x = 4 cos t y = 9sin t 0 ≤ t ≤ 2π
Since, sin 2 t + cos 2 t = 1
Use the Pythagorean Identity
to eliminate the parameter
x2
and x = 16 cos t → = cos 2 t
16
y2
2
2
y = 81sin t →
= sin 2 t
81
2
2
y
x
=1
So, +
81 16
Ellipse ( with the major axis vertical )
2
and identify the curve.
x = 4 cos t y = 9sin t 0 ≤ t ≤ 2π
2
1
1
x = sin t , y = cos t when t =
Find the slope of the line tangent to
the curve x = sin t , y = cos t when t =
π
3
.
π
3
dy
dy dt − sin t
=
=
dx dx
cos t
dt
π
3
− sin
−
dy
3
=
= 2 =− 3
π
1
dx t = π
cos
3
3
2
1
1
x = t 2 − t, y = t 2 + t
dy
dy dt 2t + 1
=
=
dx dx 2t − 1
dt
At what value of t does the curve
x = t 2 − t, y = t 2 + t
have a horizontal tangent?
Tangent lineis horizontal when
dy
=0
dx
This occurs when 2t + 1 = 0 → t =
−1
2
1
(2t + 1)
t2
x= , y=
2
3
Find the length of the curve
t
(2t + 1)
x= , y=
2
3
from 0 ≤ t ≤ 4.
2
3
1
β
3
2
2
1
dx
dy
=t
= (2t + 1) 2
dt
dt
2
⎛ dx ⎞ ⎛ dy ⎞
L = ∫ ⎜ ⎟ + ⎜ ⎟ dt = ∫
⎝ dt ⎠
α ⎝ dt ⎠
0
2
4
4
= ∫ t + 2t + 1dt = ∫
2
0
1
0
4
(t )
2
4
(
+ (2t + 1)
1
) dt
2
2
4
t2
( t + 1) dt = ∫ t + 1dt = + t = 12
2
0
0
2
1
x = 3cos t , y = 3sin t from 0 ≤ t ≤
Find the surface area when the curve
dx
= −3sin t
dt
is rotated about the x-axis.
⎛ dx ⎞ ⎛ dy ⎞
S = ∫ 2π y ⎜ ⎟ + ⎜ ⎟ dt =
⎝ dt ⎠ ⎝ dt ⎠
α
x = 3cos t , y = 3sin t from 0 ≤ t ≤
π
3
=
.
3
∫
3
dy
= 3cos t
dt
β
π
π
2
2
π
3
∫
0
2
π
( −3sin t ) + ( 3cos t )
2π (3sin t )
2
⎛ dx ⎞ ⎛ dy ⎞
2π (3sin t ) ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
2
dt =
2
0
3
∫ 2π (3sin t )
9sin 2 t + 9 cos 2 tdt
0
π
= 6π
π
3
2
2
∫ sin t 9(sin t + cos t )dt = 6π
0
π
= 18π
3
π
∫ sin tdt = − 18π [cos t ]
0
0
1
3
3
∫ sin t 9(1)dt =6π
0
π
3
∫ sin t 3dt =
0
1
= −18π [ − 1] = 9π
2
1
3
3π
(3, )
4
3π
),
4
then convert to Cartesian coordinates.
Plot the point given by (r , θ ) = (−3,
3π
)
4
x = r cos θ y = r sin θ
(r , θ ) = (−3,
2
1
-3
-2
-1
1
2
-1
-2
-3
2
( −3,
3π
)
4
3
− 2 3 2
3π
=
) = −3
4
2
2
3π
2 −3 2
=
y = −3sin( ) = −3
4
2
2
x = −3cos(
2
4r 2 cos 2 θ + 9r 2 sin 2 θ = 36
4(r cos θ )2 + 9(r sin θ ) 2 = 36
Convert the polar equation into
Cartesian and identify the curve.
4 x 2 + 9 y 2 = 36
x2 y 2
+
=1
9
4
Ellipse
4r 2 cos 2 θ + 9r 2 sin 2 θ = 36
2
2
6
Find intersection points:
6=4+4cosθ
4
2=4cosθ
2
-6
Find the area of the region
that is inside the cardiod
r = 4 + 4 cos θ
and outside the circle r = 6.
-4
-2
2
4
6
0.5=cosθ
8
-2
θ=
-4
Method 1:
π
−π
π
3
∫
=
−π
π
=
−
5π −π
=
3
3
π
3
3
π
3
∫
(16 cos θ + 8cos θ − 10)dθ =
−π
3
3
,θ =
3
3
1
1
1
(4 + 4 cos θ ) 2 dθ − ∫ (6) 2 dθ = ∫ [(4 + 4 cos θ ) 2 − 62 ]dθ
2
2
−π
−π 2
2
3
∫
π
3
3
-6
π
3
∫
A=
π
(16 cos θ + 8cos 2 θ − 10)dθ
3
π
⎛1
⎞
(16 cos θ + 8 ⎜ (1 + cos(2θ ) ⎟ − 10) dθ = ∫ (16 cos θ + 4 + 4 cos(2θ ) − 10)dθ
⎝2
⎠
−π
3
3
= 16sin θ + 2sin(2θ ) − 6θ
π
= 18 3 − 4π ≈ 18.61
3
−π
3
Method 2 (by symmetry !!)
