MA1S12: SOLUTIONS TO TUTORIAL 10
1. Find an equation for the tangent line to the ellipse
x = 3 cos t,
√
at the point P ( √32 ,
y = 2 sin t,
t ∈ [0, 2π]
2).
Solution: To find an equation for the tangent line at P we first need
to find its slope.
dy
=
dx
The point P ( √32 ,
√
dy
dt dx
dt
=−
2 cos t
3 sin t
2) has parameter value t =
3 cos
π 4
, 2 sin
π 4
π
4
since
3 √
= ( √ , 2)
2
The slope of the tangent line at P is
2 cos( π4 )
dy
2
|t= π4 = −
π = −
dx
3 sin( 4 )
3
The equation of the tangent line at P is
y−
√
2
3
2 = − (x − √ )
3
2
√
2
=⇒ y = − x + 2 2
3
2. Plot the points described by the following polar coordinates,
π
(2, ),
4
π
(1, − ),
2
1
(−1, π)
2
MA1S12: SOLUTIONS TO TUTORIAL 10
Ellipse and tangent line in Q.1.
2
1
-2
2
4
6
-1
-2
Plot of polar coordinates in question 2.
7Π
2Π
3Π
12
Π
2
5Π
Π
12
3
3
Π
4
4
5Π
Π
6
6
11 Π
Π
12
12
Π
0.
0.5
1.
1.5
0
2.5
2.
13 Π
23 Π
12
12
7Π
11 Π
6
6
5Π
4
7Π
4Π
3
4
5Π
17 Π
12
3Π
2
19 Π
3
12
Solution: In polar coordinates (r, θ), r is the directed distance from
the origin and θ is the directed angle made with the polar axis. In
our sketch the polar axis coincides with the positive x-axis. The
Cartesion coordinates for a point (r, θ) are
(x, y) = (r cos θ, r sin θ)
Polar coordinates (r, θ) Cartesian coordinates (x, y)
√ √
(2, π4 )
( 2, 2)
(1, − π2 )
(0, −1)
(−1, π)
(1, 0)
MA1S12: SOLUTIONS TO TUTORIAL 10
3
3. Find the area inside the four petal rose
r = cos 2θ
Solution: We use the area formula for a polar curve. Note that θ
varies from 0 to 2π.
Area =
Z
=
Z
b
a
2π
0
r2
dθ
2
cos2 (2θ)
dθ
2
We can apply the substitution u = 2θ,
=⇒ du =
du
dθ = 2 dθ
dθ
We will change the limits of integration,
θ = 0 =⇒ u = 0
θ = 2π =⇒ u = 4π
Area =
Z
4π
0
cos2 (u)
du
4
4π
1 1
1
(u + sin(2u))
=
4 2
2
0
π
=
2
4. Compute exactly or, alternatively, approximate the length of the
following portion of a spiral
r = θ,
θ ∈ [0, 1]
4
MA1S12: SOLUTIONS TO TUTORIAL 10
The four petal rose in question 3.
1.0
0.5
-1.0
-0.5
0.5
1.0
-0.5
-1.0
Note: You may need the integral
Z
sec3 u du =
1
1
sec u tan u + ln | sec u + tan u| + constant
2
2
Solution: Here we apply the formula for the length of a polar curve,
s
2
Z b
dr
2
Length =
r +
dθ
dθ
a
Z 1√
=
θ2 + 1 dθ
0
To compute this integral we apply a trigonometric substitution
θ = tan u
We have
√
θ2 + 1 =
dθ =
p
tan2 u + 1 = sec u
dθ
du = sec2 u du
du
MA1S12: SOLUTIONS TO TUTORIAL 10
We will change the limits of integration
θ = 1 =⇒ u = tan−1 (1) =
π
4
θ = 0 =⇒ u = tan−1 (0) = 0
Length =
Z
π
4
sec3 u du
0
π4
1
1
sec u tan u + ln | sec u + tan u|
=
2
2
0
√
1
1
= √ + ln | 2 + 1| − [0]
2 2
≈ 1.148
5