Mathematics II
Mr. Ghassan
Lecture 6
PARAMETRIC EQUATIONS AND POLAR COORDINATES
In previous study, we have studied curves as the graphs of functions or
equations involving the two variables x and y. We are now going to
introduce another way to describe a curve by expressing both coordinates
as functions of a third variable t.
𝒚 = 𝒇(𝒙)
𝒙 = 𝒇(𝒕) 𝒚 = 𝒈(𝒕)
[𝒙, 𝒚] = [𝒇 𝒕 , 𝒈(𝒕)]
The set of (x,y) produce Parametric Curve
https://www.geogebra.org/m/cAsHbXEU
Parametric Equations Plotter link
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Parametric Equations
42
Example 1: Sketch the curve defined by the parametric equations
𝒙 = 𝒕𝟐 𝒚=𝒕+𝟏
• Create t Table.
• Find x,y according to their
parametric equation.
• Each point has x,y,t values
• Identify geometrically the
curve, by finding y(x), or
x(y).
𝒙, 𝒚 = 𝒇 𝒕 , 𝒈 𝒕 = [(𝒕𝟐 ), (𝒕 + 𝟏)]
𝒙 = (𝒚 − 𝟏)𝟐
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Example 2: Graph the parametric equations
𝒙 = 𝒄𝒐𝒔(𝒕) 𝒚 = 𝒔𝒊𝒏(𝒕)
𝒘𝒉𝒆𝒓𝒆 𝟎 ≤ 𝒕 ≤ 𝟐𝞹
𝟐 𝟐
𝒄𝒐𝒔 𝒕 + 𝒔𝒊𝒏 𝒕 =𝟏
𝒙𝟐 + 𝒚𝟐 = 𝒄𝒐𝒔 𝒕 𝟐
+ 𝒔𝒊𝒏 𝒕 𝟐
=1
𝒙𝟐 + 𝒚𝟐 = 1
H.W.) Graph the Parametric Equations
𝒙 = 𝒂 · 𝒄𝒐𝒔(𝒕) 𝒚 = 𝒂 · 𝒔𝒊𝒏(𝒕) 𝒘𝒉𝒆𝒓𝒆 𝒂 > 𝟏
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Example 3: Sketch the curve defined by the parametric equations
Case 1 𝒙=𝒕 𝒚 = 𝒕𝟐 Case 2 𝒙= 𝒕 𝒚=𝒕
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Example 4: Sketch the curve defined by the parametric equations,
does this graph intersect x-axis, or y-axis?
𝟏
𝒙=𝒕+
𝒕
𝟏
𝒚=𝒕−
𝒕
𝒘𝒉𝒆𝒓𝒆 𝒕 > 𝟎
𝟏 𝟏 𝟐
𝒙 − 𝒚 =? 𝟏 𝒙 − 𝒚 = (𝒕 + ) − (𝒕 − ) 𝒙−𝒚=
𝒕 𝒕 𝒕
𝟏 𝟏
𝒙 + 𝒚 =? (𝟐) 𝒙 + 𝒚 = (𝒕 + ) + (𝒕 − ) 𝒙 + 𝒚 = 𝟐𝒕
𝒕 𝒕
𝟐
𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝟏 , 𝟐 (𝒙 − 𝒚)(𝒙 + 𝒚) = · 𝟐𝒕 𝒙𝟐 − 𝒚𝟐 = 𝟒
𝒕
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Problems: Finding Cartesian from Parametric Equations
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Tangents and Areas
𝒅𝒚 𝒚 = 𝒈(𝒕) 𝒅𝒚 = 𝒈`(𝒕) ∙ 𝒅𝒕
𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝒎𝒆𝒂𝒏 →
𝒅𝒙 𝒙 = 𝒇(𝒕) 𝒅𝒙 = 𝒇`(𝒕) ∙ 𝒅𝒕
𝒅𝒚
𝒅𝒚 𝒈` ∙ 𝒅𝒕 𝒅𝒚 𝒅𝒚 ∙ 𝒅𝒕 𝒅𝒕 𝒅𝟐 𝒚
= = = 𝐂𝐡𝐚𝐢𝐧 𝐑𝐮𝐥𝐞 =?
