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Calculus with Parametric Equations December 4, 2019 1 / 34
Calculus with ParametricEquations
December 4, 2019
1 / 34
Calculus with Parametric equations
Let C be a parametric curve described by the parametric equationsx = f(t), y = g(t). The book and the notes evoke the Chain Rule to
computedy
dxassuming it exists! Here is an approach which only
needs information aboutdx
dtand
dy
dt.
From Calc I, the slope of the tangent line is the limit of the slopes of
the chords. The slope of the cord from a to t isy(t)− y(a)x(t)− x(a)
. Hence
the slope of the tangent line (i.e. the derivative) is
dy
dx
∣∣∣∣∣t=a
= limt→a
y(t)− y(a)x(t)− x(a)
Plugging in t = a always gives 00 if x(t) and y(t) are continuous so ifthey are differentiable we can use l’Hospital’s Rule to see
dy
dx
∣∣∣t=a
= limt→a
dydtdxdt
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If x′(t) and y′(t) are continuous and x′(a) 6= 0 this gives
dy
dx
∣∣∣t=a
=y′(a)
x′(a)=
dydt
∣∣t=a
dxdt
∣∣t=a
Assuming y′(t) and x′(t) are continuous, this formula computesdy
dxat
all points where x′(t) 6= 0.
dy
dx=
dydtdxdt
ifdx
dt6= 0
At points where x′(t) = 0 we can still try to compute limt→a
y′(t)
x′(t)using
algebraic simplification or perhaps an additional application ofl’Hopital’s Rule.
Ifdx
dt
∣∣∣t=a
= 0 and ifdy
dt
∣∣∣t=a6= 0 then the limit is often ±∞ and there
is a vertical tangent at(x(a), y(a)
).
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The value ofdy
dxgives gives the slope of a tangent to the curve at any
given point.
The curve has a horizontal tangent whendy
dx= 0.
Ifdx
dt
∣∣∣t=a6= 0 but dy
dt
∣∣∣t=a
= 0 then the curve has a horizontal tangent.
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Second derivative
The second derivatived2y
dx2can also be computed which will give
information about concavity and inflection points.
d2y
dx2=
d
dx
(dydx
)=
ddt (
dydx )dxdt
ifdx
dt6= 0
This is the way to compute if asked: computedy
dxas a function of t
and simplify as much as possible. Then compute
d2y
dx2=
ddt (
dydx )dxdt
You can also proceed as follows. Computedy
dxas a function of t.
Then compute the slope of the parametrized curve
(x(t),
dy
dx(t)
).
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There is a formula in terms of the first and second derivatives of xand y:
d2y
dx2=
ddt
(dydtdxdt
)dxdt
=x′′y′ − x′y′′
(x′)3
In particular it is never the case that you want to computey′′
x′′to
discuss concavity. We bring this up becausey′
x′does compute slope
but the “obvious” generalization to concavity is false.
Formula is not really worth trying to remember: the way to
computed2y
dx2if asked is to compute
dy
dxas a function of t and simplify
as much as possible. Then compute
d2y
dx2=
ddt (
dydx )dxdt
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Example 1
The next few slides study
x = t2 − 2t y = t3 − 3t.
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Example 1: continued
x = t2 − 2t y = t3 − 3t.
Find an equation of the tangent to the curve at t = −2. First we findthe derivative
dy
dx.
I We havedx
dt= 2t− 2 and dy
dt= 3t2 − 3.
I Thereforedy
dx=y′
x′=
3t2 − 32t− 2
when 2t− 2 6= 0.
I Since we are going to work with the derivative some it pays tosimplify. Both top and bottom vanish at t = 1 so
limt→a
3t2 − 32t− 2
=3
2(a+ 1) for all a. Hence
dy
dx=
3
2(t+ 1) for all t .
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Example 1: continued
x = t2 − 2t y = t3 − 3t.Find an equation of the tangent to the curve at t = −2.
I When t = −2, the corresponding point on the curve isP = (4 + 4,−8 + 6) = (8,−2).
