MA 22S1
Assignment 1
Due 12-14 October 2015
Id:
22S1-f2015-1.m4,v 1.3 2015/10/11 19:11:54 john Exp john
1. Convert
x = 2t + 3,
y = 4t + 5
from paramatric to implicit form and
x + 2y = 3
from implicit to parametric.
Solution:
For the first curve we multiply the first equation by 2 and substract
the second equation to eliminate the parameter t,
x − 2y = 1.
For the second curve we solve for x as a function of y, obtaining x =
3 − 2y, and use y as the parameter:
x = 3 − 2t,
y = t.
2. Find the tangent lines to the parametric curve
x=
1 + t2
,
1 − t2
2t
1 − t2
at the points corresponding to the parameter values t = 0, t = 2 and
t = −3.
Solution: First we find the derivatives,
dx
4t
=
,
dt
(1 − t2 )2
1
dy
2 + 2t2
=
,
dt
(1 − t2 )2
Id:
22S1-f2015-1.m4,v 1.3 2015/10/11 19:11:54 john Exp john 2
and then their quotient
m=
dy/dt
1 + t2
=
.
dx/dt
2t
This is the slope of the tangent line at the point with parameter value t,
except in the exceptional case where the denominator is zero.
For t = 0 the point is (1, 0) and the tangent is vertical, since dx/dt = 0
and dy/dt 6= 0, so the tangent line is
x = 1.
For t = 2 the point is (−5/3, −4/3) and the tangent line has slope
m = 5/4, so the line is
5
y + 4/3
=
4
x + 5/3
or
15x − 12y + 9 = 0.
For t = 3 the point is (−5/4, 3/4) and the tangent line has slope m =
−5/3, so the line is
y − 3/4
5
− =
3
x + 5/4
or
10x + 6y + 8 = 0.
3. Find the length of the parametric curve
x=
1 + t2
,
1 − t2
y=
2t
,
1 − t2
1
1
− <t< .
2
2
Note: the integral you get is not one with elementary indefinite integral,
so you can’t get a closed form solution. So just simplify the integrand
as much as you can, and leave it unintegrated, as in the last arc length
example I did in lecture.
Solution:
The derivatives are still
dx
4t
=
,
dt
(1 − t2 )2
dy
2 + 2t2
=
,
dt
(1 − t2 )2
Then
dx
dt
!2
dy
+
dt
!2
=
4(t4 + 6t2 + 1)
(4t)2 + (2 + 2t2 )2
=
,
(1 − t2 )4
(1 − t2 )4
Id:
22S1-f2015-1.m4,v 1.3 2015/10/11 19:11:54 john Exp john 3
and
v
u
u
t
dx
dt
!2
L=
dy
+
dt
Z
1/2
2
−1/2
!2
√
=2
√
t4 + 6t2 + 1
(1 − t2 )2
t4 + 6t2 + 1
dt.
(1 − t2 )2
4. Convert
r 4 sin2 (2θ) = 4
from polar to Cartesian coordinates and
1
1
y = x2 −
2
2
from Cartesian to polar coordinates. Solution:
For the first curve we use the angle addition formula sin(2θ) = 2 sin θ cos θ
to get
4r 4 sin2 θ cos2 θ = 4
Then, because x = r cos θ, y = r sin θ,
4x2 y 2 = 4,
or, more simply,
x2 y 2 = 1.
For the second curve we use again the equations x = r cos θ, y = r sin θ
to get
1
1
r sin θ = r 2 cos2 θ − .
2
2
In principle this is already a correct answer, but it can be simplified
somewhat.
1
1
1
r sin θ = r 2 − r 2 sin2 θ −
2
2
2
1
1 2
1 2 2
r sin θ + r sin θ + = r
2
2
2
2
2
r sin θ + 2r sin θ + 1 = r 2 ,
(r sin θ + 1)2 = r 2 ,
r sin θ + 1 = r.
The last step might look dodgy. How do we know to take the positive
square root? Well, r ≥ 0 by definition and r sin θ ≥ −r because sin θ ≥
−1. But then r sin θ + 1 > −r, so we can’t have r sin θ + 1 = −r.