

Find the area inside the cardioid r = 2 + 2 sin θ.

Find the area of one loop (petal) of the rose r = 3 sin(5θ).

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Set up an integral to find the area of one loop (petal) of the rose r = 2 cos(4θ).

The area of the region bounded by two polar curves r1 and r2 where r1 is the “inner” polar curve and r2 isthe “outer” polar curve is given by

A =

∫ b

a

1

2[r22 − r21] dθ

Find the area inside the circle r = 6 sin θ and oustide the circle r = 3.

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Find the area inside the circle r = 6 cos θ and outside the cardioid r = 2 + 2 cos θ.

The arc length of a polar curve r = f(θ), a ≤ θ ≤ b is given by

L =

∫ b

a

√r2 +

(dr

dθ

)2

dθ

Find the length of the polar curve r = θ2, 0 ≤ θ ≤ π/2.

Find the length of the polar curve r = e3θ, 0 ≤ θ ≤ π.

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10.5-10.6 Conic Sections in Cartesian and Polar Coordinates

There are three types of conic sections: parabolas, ellipses, and hyperbolas. (Keep in mind a circle is aspecial case of an ellipse.)

The general equation of a parabola with vertex (h, k) that opens up/down is given below. If a is positive,the parabola opens upward. If a is negative, the parabola opens downward.

y = a(x− h)2 + k

The general equation of a parabola with vertex (h, k) that opens right/left is given below. If a is positive,the parabola opens to the right. If a is negative, the parabola opens to the left.

x = a(y − k)2 + h

Find the vertex of the parabola and sketch a graph.

x− 3 = −2(y + 3)2

The general form of an ellipse centered at (h, k) is given by:

(x− h)2

a2+

(y − k)2

b2= 1

Sketch a graph of the ellipse 16x2 + 4y2 = 64 by first finding x and y-intercepts.

Find the center of the ellipse 3x2 + y2 + 18x− 10y = −49.

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The general equation of a hyperbola centered at the origin is

x2

a2− y2

b2= 1 or

y2

b2− x2

a2= 1

Sketch a graph of the hyperbolax2

36− y2

25= 1 by finding intercepts.

Sketch a graph of the hyperbola 9y2 − x2 = 81 by finding intercepts.

Identify the type of conic given by the following equations.

• x2 − 3x = 6y − y2

• x+ 3 = −y2 + y

• 5x2 + 6y = 3x+ 4y2

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Conic sections can also be represented in polar form. The general form of a conic in polar form is one ofthe following:

r =a

1± e cos(θ)or r =

a

1± e sin(θ)

The number e is called the eccentricity and determines which type of conic the polar equation represents.

• If e < 1, the conic is an ellipse.

• If e = 1, the conic is a parabola.

• If e > 1, the conic is a hyperbola.

Notes:• The number a is a constant that has geometric meaning (which we don’t discuss).• The ± and the cos(θ) vs. sin(θ) determine the general shape and properties of the conic. We won’t getinto that in this class, but you can see how these affect the shape by converting from a polar equation to aCartesian equation and completing the square.

Identify the conic represented by the polar equations below.

• r =2

3 + 4 cos(θ)

• r =4

5− 5 sin(θ)

• r =8

7− 3 cos(θ)

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10.1 review of parametric equations - math.tamu.edubaurispa/math152/152ch10.pdf · math 152, c...

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