𝒅𝒙 𝒇` ∙ 𝒅𝒕 𝒅𝒙 𝒅𝒙 ∙ 𝒅𝒕 𝒅𝒙 𝒅𝒙 𝟐
𝒅𝒕
𝑨𝒓𝒆𝒂 𝒎𝒆𝒂𝒏 → 𝑨 = 𝒚 ∙ 𝒅𝒙 𝑨= 𝒈(𝒕) ∙ 𝒇`(𝒕) ∙ 𝒅𝒕
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Example 1:
𝑑𝑦 𝑑𝑦 𝑑𝑡 𝑠𝑒𝑐 2 𝑡 𝑠𝑒𝑐 𝑡
= = =
𝑑𝑥 𝑑𝑥 𝑑𝑡 𝑠𝑒𝑐 𝑡 𝑡𝑎𝑛(𝑡) 𝑡𝑎𝑛 𝑡
π
𝑑𝑦 𝑠𝑒𝑐 4 2 π π
= π = − <𝑡<
𝑑𝑥 𝑡𝑎𝑛 1 2 2
4
𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 → 𝑦 − 𝑏 = 𝑚(𝑥 − 𝑎)
𝑎= 2 ,𝑏 = 1 ,𝑚 = 2
𝑦 = 2𝑥 − 1
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Example 2:
Be Careful it is NOT
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Example 3: 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒂𝒓𝒆𝒂 𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒔:
𝒙 = 𝒄𝒐𝒔 𝒕 , 𝒚 = 𝒔𝒊𝒏 𝒕 , 𝟎 ≤ 𝒕 ≤ 𝟐𝝅
𝟏
𝑨= 𝒚 ∙ 𝒅𝒙 → 𝑨=𝟒 𝒚 ∙ 𝒅𝒙 → 𝒘𝒉𝒚?
𝟎
π
𝟐
𝑨=𝟒 𝒔𝒊𝒏(𝒕) ∙ −𝒔𝒊𝒏(𝒕) ∙ 𝒅𝒕
𝟎
π π
𝟐 𝟐𝟏 − 𝒄𝒐𝒔(𝟐𝒕)
𝑨 = −𝟒 𝒔𝒊𝒏𝟐 𝒕 ∙ 𝒅𝒕 → −𝟒 ∙ 𝒅𝒕
𝟎 𝟎 𝟐
π
𝟐
𝟐 (𝒄𝒐𝒔 𝟐𝒕 − 𝟏) ∙ 𝒅𝒕 → ???
𝟎
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Example 4:
t x y 𝒙 = 𝒄𝒐𝒔𝟐 𝒕 , 𝒚 = 𝒔𝒊𝒏𝟐 𝒕 , 𝟎 ≤ 𝒕 ≤ 𝟐𝝅
0
π 𝒙 = 𝒄𝒐𝒔𝟐 𝒕
𝟏
𝟒 𝒚 = 𝒔𝒊𝒏𝟐 𝒕
𝟐𝝅 𝟎 ≤ 𝒕 ≤ 𝟐𝝅 𝑨= 𝒚 ∙ 𝒅𝒙 → 𝑨= 𝒚 ∙ 𝒅𝒙
𝟎
𝟒
𝟑π π
𝟒 𝟐
𝟒π 𝑨= 𝒔𝒊𝒏𝟐 (𝒕) ∙ 𝟐𝒄𝒐𝒔(𝒕) ∙ −𝒔𝒊𝒏(𝒕) ∙ 𝒅𝒕
𝟎
𝟒
𝟓π π
𝟐
𝟒 𝑨 = −𝟐 𝒔𝒊𝒏𝟑 (𝒕) ∙ 𝒄𝒐𝒔(𝒕) ∙ 𝒅𝒕
𝟔π 𝟎
𝟒
𝟕π 𝑨 =? ? ?
