This set of Differential Calculus Multiple Choice Questions & Answers focuses on “Polar Curves”.

1. For the below mentioned figure the angle between radius vector (op) ⃗ and tangent to the polar curve where r=f(θ) has the one among the following relation?

a) \(tan\,ω=\frac{tan∅+tanθ}{1-tan∅tanθ}\)

b) \(tan\,ω=\frac{tan∅-tanθ}{1+tan∅tanθ}\)

c) \(tan\,∅=r(\frac{dr}{dθ)}\)

d) \(tan\,ω=tan∅+tanθ \)

View Answer

Explanation: From the fig (1) ω=∅+θ…….. (triangle rule & \hat{opt} = ∅)

tan\,ω=tan(θ+∅)

\(tan\,ω=\frac{tan∅+tanθ}{1-tan∅tanθ}\) ……(1) further more w.k.t x=rcosθ, y=rsinθ

But \(tan\,ω = \frac{dy}{dx}\)……(slope of the tangent TT’)

\(tan\,ω = \frac{\frac{dy}{dθ}} {\frac{dx}{dθ}}\)……..since x & y are functions of θ

i.e \(\displaystyle tan\,ω = \frac{\frac{d(r sinθ)}{dθ}}{\frac{d(r cosθ)}{dθ}} = \frac{r cosθ + \frac{dr}{dθ} sinθ}{-r sinθ + \frac{dr}{dθ} cosθ}\)

\(\displaystyle tan\,ω = \frac{\frac{r cosθ} {\frac{dr}{dθ} cosθ} + \frac{\frac{dr}{dθ} sinθ}{\frac{dr}{dθ} cosθ}} {\frac{-r sinθ}{\frac{dr}{dθ} cosθ}+\frac{\frac{dr}{dθ} cosθ}{\frac{dr}{dθ} cosθ}} = \frac{\frac{r}{\frac{dr}{dθ}}+tanθ}{\frac{r}{\frac{dr}{dθ}}tanθ}\)…………(2)

from (1) &(2)

\(tan∅ = \frac{r}{\frac{dr}{dθ}} = r (\frac{dθ}{dr}).\)

2. The angle between Radius vector r=a(1-cosθ)and tangent to the curve is ∅ given by _______

a) ∅=\(\frac{π}{2}\)

b) ∅=π

c) ∅=\(-\frac{π}{2}\)

d) ∅=0

View Answer

Explanation: r= a(1-cosθ)

taking logarithms on both sides we get,

logr = loga + log(1-cosθ)

differentiating w.r.t θ we get,

\( \frac{1}{r} \frac{dr}{dθ} = 0 + \frac{sinθ}{1-cosθ}\)

\(\frac{1}{r} \frac{dr}{dθ} = \frac{2 sin \frac{θ}{2} cos\frac{θ}{2}}{2sin^2 \frac{θ}{2}} = cot\frac{θ}{2}\) ..(1),

but \(cot∅ = \frac{1}{r} \frac{dr}{dθ}\) ….(2)

From (1)&(2)

∅=\(\frac{π}{2}\).

3. Angle of intersection of two polar curves is equal to the angle between the tangents drawn at the point of intersection of the two curves then What is the condition for the two curves intersecting orthogonally for the below mentioned figure?

a) tan φ1 . tanφ3=1

b) tanφ1 . tanφ3=-1

c) tan(φ1+φ3)=\(\frac{1}{\sqrt{2}}\)

d) tan(φ1-φ3)=\(\frac{1}{\sqrt{2}}\)

View Answer

Explanation: From the figure & according to orthogonal principle |φ1-φ3|=\(\frac{π}{2}\)

further φ1=\(\frac{π}{2}\)+φ3……taking tangent of the angle

\(tan φ1 = tan(\frac{π}{2}+φ3) = -cotφ3 = -\frac{1}{tanφ3} \)

tanφ1 .tanφ3 = -1.

