Parametric Equations Calculus of Parametric Equations
Parametric Curves
Institute of Mathematics, University of the Philippines Diliman
Mathematics 54–Elementary Analysis 2
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Parametric Equations Calculus of Parametric Equations
Introduction
Consider a particle moving along the curve below
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Parametric Equations Calculus of Parametric Equations
Introduction
The curve cannot be expressed as the graph of an equation of the form y = f (x).
Sometimes, it’s impossible to find an equation relating the x− and y−coordinatesof the particle.
But we can express the x− and the y−component of the particle’s position asfunctions of time t, i.e.
x = x(t)
y = y(t)
In this process, we say that we are parametrizing the curve above.
The equations defining x and y as functions of t are called parametric equations,and t is called a parameter.
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Parametric Equations Calculus of Parametric Equations
Parametric Curves
Examples.
Sketch the curve defined by C : x = t, y = t2, where t ∈R.
t → (x,y)−2 → (−2,4)−1 → (−1,1)0 → (0,0)1 → (1,1)2 → (2,4)3 → (3,9)
Plotting these points, we have
Indeed, if we relate x and y directly, eliminating t, we have y = x2.
We have just obtained a cartesian equivalent of the parametric curve.
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Parametric Equations Calculus of Parametric Equations
Examples
Example.
Sketch the curve defined by x = cos t and y = sin t, where 0 É t É 2π.
t → (x,y)0 → (1,0)π4 →
(p2
2 ,p
22
)π2 → (0,1)π → (−1,0)3π2 → (0,−1)
2π → (1,0)
Indeed, if we relate x and y directly, eliminating t, we have x2 +y2 = 1.
The direction of trace is known as the direction of increasing parameter. In thiscase, it’s counterclockwise.
The point corresponding to t = 0 is the initial point.The point corresponding to t = 2π is the terminal point.
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Parametric Equations Calculus of Parametric Equations
Examples
Parametrization of circles...
x = sin t
y = cos t
0 É t É 2π
’clockwise’ circle
x = 2cos t
y = 2sin t
0 É t É 2π
’counterclockwise’ circle of radius2 units
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Parametric Equations Calculus of Parametric Equations
Examples
x = h+ r cos t
y = k+ r sin t
0 É t É 2π
circle of radius r centered at (h,k)
x = cos2t
y = sin2t
0 É t Éπ
’faster’ circle...
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Parametric Equations Calculus of Parametric Equations
Interesting Examples
x = sin2t
y = sin3t
Lissajous curves
x = cos t +0.5cos7t +(
1
3
)sin17t
y = sin t +0.5sin7t +(
1
3
)cos17t
0 É t É 2π
(?)
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Parametric Equations Calculus of Parametric Equations
Examples
Consider the curvedefined by
x = sec t
y = tan t
−π2É t É π
2.
Note that
x2 = sec2 t
y2 = tan2 t
Thus,
x2 −y2 = 1
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Parametric Equations Calculus of Parametric Equations
Exercises
1 Describe the curve given by the parametric equations1 x = t2 and y = t4.2 x = sin t and y = sin2 t.
2 What can be said about the parametric equations1 x = cos t, y =−sin t2 x = cos2t, y =−sin2t
3 Describe the curve given by the parametric equations x = 2cos t andy = 3sin t.
4 Eliminate the parameter t, i.e., transform the following into theircorresponding cartesian form
1 x = 3− t, y = 2t −3
2 x = t2, y = 1−2t
3 x =pt +1, y = 6− t
4 x = t2, y = t3
5 x = et , y = e−t
6 x = cos t, y = cos2t
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Parametric Equations Calculus of Parametric Equations
Slope of Parametric Curves
Smooth Curve
A parametric curve C : x = f (t),y = g(t) is said to be smooth if f and g aredifferentiable and no value of t satisfies f ′(t) = g′(t) = 0.