I When t = −2, dydx
= −32
.
I The equation of the tangent line at the point P is(y + 2) = − 32 (x− 8).
The equation of the tangent line at the point P (t = −2) inslope-intercept form is
y = −32x+ 10 x = t2 − 2t y = t3 − 3t.
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Example 1: continued
Find the point on the parametric curve where the tangent ishorizontal. x = t2 − 2t y = t3 − 3t
I From above, we have thatdy
dx=
3
2(t+ 1).
Idy
dx= 0 if t+ 1 = 0 if t = −1.
I The graph has a horizontal tangent at t = −1.I The corresponding point on the curve is Q = (3, 2).
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Example 1: continued
It is clear from the graph that (3, 2) is a local maximum. You can
prove this by showing3
2(t+ 1) is positive for t > −1 and negative for
t < −1.Another proof is to compute
d2y
dx2=
d 32 (t+1)
dtdxdt
=32
2t− 2=
3
4(t− 1)
which is negative at t = −1 so we have a local maximum by theSecond Derivative test. Note the curve is concave down at the localmax. 11 / 34
Example 1: continued
y = t3 − 3tx = t2 − 2t
I limt→−∞
x =∞; limt→−∞
y = −∞
I limt→∞
x =∞; limt→∞
y =∞I No horizontal asymptotes.
I Since limt→a±
x = x(a) and limt→a±
y = y(a) there are no vertical
asymptotes.
I Sincedy
dx=
3
2(t+ 1) is never ±∞, there are no vertical tangents.
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Example 1: continued
y = t3 − 3tx = t2 − 2t
The point where the curve makes a sharp turn is interesting. It iscalled a cusp. It occurs when t = 1 so its coordinates are (−1,−2).
Sincedx
dt= 2t− 2 is negative for t < 1 and positive for t > 1, the
motion of a particle described by the parametrization is right to leftfor t < 1 and left to right for t > 1. Changes direction when t = 1.
Sincedy
dt= 3t2 − 3 is positive for t < −1 and t > 1 and negative for
t ∈ (−1, 1), the motion of a particle described by the parametrizationis down for t ∈ (−1, 1) and up otherwise. Changes direction whent = ±1.Hence the particle starts at (∞,−∞) and moves left until it hits thecusp. Then it moves up and to the right until it ends at (∞,∞). 13 / 34
Example 1: continued
y = t3 − 3tx = t2 − 2t
Sinced2y
dx2=
3
4(t− 1)the curve is concave down for t < 1 and concave
up for t > 1.
The slope at the cusp isdy
dx
∣∣∣∣∣t=1
=3
2(t+ 1)
∣∣∣t=1
= 3. The equation of
the tangent line at the cusp is y − (−2) = 3(x− (−1)
), or y = 3x+ 1.
The concavity switches at the cusp and there is a tangent line so thecusp is an inflection point.
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Example 1: continuedy = t3 − 3tx = t2 − 2t
dy
dx= 3
2(t + 1)
d2y
dx2=
3
4(t− 1)
This picture shows all the “interesting” features of this curve.
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Summary
I limt→±∞
x(t) and limt→±∞
y(t) long term future/past behavior.
I There is a horizontal asymptote if and only if a y limit is anumber.
I There is a vertical asymptote if x limit is a number and a y limitis ±∞.
I Points where y(t) are not defined may indicate verticalasymptotes.
I DerivativesI
dy
dxincreasing/decreasing; local max/min
Idx
dtleft/right velocity; possible direction change
Idy
dtup/down velocity; possible direction change
I Second derivative
Id2y
dx2concave up/down. Possible inflection points.
Id2x
dt2left/right acceleration
Id2y
dt2up/down acceleration
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Example 2
Consider the curve C defined by the parametric equations
x = t cos t y = t sin t − π 6 t 6 π
This is a spiral x2 + y2 = t2 but it has an interesting point where thecurve crosses itself. Because of the limited parameter interval, thegraph does not show the spiral structure.Since
(x(−t), y(−t)
)=(−x(t), y(t)
)the curve is symmetric about the
y-axis, so the crossing point must have x = 0 (as the graph appears toshow).
