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AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 67 of 160

Marks: +1, -0

If x = sin θ and y = cos pθ , then

(1-x^2)y_2

=

$xy_1 - p^2 y$
$p^y - xy_1$
$xy_1$
$p^2 y$

Solution:

It is given that,

x = \sin \theta

and

y = \cos p \theta

Then,

\frac{dy}{d\theta} = -p \sin p \theta

and

\frac{dx}{d\theta} = \cos \theta

This implies,

\frac{dy}{dx} = - \frac{p \sin p \theta}{\cos \theta}

= -p \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}

\sqrt{1-x^2} \left( \frac{dy}{dx} \right)^2 = p^2 (1-y^2)

Square both sides,

(1-x^2) \left( \frac{dy}{dx} \right)^2 = p^2 (1-y^2)

Differentiate above,

-2x \left( \frac{dy}{dx} \right)^2 + (1-x^2) 2 \frac{dy}{dx} \frac{d^2 y}{dx^2}

= -p^2 2 y \frac{dy}{dx}

-x \frac{dy}{dx} + (1-x^2) \frac{d^2 y}{dx^2} = -p^2 y

(1-x^2) y_2 = x y_1 - p^2 y

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