π
2
3
A=2∫
0
π
1
[(4 + 4 cos θ )2 − 62 ]dθ =
2
3
∫ [(4 + 4 cos θ )
0
2
− 62 ]dθ = 18 3 − 4π ≈ 18.61
2
r
3
Sketch the polar curve using the graph below.
p
2p
q
-3
r
3
2
1
p
2p
q
-2
-1
1
2
-1
-3
-2
2
2
r = sin(3θ )
Find the slope of the line tangent
to the curve r = sin(3θ ) at θ =
π
4
.
2
dr
= 3cos(3θ )
dθ
dr
sin θ + r cos θ
dy dθ
3cos(3θ ) sin θ + sin(3θ ) cos θ
=
=
dx dr cos θ − r sin θ 3cos(3θ ) cos θ − sin(3θ ) sin θ
dθ
π
π
π
π
− 2 2
2 2
3cos(3 ) sin + sin(3 ) cos
3
+
dy
4
4
4
4 =
2 2
2 2
=
dx θ = π 3cos(3 π ) cos π − sin(3 π ) sin π
− 2 2
2 2
3
−
4
4
4
4
4
2 2
2 2
−2 2 −3 1
+
3 +
−1 1
=
= 4 4= 2 2=
−2 2 −3 1 −2 2
3 −
−
4 4
2 2
2
y 2 = 12 x
Sketch the conic section y 2 = 12 x,
including the focus and directrix.
General form : y 2 = 4 px
4 p = 12
p=3
Focus : (3, 0)
Directrix : x = −3
5
-4
-2
2
4
-5
3
Find an equation for the ellipse that
3
The center of the ellipse is midway between
the foci at (-1,2).
The major axis is vertical and c = 1 and b = 2
has foci at (−1,1) and ( −1,3) and a
Thus, c = a 2 − b 2 → 1 = a 2 − 22 → a = 5
minor axis of length 4.
( x + 1) 2 ( y − 2) 2
+
=1
4
5
3
3
4 x 2 − 9 y 2 − 16 x − 20 = 0
4 x 2 − 16 x − 9 y 2 = 20
4( x 2 − 4 x) − 9 y 2 = 20
4( x 2 − 4 x + 4) − 9 y 2 = 20 + 16
4( x − 2) 2 − 9 y 2 = 36
Find the center, foci and vertices of
y
( x − 2) 2 y 2
−
= 1 a = 3, b = 2
9
4
Center (2, 0) (major axis horizontal )
Vertices are a units from the center.
4 x 2 − 9 y 2 − 16 x − 20 = 0.
Sketch the graph.
-5
V1 (−1, 0) and V2 (5, 0)
c = a + b = 3 + 2 = 13
3
2
2
4
2
-3
-1
1
3
5
7
9
-2
Foci are c units from the center.
2
6
-4
2
-6
F1 (2 + 13, 0) and F2 (2 − 13, 0)
3
Since the directrix is vertical and
to the right of the focus the form is
Find a polar equation for the ellipse
with a focus at the origin, a directrix
3
at x=2 and e = .
4
3
ed
1 + e cos θ
3
2
4
r=
3
1 + cos θ
4
6
r=
4 + 3cos θ
r=
3
x
3
3 − 6sin θ
1
r=
1 − 2sin θ
e = 2 >1
r=
Find the eccentricity, identify the conic
and the equation of the directrix for
3
r=
.
3 − 6sin θ
The conic is a hyperbola
ed = 1 and e = 2.
1
2
Due to the form '− e sin θ '
So, d =
1
the directrix is y = − .
2
3
3
f ( x, y ) =
Need : y − 1 > 0
Sketch the domain of
f ( x, y ) =
2x2
+5
y −1
y >1
2
Notice there are no constraints on x.
D = {( x, y ) | x ∈ℜ, y > 1}
2x
+ 5.
y −1
4
4
f ( x, y ) = ln( x 2 − y )
Need : x 2 − y > 0
y < x2
Sketch the domain and find range of f ( x, y ) = ln( x 2 − y ).
D = {( x, y ) | y < x 2 }
Range : Sincethe range of y = ln x is (−∞, ∞),
the range of f ( x, y ) = ln( x 2 − y ) is (−∞, ∞) = ℜ.
R = {z | ℜ}
4
4
f ( x, y ) = 4 x 2 + y 2
Level curves are defined by : k = 4 x 2 + y 2
These are all ellipses with a vertical major axis.
Given f ( x, y ) = 4 x 2 + y 2 .