𝟒
𝟖π
𝟒 SIE dept, Mathematics 2019-2020, Mr. Ghassan 52
Length of a Parametrically Defined Curve
𝒙=𝒇 𝒕 ,𝒚 = 𝒈 𝒕 ,𝒂 ≤ 𝒕 ≤ 𝒃
𝑳𝒌 = 𝒅𝒙𝟐 + 𝒅𝒚𝟐
𝒃
𝑳= 𝒅𝒙𝟐 + 𝒅𝒚𝟐
𝒂
𝒃
𝑳= 𝒇` 𝒕 𝟐 + 𝒈` 𝒕 𝟐 ∙ 𝒅𝒕
𝒂
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Example 1: Using the definition, find the length of the circle of radius r defined
parametrically by
𝒙 = 𝒓 ∙ 𝒄𝒐𝒔 𝒕 , 𝒚 = 𝒓 ∙ 𝒔𝒊𝒏 𝒕 , 𝟎 ≤ 𝒕 ≤ 𝟐𝝅
𝑟2
𝒅𝒙 = 𝒓 ∙ −𝒔𝒊𝒏 𝒕 ∙ 𝒅𝒕 , 𝒅𝒚 = 𝒓 ∙ 𝒄𝒐𝒔 𝒕 ∙ 𝒅𝒕
𝒃
𝑳= 𝒇` 𝒕 𝟐 + 𝒈` 𝒕 𝟐 ∙ 𝒅𝒕
𝒂 (𝑟, 0)
𝟐𝝅
𝑳= 𝒓 − 𝒔𝒊𝒏 𝒕 𝒅𝒕 𝟐 + 𝒓𝒄𝒐𝒔 𝒕 𝒅𝒕 𝟐
𝟎
𝟐𝝅
𝟐 𝟐
𝑳= 𝒓 𝒔𝒊𝒏 𝒕 + 𝒄𝒐𝒔 𝒕 𝒅𝒕 𝑳 =? ? ?
𝟎
SIE dept, Mathematics 2019-2020, Mr. Ghassan 54
Example 2: Find the length of astroid function
𝒙 = 𝒄𝒐𝒔𝟑 𝒕 , 𝒚 = 𝒔𝒊𝒏𝟑 𝒕 , 𝟎 ≤ 𝒕 ≤ 𝟐𝝅
𝒅𝒙 = −𝟑𝒄𝒐𝒔𝟐 𝒕 𝒔𝒊𝒏 𝒕 𝒅𝒕 , 𝒅𝒚 = 𝟑𝒔𝒊𝒏𝟐 𝒕 𝒄𝒐𝒔(𝒕)𝒅𝒕
𝒃
𝑳= 𝒇` 𝒕 𝟐 + 𝒈` 𝒕 𝟐 ∙ 𝒅𝒕
𝒂
𝟐𝝅
𝑳= 𝟗𝒄𝒐𝒔𝟒 𝒕 𝒔𝒊𝒏𝟐 (𝒕) + 𝟗𝒔𝒊𝒏𝟒 𝒕 𝒄𝒐𝒔𝟐 (𝒕) ∙ 𝒅𝒕
𝟎
𝟐𝝅
𝑳= 𝟗𝒄𝒐𝒔𝟐 𝒕 𝒔𝒊𝒏𝟐 (𝒕)(𝒄𝒐𝒔𝟐 (𝒕) + 𝒔𝒊𝒏𝟐 𝒕 ) ∙ 𝒅𝒕
𝟎
𝝅
𝟐
𝑳=𝟒 𝟑𝒄𝒐𝒔 𝒕 𝒔𝒊𝒏(𝒕) 𝒅𝒕 𝑳 =? ? ?
𝟎
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Problems:
Find the Slope
Find the Area
Find the Length
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