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Example 2: continuedy = t sin(t)x = t cos(t)
At x = 0, t cos(t) = 0 so t = 0 or cos(t) = 0. In the parameter
interval, cos(t) = 0 if and only if t = ±π2
. At t = 0, we are at (0, 0)
and if we are at (0, 0) t = 0 is the only solution so t = 0 is not thecrossing point.
At t = ±π2
, y =π
2so(
0,π
2
)is the crossing point.
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Example 2: continuedy = t sin(t)x = t cos(t)
Compute x′(t) = cos(t)− t sin(t) so x′(±π) = −1; x′(0) = 1 so theleft/right motion changes at x′(t) = 0; cos(t)− t sin(t) = 0; t = cot(t).There are two solutions in the parameter interval, t0 > 0 and −t0where t0 ≈ 0.860333589019380.
Compute y′(t) = sin(t) + t cos(t) so at t0 = cot(t0),y′(t0) = sin(t0) + cot(t0) cos(t0) = csc(t0) which is never 0 so thecurve has vertical tangents when t = ±t0. The points areapproximately (±0.5610963382, 0.6521846239).
At t = −π we are at (π, 0) so the particle moves left to right startingat (π, 0) at the right of the graph. It keeps moving left until we get to(x(−t0), y(−t0)
). Then it moves right until it gets to
(x(t0), y(t0)
)and then it moves right again until it gets to (−π, 0) at t = π.
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Example 2: continuedy = t sin(t)x = t cos(t)
Find the equations of both tangents to C at (0, π2 )dy
dx=y′
x′=
sin t+ t cos t
cos t− t sin t.
The calculation on the slides is wrong at this point as indicated bythe Woops! but I can’t fix it. Thanks to a sequence of errors the finalanswers on the slides are correct.
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Example 2: continuedy = t sin(t)x = t cos(t) dy
dx=
sin t + t cos t
cos t− t sin t
The first tangent occurs at t = −π2
so the slope there is
dy
dx
∣∣∣∣∣t=−π2
=−1−π2
=2
πand the slope-intercept tangent line equation is
y =2
πx+
π
2.
When t =π
2the slope is − 2
πand the slope-intercept tangent line
equation is
y = − 2πx+
π
2.
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Example 2: continued
y = t sin(t)
x = t cos(t) dy
dx=
sin t + t cos t
cos t− t sin t
Discussing concavity will be left to the interested reader.
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Area under a curve
Recall why the area under the curve y = F (x) where a 6 x 6 b andF (x) > 0 is given by ∫ b
a
F (x) dx
If the curve is given as x = g(t), y = f(t), t ∈ [α, β] then in the limitdx = ±x′(t) dt and ∫ β
α
y(t)x′(t) dt
is the area if a = x(α) and is minus the area if b = x(α).
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Area continued
With the same notation as the last slide, x = g(t), y = f(t), t ∈ [α, β]and assuming y(t) > 0 for t ∈ [α, β] the area A is given by
A =
∫ βα
y(t)x′(t) dt x(α) < x(β)∫ αβ
y(t)x′(t) dt x(β) < x(α)
Or in one formula
A =
∫ t of right-hand end pointt of left-hand end point
y(t)x′(t) dt
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Variations on area
Suppose you want the area to the right of the y-axis and left of agraph such as
∆yx
The ∆y is the height of the rectan-gle and the x is its length. Henceif the curve is given as x = g(t),y = f(t), t ∈ [α, β] then in the limitdy = ±y′(t) dt and∫ β
α
x(t)y′(t) dt
is the area if a = y(α) and is minus the area if b = y(α).Or
A =
∫ t of upper end pointt of lower end point
x(t)y′(t) dt
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The interpretation of the integrals on the last two slides as an areadepends on the curve lying above the x-axis (
∫yx′dt) or the the right
of the y-axis (∫xy′dt).