If C is a smooth curve, then its slope dydx is given by
dy
dx=
dy
dtdx
dt
Also, the concavity d2ydx2 is given by
d2y
dx2=
d(
dydx
)dtdx
dt
6=d2y
dt2
d2x
dt2
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Parametric Equations Calculus of Parametric Equations
Slope of Parametric Curves
Example.
Find the slope of the astroid having the equation x = cos3 t, y = sin3 t, where t = 3π4 .
Solution:dy
dx= dy/dt
dx/dt= 3sin2 t cos t
−3cos2 t sin t=− tan t
So,dy
dx
∣∣∣t= 3π
4
=− tan3π
4= 1
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Parametric Equations Calculus of Parametric Equations
Tangent Lines to Parametric Curves
Example.
Find the equation of the tangent line to the right branch of the hyperbolax = sec t,y = tan t, where −π
2 < t < π2 , at the point (
p2,1).
Solution: Note that the point (p
2,1) corresponds to t = π4 . Also,
dy
dx= dy/dt
dx/dt= sec2 t
sec t tan t= csc t
Thus,
dy
dx
∣∣∣(p
2,1)= csc t
∣∣∣t=π/4
=p2
Hence, the equation is
y−1 =p2(x−p
2)
y =p2x−1
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Parametric Equations Calculus of Parametric Equations
Concavity
Example.
Find d2ydx2 if x = t − t2,y = t − t3.
Solution: y ′ = dy
dx= 1−3t2
1−2tThus,
d2y
dx2=
d(
1−3t21−2t
)dtdxdt
=(1−2t)(−6t)−(1−3t2)(−2)
(1−2t)2
1−2t
= 6t2 −6t +2
(1−2t)3
t < 12 t > 1
2
Caution :d2y
dt2
d2xdt2
= −6t1 =−6t 6= d2y
dx2 .
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Parametric Equations Calculus of Parametric Equations
Length of a Parametric Curve
Recall that the length of arc of the smooth curve y = F(x), from t = a to t = b, isgiven by
L =∫ b
a
√1+
(dy
dx
)2dx.
Such a curve can be parametrized as x = t, y = F(t). Hence,
√1+
(dy
dx
)2dx =
√√√√√1+ dy
dtdxdt
2dx
dt·dt =
√(dx
dt
)2+
(dy
dt
)2dt.
For a general parametric curve, we prove later that the formula still works.
Length of Arc
Let C : x = x(t),y = y(t),a É t É b, be a smooth curve traced exactly once. Then thelength of C is given by
L =∫ b
a
√(dx
dt
)2+
(dy
dt
)2dt
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Parametric Equations Calculus of Parametric Equations
Arc Length
Example.
Find the length of thecycloid
x = 4(t − sin t)
y = 4(1−cos t)
0 É t É 2π
dx
dt= 4(1−cos t) and
dy
dt= 4sin t
L =∫ 2π
0
√16(1−cos t)2 +16sin2 t dt
=∫ 2π
0
√16−32cos t +16cos2 t +16sin2 t dt
=∫ 2π
0
p32−32cos t dt =
∫ 2π
04√
2(1−cos t)dt
=∫ 2π
04
√4sin2
(t
2
)dt
=∫ 2π
08sin
(t
2
)dt
= −16cos
(t
2
)∣∣∣2π
0
= 32
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Parametric Equations Calculus of Parametric Equations
Arc Length
Example.
Find the length of theastroid
x = cos3 t
y = sin3 t
0 É t É 2π
L = 8∫ π/4
0
√(−3cos2 t sin t)2 + (3sin2 t cos t)2 dt
= 12∫ π/4
02cos t sin t dt
= 12∫ π/4
0sin2t dt
= 6[−cos2t]π/40
= 6
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Parametric Equations Calculus of Parametric Equations
Exercises
1 Findd2y
dx2of the curve with parametric equation x = t2 and y = t3 −3t.
2 Find the length of the curve given by x = et cos t and y = et sin t where0 ≤ t ≤π.
3 Set-up the iterated integral that will give the distance travalled by a particle
with position (x,y) as time t varies where x = t
1+ tand y = ln(1+ t), with
t ∈ [0,2].
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