Plot the level curves f ( x, y ) = 1, f ( x, y ) = 2,
f ( x, y ) = 3, f ( x, y ) = 4and f ( x, y ) = 5
to construct a contour map.
y
3
2
1
-3
-2
-1
1
2
3
x
-1
-2
-3
4
4
a)
⎛x −y ⎞
a ) lim ⎜ 2
⎟
( x , y ) →(0,0) x + y 2
⎝
⎠
2
⎛ x2 − y2 ⎞
b) lim ⎜ 2
⎟
( x , y ) →(0,1) x + y 2
⎝
⎠
2
approach along the x − axis ( y = 0)
Evaluate the limit,if it exists.
2
⎛ x2 − y 2 ⎞
⎜
⎟
( x , y ) → (0,0) x 2 + y 2
⎝
⎠
lim
2
2
2
⎛ x 2 − 02 ⎞ ⎛ x 2 ⎞
= ⎜ 2 ⎟ =1
f ( x, 0) = ⎜ 2
2 ⎟
⎝ x +0 ⎠ ⎝ x ⎠
So, f ( x, y )→ 1 as ( x, y )→ (0, 0) along y = 0.
approach along the line y = x
2
2
⎛ x2 − x2 ⎞ ⎛ 0 ⎞
=⎜ 2 ⎟ =0
f ( x, x ) = ⎜ 2
2 ⎟
⎝ x + x ⎠ ⎝ 2x ⎠
So, f ( x, y )→ 0 as ( x, y )→ (0, 0) along y = x.
2
2
2
⎛ x 2 − y 2 ⎞ ⎛ 02 − 12 ⎞ ⎛ −1 ⎞
b) lim ⎜ 2
⎟ = ⎜ 2 2 ⎟ = ⎜ ⎟ =1
( x , y ) →(0,1) x + y 2
⎝
⎠ ⎝ 0 +1 ⎠ ⎝ 1 ⎠
2
4
Discuss the continuity of f ( x, y ) =
4
x − 2y
is a rational function.
x2 + y 2
Thus,it is continuous on its domain.
D = {( x, y ) | ( x, y ) ≠ (0, 0)}
So,it is continuous for all points except ( x, y ) = (0, 0).
f ( x, y ) =
x − 2y
.
x2 + y 2
4
4
1
f ( x, y ) =
4 − x2 − y 2
1
f ( x, y ) =
Determine the largest set on which
1
f ( x, y ) =
is continuous.
4 − x2 − y2
is a composite function.
= g (h( x, y ))
4 − x2 − y 2
where h( x, y ) = 4 − x 2 − y 2 and g (t ) =
1
t
Domain of h( x, y ) = {( x, y ) | ( x, y ) ∈ℜ2 }
and the domain of g (t ) = {t | t > 0}
Thus, the domain of f ( x, y ) = {( x, y ) | 4 − x 2 − y 2 > 0}
So,it is continuous for all points within a circle of radius 2.
5
5
f ( x, y ) = xe x y
2
Given f ( x, y ) = xe x y
Find f x and f yx .
2
f x = ( x)(e x y ⋅ 2 xy ) + (e x y )(1) = 2 x 2 ye x y + e x y
2
2
2
2
f y = xe x y ⋅ x 2 = x 3e x y
2
2
f yx = ( x 3 )(e x y ⋅ 2 xy ) + (e x y )(3 x 2 ) = 2 x 4 ye x y + 3 x 2e x y
2
5
2
2
2
5
1 + x2 y
f ( x, y , z ) =
z
∂f 2 xy
=
z
∂x
2
∂f x
=
∂y z
1 + x2 y
Given f ( x, y, z ) =
z
∂f ∂f
∂f
Find , , and .
∂x ∂y
∂z
∂f −(1 + x 2 y )
=
z2
∂z
5
5
w = x 2 + y 2 with x = cos t and y = sin t
dw ∂w dx ∂w dy
=
+
dt ∂x dt ∂y dt
dw
= 2 x(− sin t ) + 2 y (cos t )
dt
dw
= 2cos t (− sin t ) + 2sin t (cos t )
dt
dw
= 0.
dt
dw
when t = π ,given
dt
w = x 2 + y 2 with x = cos t and y = sin t.
Evaluate
5
∂z
∂z
and ,given
∂s
∂t
z = ln( xy 2 ) with x = 2t + s 2 and y = s 3 cos t
1 ∂z
1
2
1
∂z
=
y2 =
2 xy =
=
∂x x y 2
x ∂y xy 2
y
Find
∂z
∂z
and ,given
∂s
∂t
2
z = ln( xy ) with x = 2t + s 2 and y = s 3 cos t.
Find
∂x
∂x
∂y
∂y
= 2s
=2
= 3s 2 cos t
= − s 3 sin t
∂s
∂t
∂s
∂t
∂z ∂z ∂x ∂z ∂y 1
2
2s
6s 2 cos t
2s
6
=
+
= 2s + 3s 2 cos t =
+ 3
=
+
2
∂s ∂x ∂s ∂y ∂s x
2t + s
y
s cos t 2t + s 2 s
∂z ∂z ∂x ∂z ∂y 1
2
2
2s 3 sin t
2
=
+
= 2 + (− s 3 sin t ) =
− 3
=
− 2 tan t
2
y
s cos t 2t + s 2
∂t ∂x ∂t ∂y ∂t x
2t + s
5