Both integrals make sense without this assumption and what you getin either case is the net or signed area. See Chapter 4.2 for moreinformation.
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Example
Find the area under the curve
x = 2 cos t y = 3 sin t 0 6 t 6π
2
I The graph of this curve is a quarter ellipse, starting at (2, 0) andmoving counterclockwise to the point (0, 3).
I From the formula, we get that
∫ π2
0
y(t)x′(t) dt is minus the area.
I
∫ π/20
3 sin t(2(− sin t))dt = −6∫ π/20
sin2 tdt =
−612
∫ π2
0
(1− cos(2t))dt = −3(t− sin(2t)
2
)∣∣∣π20
= −3π2
.
We got the negative of thearea because we started atthe right-hand end pointand ended at the left-handend point.
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Example: same problem
Find the area under the curve
x = 2 cos t y = 3 sin t 0 6 t 6π
2
I The graph of this curve is a quarter ellipse, starting at (2, 0) andmoving counterclockwise to the point (0, 3).
I From the formula, we get that
∫ π2
0
x(t)y′(t) dt is the area.
I
∫ π/20
2 cos(t)3(cos(t) dt = 6
∫ π/20
cos2 tdt = 61
2
∫ π2
0
(1 +
cos(2t))dt = 3(t+
sin(2t)
2
)∣∣∣π20
= 3(π
2+
sinπ
2− 0− sin 0
2
)=
3π
2.
I Therefore the area under the curve is3π
2.
This time we started at thelower end point and went tothe upper end point so wegot the area.
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Arc Length: Length of a curve
If a curve C is given by parametric equations x = f(t), y = g(t),α 6 t 6 β, where the derivatives of f and g are continuous in theinterval α 6 t 6 β and C is traversed exactly once as t increases fromα to β, then we can compute the length of the curve with thefollowing integral:
L =
∫ βα
√(dxdt
)2+(dydt
)2dt =
∫ βα
√(x′(t)
)2+(y′(t)
)2dt
It is always the case that L is the distance travelled by the particle.
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It is always the case that L =
∫ βα
√(x′(t)
)2+(y′(t)
)2dt is the
distance travelled by the particle.
If you want the length of the curve you need to insure that theparticle moves from one end of the curve to the other without backingup. The only points where you could back up are points
(x(t), y(t)
)for which x′(t) = 0 = y′(t). If there are such points you need tocarefully study the behavior of the curve around such a point.
You also need to be sure you are not going around a closed curvemore than once.
Once you have such a parametrization, you can define an arc lengthfunction
s(T ) =
∫ Tα
√x′(t)2 + y′(t)2 dt for T ∈ [α, β]
so s(T ) is the distance along the curve from(x(α), y(α)
)to(
x(T ), y(T )).
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On the unit circle, starting at (1, 0) and going counterclockwise, thearc length function is radians.
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Example
Find the arc length of the spiral defined by
x = et cos t y = et sin t 0 6 t 6 2π
I x′(t) = et cos t− et sin t, y′(t) = et sin t+ et cos t.
I L =
∫ 2π0
√e2t(cos t− sin t)2 + e2t(sin t+ cos t)2 dt
I e2t(cos t− sin t)2 + e2t(sin t+ cos t)2 =I e2t
(cos2 t−2 cos t sin t+sin2 t+sin2 t+2 sin t cos t+cos2 t
)= e2t2.
I Hence L =
∫ 2π0
et√
2 dt =√
2et∣∣∣2π0
=√
2(e2π − 1).
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Example
Find the arc length of the circle defined by
x = cos 2t y = sin 2t 0 6 t 6 2π
Do you see any problems?
I If we apply the formula L =
∫ βα
√(x′(t)
)2+(y′(t)
)2dt, then, we
get
I L =
∫ 2π0
√4 sin2(2t) + 4 cos2(2t) dt = 2
∫ 2π0
√1 dt = 4π
I The problem is that this parametric curve traces out the circletwice, so we get twice the circumference of the circle as ouranswer.
I But notice that this is the distance that the particle travelled!
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