Question

The length of the perpendicular from the pole to the tangent at the point θ=\(\frac{π}{2}\) on the curve. r=a sec^2(\(\frac{π}{2}\)) is _____

(a) \(p = \frac{2a}{\sqrt{3}}\)

(b) \(p = \frac{4a}{\sqrt{3}}\)

(c) \(p = 2a\sqrt{2}\)

(d) p = 4a

I had been asked this question by my college professor while I was bunking the class.

My doubt stems from Polar Curves in section Differential Calculus of Engineering Mathematics

Answer 1

The correct choice is (c) \(p = 2a\sqrt{2}\)

The best explanation: Taking Logarithm on both side of the polar curve

 we get log⁡r = log⁡a + 2 log⁡sec⁡(\(\frac{θ}{2}\))

differentiating w.r.t θ we get

\(\frac{1}{r} \frac{dr}{dθ} = \frac{2 sec⁡(\frac{θ}{2}).tan⁡(\frac{θ}{2}) }{2 sec⁡(\frac{θ}{2})} = tan⁡(\frac{θ}{2})\)

\(cot⁡∅ = cot⁡(\frac{π}{2} – \frac{θ}{2}) \rightarrow ∅ = \frac{π}{2} -\frac{θ}{2}\)

w.k.t length of the perpendicular is given by p=r sin ∅

thus substituting ∅ value we get \(p = r \,sin(\frac{π}{2} – \frac{θ}{2}) = r \,cos(\frac{θ}{2})\)

at \( θ=\frac{π}{4}, p = r \,cos \frac{π}{4} = \frac{r}{\sqrt{2}}….(1)\)

but \( r=a \,sec^2 (\frac{θ}{2}) \,at\, θ = \frac{π}{4}, r=a \,sec^2 (\frac{π}{4}) = 4a…(2)\)

from (1) & (2) \(p=\frac{4a}{\sqrt{2}} = 2a\sqrt{2}\).