EXERCISES 1. Solutions are now posted on the website. Check your work
yourself. Don't hand it in. If you still have difficulties after working through the.
MA100: Mathematical Methods Calculus Lecture 2
EXERCISES 1 Solutions are now posted on the website. Check your work yourself. Don’t hand it in. If you still have difficulties after working through the solutions, go to the special one-off ‘class’ on Friday, 4-5, E171 (New Theatre).
Extra Examples Session (Optional) Thursdays 11-12, Old Theatre. This week:
• Matrices
• Planes
• Conics
PLANES Vector v is normal (or is a normal) to a plane if it is orthogonal to (every vector in the plane). If ξ is a point in the plane, then for any other point x in the plane, v will be orthogonal to x − ξ. (And conversely.) So the plane has equation hv, x − ξi = 0.
hv, x − ξi = 0 can be rewritten as hv, xi = hv, ξi = c. So we have, v1x + v2y + v3z = c, the Cartesian equation of the plane.
Example: What isthe Cartesian equation of the 1 3
plane through ξ =
0
−2
with normal v =
x − y + z = 3 − 0 + (−2) = 1.
−1 ?
1
If
2
1
is normal to a plane, so is
4
2
6
3
or
−4
−2 ,
−6
Sometimes it’s helpful to choose a unit vector 2 (length 1). The length of
1
is
3 22 + 12 + 32 = 14, so a unit vector in this 2 1 direction is √ 1 . 14 3 q
√
etc.
Distance of plane from origin Suppose v a unit normal and that ξ = cv is on the plane (the closest point of the plane to the origin). Equation of plane is hv, x − ξi = 0, so hv, xi = hv, ξi = hv, cvi = chv, vi = ckvk2 = c.
So if v is a unit normal to the plane and c is the distance of the plane from 0, then the plane has equation v1x + v2y + v3z = c. A plane v1x + v2y + v3z = 0 is a plane through 0.
Intersection of planes Example: Plane 1 has equation 2x − y + z = 6 and plane 2 has equation x + y + 2z = 1. These planes intersect in a line. To find the equation of the line, use Linear Algebra to solve the equations simultaneously. Find that the line is
x
y
z
=
7/3
−4/3
0
+
−1
t −1 .
1
The angle between two planes The angle between two planes is the angle between their normal vectors. (See picture.) Example: The angle θ between planes 2x − y + z = 6 and x + y + 2z = 1 is the angle between the normal 2 1
vectors v1 =
−1
1
and v2 =
1 .
Now,
2
hv1, v2i = kv1kkv2k cos θ,
so
cos θ = and θ = π/3.
2 1 + * −1 , 1
1 2 kv1kkv2k
1 3 =√ √ = , 2 6 6
PARAMETRIC EQUATION OF A PLANE
x = ξ + su + tv
where
• ξ is one (any) point in the plane,
• u and v are non-parallel direction vectors in the plane, orthogonal to the normal vector,
• s, t are arbitrary scalars (real numbers).
Example: Find a parametric equation for the plane 2x − y + z = 6.
This has normal vector
2
−1
and a point on the
1
plane is ξ =
3
0 .
Two non-parallel vectors
0
orthogonal to the normal are u =
0
1 ,
1
v=
1 0 −2
.
So the plane is
x=
3
0
0
+
0
t 1
1
+
s
1 0 −2
.
CURVES Now we think about curves in R2. First, a line is the set of solutions to a linear equation ax + by = c. (Special cases: x = c is a vertical line and y = c is a horizontal line.)
More complicated equations give rise to more complicated curves as the set of solutions. Examples: y − x2 = 0
x2 + y 2 = 1 xy = 1.
Parabolas
100 80 60 40 20
–10 –8
–6
–4
–2 0
The parabolas y =
2
x2
4 x 6
8
10
1 2 and y = x . 2
4
2
–2
–1
0
1
2 x
3
4
–2
–4
The parabola with equation y + 4 = (x − 1)2.
Ellipses
x2 y2 + 2 =1 2 a b is the equation of an (axis-alligned) ellipse, centred at the origin.
1
y 0.5
–2
–1
0
1 x
2
–0.5
–1
x2 Example: the ellipse + y 2 = 1. 4
100
y50
–1
–0.5
0.5 x
1
–50
–100
yx = 1 (or y = 1/x). A hyperbola with the axes as asymptotes.
4 y 2
–10
–5
0
5 x
10
–2 –4
x2 − 4y 2 = 9 is also a hyperbola. Its asymptotes are y = x/2 and y = −x/2.
4 y 2
–10
–5
0 –2 –4
5 x
10
Why does x2 − 4y 2 = 9 have asymptotes y = x/2 and y = −x/2? Well, if x and y are both very large, then to ensure that the difference between x2 and 4y 2 is as small as 9, we need x2 ≈ 4y 2, so y ≈ x/2 or y ≈ −x/2.
FACT: For any second-degree equation Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0. the graph is either:
• a line
• a parabola
• an ellipse (of which circle is a special case)
• a hyperbola
Actually, there are a few (degenerate) cases where this isn’t true, but these aren’t interesting or important.
All of these are conic sections: obtained by cutting a double cone with a plane.
