Bonus Problem 11:
Consider the angle ψ between the radius vector and the tangent line to a curve, r = f (θ ), given in
polar coordinates, as shown in Fig. 1. Show that ψ = tan−1 (r/(dr/dθ )).
ψ
r
θ
φ
x
r = f (θ )
Figure 1: The tangent line to the curve r = f (θ ) makes an angle of ψ with respect to the radial line at
the point of tangency, and an angle φ with respect to the x-axis.
Proof :
• Consider φ = θ + ψ. Then r = f (θ ) is given in polar coordinates by
x = r cos θ ,
y = r sin θ ,
(1)
with associated derivatives given by.
dx
dr
= −r sin θ + cos θ
.
dθ
dθ
dy
dr
= r cos θ + sin θ
.
dθ
dθ
(2)
• From the geometry of the problem in which it is evident that 2π − ψ − θ = 2π − φ , we have that
ψ = φ − θ , and consequently, using a familiar multiple angle formula from trigonometry, that
tan ψ = tan(φ − θ ) =
tan φ − tan θ
.
1 + tan φ tan θ
(3)
• Since the tangent line to the curve f (θ ) makes an angle φ with respect to the x−axis we have
dy/dθ
that tan φ =
, and trivially from the geometry that tan θ = y/x. Substituting these into (3)
dx/dθ
gives,
dy/dθ y
−
dx/dθ x
tan ψ =
y dy/dθ
1+
x dx/dθ
dy
dx
−y
dθ
= dθ
dx
dy
x
+y
dθ
dθ
dr
dr
x r cos θ + sin θ
− y −r sin θ + cos θ
dθ
dθ
=
dr
dr
x −r sin θ + cos θ
+ y r cos θ + sin θ
dθ
dθ
x
(4)
(5)
Substituting (1) into (5) yields,
r cos θ (r cos θ + sin θ (dr/dθ )) − r sin θ (−r sin θ + cos θ (dr/dθ ))
r cos θ (−r sin θ + cos θ (dr/dθ )) + r sin θ (r cos θ + sin θ (dr/dθ ))
r2
=
rdr/dθ
r
=
dr/dθ
tan ψ =
(6)
(7)
(8)
2
which is the desired result.
2