250+ TOP MCQs on Derivative of Arc Length and Answers

Differential Calculus Question Bank focuses on “Derivative of Arc Length”.

1. For the cartesian curve y=f(x) with ‘s’ as arc length which of the following condition holds good?
a) (frac{ds}{dx} = sqrt{1+(frac{dy}{dx})^2})
b) (frac{d^2s}{dx^2} = sqrt{1-(frac{dy}{dx})^2})
c) (frac{dy}{dx} = sqrt{1+(frac{ds}{dx})^2})
d) (sqrt{(frac{ds}{dx})^2 + (frac{dy}{dx})^2} = 1)
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2. For the curve (y=a ,log ,sec(frac{x}{a})) what is the value of ( frac{ds}{dx})? (where φ is the angle made by tangent to the curve with x-axis)?
a) cos φ
b) sec φ
c) tan φ
d) cot φ
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Answer: b
Explanation: w.k.t (frac{ds}{dx} = sqrt{1+(frac{dy}{dx})^2})
(frac{dy}{dx} = a frac{tan frac{x}{a}.secfrac{x}{a}}{secfrac{x}{a}} frac{1}{a} = tan frac{x}{a} )
substituting we get
(frac{ds}{dx} = sqrt{1+(tan frac{x}{a})^2}=secfrac{x}{a}, ,but ,w.k.t, frac{dy}{dx} = tan φ = tanfrac{x}{a} ,thus, φ=frac{x}{a})
(frac{ds}{dx} = sec⁡φ.)

3. If the parametric equation of the curve is given by x=ae^t sin⁡t & y=ae^t cos⁡ t the value of (frac{ds}{dt}) is given by _____
a) ae^t
b) 2ae^t
c) √3 ae^t
d) √2 ae^t
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Answer: d
Explanation: w.k.t (frac{ds}{dt} = sqrt{(frac{dx}{dt})^2 + (frac{dy}{dt})^2}…(1))
(frac{dx}{dt} = ae^t (cos t + sin t), frac{dy}{dt} = ae^t (-sin⁡ t + cos⁡t))
substituting in (1) we get
(frac{ds}{dt} = ae^t sqrt{(cos t + sin t)^2+(-sin t+cos⁡t)^2})
(frac{ds}{dt} = ae^t sqrt{1+2 sin t cos + 1 – 2 sin⁡ t cos⁡t} = sqrt{2} ae^t…(cos^{2} t + sin{2} t = 1).)

4. For the curve in polar form (sqrt{frac{r}{a}} = sec⁡(frac{θ}{2}) ,the, ,value, ,of, frac{ds}{dθ}) is _____
a) r sec θ
b) r sec ((frac{θ}{2}))
c) r sec (2θ)
d) r cosec ((frac{θ}{2}))
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Answer: c
Explanation: Squaring the given curve on both side i.e (r=a sec ^2 (frac{θ}{2}))…(1)
(frac{dr}{dθ} = a.2 sec⁡(frac{θ}{2}) sec⁡(frac{θ}{2}) tan (frac{θ}{2}).1/2 = a sec^2 (frac{θ}{2}) tan(frac{θ}{2}) )
from (1)
(frac{dr}{dθ} = r tan (frac{θ}{2}) ,the, ,equation, ,for, frac{ds}{dθ} ,is, = sqrt{(r^2+(frac{dr}{dθ})^2)} = sqrt{r^2(1+ tan^2 (frac{θ}{2}))} )
(frac{ds}{dθ} = r sec (frac{θ}{2}).)

5. If r=b e^{θ cot⁡a}, where a, b are constants then ‘s’ is represented by which one of the following equation?
a) s=r sec(a)+c
b) s=r cos(a)+c
c) s=r+c
d) s=r tan(a)+c
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Answer: a
Explanation: r=b e^{θ cot⁡a}
(frac{dr}{dθ} = b e^{θ cot⁡a}(cot a)=r cot(a) )
w.k.t (frac{ds}{dr} = sqrt{1+r^2 (frac{dθ}{dr})^2} ,where, frac{dθ}{dr}=frac{1}{r} ,tan⁡a)
(frac{ds}{dr} = sqrt{1+tan^2 a} = sec ,a rightarrow s = int sec ,a ,dr , = r sec(a)+c,) where c is constant of integration.