Find the angle between the radius vector and tangent to curve rcos^2(t/2)=a
Answers
Answer:
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)).
θ
ψ
φ
r
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 cosθ, y = rsinθ, (1)
with associated derivatives given by.
dx
dθ
= −rsinθ +cosθ
dr
dθ
.
dy
dθ
= r cosθ +sinθ
dr
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
tanψ = tan(φ −θ) = tanφ −tanθ
1+tanφ tanθ
. (3)
• Since the tangent line to the curve f(θ) makes an angle φ with respect to the x−axis we have
that tanφ =
dy/dθ
dx/dθ
, and trivially from the geometry that tanθ = y/x. Substituting these into (3)
gives,
tanψ =
dy/dθ
dx/dθ
−
y
x
1+
y
x
dy/dθ
dx/dθ
=
x
dy
dθ
−y
dx
dθ
x
dx
dθ
+y
dy
dθ
(4)
=
x
r cosθ +sinθ
dr
dθ
−y
−rsinθ +cosθ
dr
dθ
x
−rsinθ +cosθ
dr
dθ
+y
r cosθ +sinθ
dr
dθSubstituting (1) into (5) yields,
tanψ =
r cosθ (r cosθ +sinθ(dr/dθ))−rsinθ (−rsinθ +cosθ(dr/dθ))
r cosθ (−rsinθ +cosθ(dr/dθ)) +rsinθ (r cosθ +sinθ(dr/dθ)) (6)
=
r
2
rdr/dθ
(7)
=
r
dr/dθ
(8)
which is the desired result.
Step-by-step explanation:
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