2.9. EXERCISES 75
this will suffice to explain the above question. Then the acceleration observed by a personon the earth relative to the apparently fixed vectors, i,k, j, is
aB = −a (rB) (xi+yj+zk)− 2ω [−y′ cos (ϕ) i+(x′ cos (ϕ) + z′ sin (ϕ)) j− (y′ sin (ϕ)k)]
Therefore, one obtains some differential equations from aB = x′′i+ y′′j+ z′′k by matchingthe components. These are
x′′ + a (rB)x = 2ωy′ cosϕ
y′′ + a (rB) y = −2ωx′ cosϕ− 2ωz′ sin (ϕ)
z′′ + a (rB) z = 2ωy′ sinϕ
Now remember, the vectors, i, j,k are fixed relative to the earth and so are constant vectors.Therefore, from the properties of the determinant and the above differential equations,
(r′B × rB)′=
∣∣∣∣∣∣∣i j k
x′ y′ z′
x y z
∣∣∣∣∣∣∣′
=
∣∣∣∣∣∣∣i j k
x′′ y′′ z′′
x y z
∣∣∣∣∣∣∣=
∣∣∣∣∣∣∣i j k
−a (rB)x+ 2ωy′ cosϕ −a (rB) y − 2ωx′ cosϕ− 2ωz′ sin (ϕ) −a (rB) z + 2ωy′ sinϕ
x y z
∣∣∣∣∣∣∣Then the kth component of this cross product equals
ω cos (ϕ)(y2 + x2
)′+ 2ωxz′ sin (ϕ) .
The first term will be negative because it is assumed p (t) is the location of low pressurecausing y2+x2 to be a decreasing function. If it is assumed there is not a substantial motionin the k direction, so that z is fairly constant and the last term can be neglected, then thekth component of (r′B × rB)
′is negative provided ϕ ∈
(0, π2
)and positive if ϕ ∈
(π2 , π
).
Beginning with a point at rest, this implies r′B ×rB = 0 initially and then the above impliesits kth component is negative in the upper hemisphere when ϕ < π/2 and positive in thelower hemisphere when ϕ > π/2. Using the right hand and the geometric definition of thecross product, this shows clockwise rotation in the lower hemisphere and counter clockwiserotation in the upper hemisphere.
Note also that as ϕ gets close to π/2 near the equator, the above reasoning tends tobreak down because cos (ϕ) becomes close to zero. Therefore, the motion towards the lowpressure has to be more pronounced in comparison with the motion in the k direction inorder to draw this conclusion.
2.9 Exercises
1. Show the map T : Rn → Rm defined by T (x) = Ax where A is an m× n matrix andx is an m× 1 column vector is a linear transformation.
2. Find the matrix for the linear transformation which rotates every vector in R2 throughan angle of π/3.
3. Find the matrix for the linear transformation which rotates every vector in R2 throughan angle of π/4.