DERIVATIVE OF ARC LENGTH. Let y = f(x) be the equation of a given curve. Let A be some fixed point on the curve and P(x,y) and Q(x+ Δ x, y+ Δ y) be two neighbouring points on the curve as shown in the Figure 1 . Let arc AP = S, arc PQ = Δ S and the chord PQ = Δ c.
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Let y = f(x) be the equation of a given curve. Let A be some fixed point on the curve and P(x,y) and Q(x+Δx, y+ Δy) be two neighbouring points on the curve as shown in the Figure 1.
Let arc AP = S, arc PQ = ΔS and the chord PQ = Δc.
Draw PL, QM perpendiculars on the x-axis & also PN perpendicular to QM ( see Figure 1)
From the right angled triangle PQN, we have
PQ2 = PN2 + NQ2
Δc2 = Δx2 + Δy2
If s increases with x as shown in Figure, then is
Thus we have
See Figure 1
Cor.2 : For parametric cartesian equations with parameter
i.e. for x = f(t) and y = g(t), we have
Cor. 3 :
1. Find for the curves :
(i) y = c cosh x/c (ii) y = a log[a2 / (a2 – x2)]
2. Find for the curves :
(i) x = a cos3 t, y = b sin3 t
(ii) x = a et sin t, y = a et cos t
(iii) x = a( cos t + t sin t), y = a ( sin t – t cos t)
(iv) x = a ( t – sin t ), y = a ( 1 – cos t )
Let r = f(θ) be the equation of a given curve. Let P(r, θ)
and Q(r+Δr, θ+Δ θ) be two adjacent points on the
curve AB as shown in the Figure 2.
Let arc AP = S, arc PQ= ΔS & the chord PQ = Δc.
Drop PM perpendicular to OQ
From the right angled triangle PMQ, we have
PQ2 = PM2 + MQ2 --------------- (i)
PM = r sin Δθ. --------------- (ii)
From the figure 2., we have
MQ = OQ – OM
= r + Δr – r cos Δθ
= Δr + r (1– cos Δθ)
= Δr + 2r sin2Δθ⁄ 2) --------------- (iii)
From (i), (ii) and (iii) we have
Δc2 = (r sin Δθ)2 + (Δr + 2r sin2Δθ ⁄ 2)2
above equation reduces to
Cor. 1 : If the equation of the curve is θ = f(r), then
Example 1 :Find ds/dθ for r = a ( 1 + cos θ )
Example 3 :Show that r (ds/dr) is constant for the curve
Example 4 : If the tangent at a point P(r,θ) on the curve
r2 sin 2θ = 2 a2 meets the initial line in T,
(i) ds/dθ = r3/ 2 a2 (ii) PT = r
END OF SHOW
Differentiating with respect to θ, we obtain
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