4. Angle of intersection between two polar curves given by r=a(1+sinθ) & r=a(1-sinθ) is given by ________

a) \(\frac{π}{4}\)

b) \(\frac{π}{2}\)

c) Π

d) 0

View Answer

Explanation: r=a(1+sinθ) : r=a(1-sinθ)

taking logarithm on both the equations

logr = loga + log(1+sinθ) : logr = loga + log(1-sinθ)

differentiating on both side we get

\( \frac{1}{r} \frac{dr}{dθ} = \frac{cosθ}{1+sinθ} : \frac{1}{r} \frac{dr}{dθ} = \frac{-cosθ}{1-sinθ}\)

\(cot∅1 = \frac{cosθ}{1+sinθ} : cot∅2 = \frac{-cosθ}{1-sinθ}\)

where ∅1&∅2 are the angle between tangent & the vector respectively

\(tan∅1 = \frac{1+sinθ}{cosθ} : tan∅2 = \frac{1-sinθ}{-cosθ}\)

\(tan∅1 . tan∅2 = \frac{1+sinθ}{cosθ} . \frac{1-sinθ}{-cosθ} = \frac{1-sin^2 θ}{-cos^2 θ} = -\frac{cos^2 θ}{cos^2 θ} = -1\)

above is the condition of orthogonality of two polar curves thus

|∅1-∅2|=\(\frac{π}{2}\).

5. One among the following is the correct explanation of pedal equation of an polar curve, r=f (θ), p=r sin(∅) (where p is the length of the perpendicular from the pole to the tangent & ∅ is the angle made by tangent to the curve with vector drawn to curve from pole)is _______

a) It is expressed in terms of p & θ only

b) It is expressed in terms of p & ∅ only

c) It is expressed in terms of r & θ only

d) It is expressed in terms of p& r only

View Answer

Explanation: It is expressed in terms of p& r only

where p=r sin(∅) & \(tan∅ = \frac{r}{\frac{dr}{dθ}} = r (\frac{dr}{dθ})\)

& r=f (θ) or after solving we get direct relationship between p & r as

\(\frac{1}{p^2} = \frac{1}{r^2} cosec^2∅.\)

6. The pedal Equation of the polar curve r^{n}=a^{n} cosnθ is given by ______

a) r^{n}=pa^{n}

b) r^{n-1}=pa^{n}

c) r^{n+1}=pa^{n+1}

d) r^{n+1}=pa^{n}

View Answer

Explanation: Taking logarithm for the given curve we get

n logr = n loga + log(cosnθ)

differentiating w.r.t θ, we get

\(\frac{n}{r} \frac{dr}{dθ} = \frac{-n sinnθ}{cosnθ} \rightarrow \frac{1}{r} \frac{dr}{dθ} = -tanθ\)

thus \(cot∅ = cot(\frac{π}{2} + nθ)\rightarrow ∅ = \frac{π}{2} + nθ……(1)\)

from the eqn w.k.t p=r sin ∅

substituting from (1)

p = r sin (\(\frac{π}{2}\) + nθ) = r cos (nθ), but we have r

^{n}= a

^{n}cosnθ

hence dividing them we get \(\frac{p}{r^n} = \frac{r cos (nθ)}{a^n cosnθ}\)

r

^{n+1}=pa

^{n}.

7. 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

View Answer

Explanation: Taking Logarithm on both side of the polar curve

we get logr = loga + 2 logsec(\(\frac{θ}{2}\))

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

8. Polar equations of the circle for the given coordinate (x,y) which satisfies the equation given by (x-a)^{2}+(y-b)^{2}=r^{2} where (a,b) is the coordinates of the centre of the circle &r is the radius.

a) x = r cosθ, y = r sinθ

b) x = a+ r cosθ, y = b + r sinθ

c) y = a+r cosθ, x = b + r sinθ

d) x = r sinθ, y = r cosθ

View Answer

Explanation: option x = a+ r cosθ, y = b + r sinθ satisfies the equation (x-a)

^{2}+(y-b)

^{2}=r

^{2}

because LHS=(a+r cosθ-a)

^{2}+ (b+ r sinθ-b)

^{2}= r

^{2}(cos

^{2}θ + sin

^{2}θ)= r

^{2}=RHS.

9. In an polar curve r=f (θ) what is the relation between θ & the coordinates (x,y)?

a) tanθ = \(\frac{x}{y} \)

b) (1+sinθ) = \(\frac{y}{x} \)

c) (1+sec^{2} θ) = \(\frac{y^2}{x^2} \)

d) (1+cosθ) = \(\frac{x}{y} \)

View Answer

Explanation: w.k.t for the polar curve r=f (θ) x=rcosθ, y=rsinθ

dividing them we get \(\frac{y}{x} = \frac{r sinθ}{r cosθ} = tanθ\)

squaring on both side \(\frac{y^2}{x^2} = tan^2 θ = (1+sec^2 θ)\).

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