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UPSC NDA Math Model Paper 1

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Question : 84 of 120

Marks: +1, -0

If

x=a \sin p t

y=b \cos p t,

then the value of

\frac{d^{2} y}{d x^{2}}

t=0

$-\frac{b}{a^{2}}$
$-\frac{a}{b^{2}}$
$\frac{a}{b^{2}}$
$\frac{b}{a^{2}}$

Solution:

x=a \sin pt

\Rightarrow \frac{dx}{dt}=a \frac{d}{d(pt)} \sin(pt) \times \frac{d}{dt}(pt)

=a p \cos(pt)

y=b \cos pt

\Rightarrow \frac{dy}{dt}=b \frac{d}{d(pt)} \cos(pt) \times \frac{d}{dt}(pt)

=-b p \sin(pt)

=-b p \sin(pt) \times \frac{1}{a p \cos(pt)}

=-\frac{b}{a} \times \tan(pt)

Now,

\frac{d^{2} y}{dx^{2}} = \frac{d}{dx} \left( \frac{dy}{dx} \right)

= \frac{d}{dx} \left[ -\frac{b}{a} \tan(pt) \right]

=-\frac{b}{a} \times \frac{d}{d(pt)} \tan(pt) \times \frac{d}{dt}(pt) \times \frac{dt}{d x}

=-\frac{b}{a} \times \sec^{2}(pt) \times p \times \frac{1}{a p \cos(pt)}

=-\frac{b}{a^{2}} \times \sec^{3}(pt)

At

t=0, \frac{d^{2} y}{dx^{2}} = -\frac{b}{a^{2}} \times \sec^{3} 0^{\circ}

=-\frac{b}{a^{2}} \times 1 = -\frac{b}{a^{2}} .

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