If x - a cos-sin- and y - a sin-cos- then a - d 2 y - dx 2-A- cos-B- 2-C- 3-D- 0

The correct option is C ^3 θdydx=dydθdxdθ=a(cosθ (cosθ θsinθ))a( sinθ+(sinθ+θcosθ)) dydx=aθsinθaθcosθ=tanθ d^2ydx^2=dtanθdθ·dθdx=^2 θaθcosθ nbsp;∴ aθd^2ydx^ ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

If x = a cosθ...

Question

If $x = a (cos θ + θ sin θ)$ and $y = a (sin θ - θ cos θ),$ then $a θ \frac{d^{2} y}{d x^{2}} =$

0

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cos θ

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{sec}^{2} θ

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{sec}^{3} θ

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Solution

The correct option is D ${sec}^{3} θ$
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (cos θ - (cos θ - θ sin θ))}{a (- sin θ + (sin θ + θ cos θ))}$
$\frac{d y}{d x} = \frac{a θ sin θ}{a θ cos θ} = tan θ$
$\frac{d^{2} y}{d x^{2}} = \frac{d tan θ}{d θ} \cdot \frac{d θ}{d x} = \frac{{sec}^{2} θ}{a θ cos θ}$
$∴ a θ \frac{d^{2} y}{d x^{2}} = \frac{a θ {sec}^{2} θ}{a θ cos θ} = {sec}^{3} θ$

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Similar questions

Q. If

x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ),

then

\frac{d^{2} x}{d θ^{2}} = a (cos θ - θ sin θ), \frac{d^{2} y}{d θ^{2}} = a (sin θ + θ cos θ)

Q. If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), prove that

\frac{d^{2} x}{d θ^{2}} = a (\cos θ - θ \sin θ), \frac{d^{2} y}{d θ^{2}} = a (\sin θ + θ \cos θ) and \frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} θ}{a θ} .

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x = a (cos θ + θ sin θ)

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y = a (sin θ - θ cos θ)

What is

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

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Q. Find

\frac{d y}{d x}

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. and

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Q. Find

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If x=a(cosθ+θsinθ) and y=a(sinθ−θcosθ), then aθd2ydx2=

The correct option is D sec3θdydx=dydθdxdθ=a(cosθ−(cosθ−θsinθ))a(−sinθ+(sinθ+θcosθ)) dydx=aθsinθaθcosθ=tanθ d2ydx2=dtanθdθ⋅dθdx=sec2θaθcosθ ∴aθd2ydx2=aθsec2θaθcosθ=sec3θ

If $x = a (cos θ + θ sin θ)$ and $y = a (sin θ - θ cos θ),$ then $a θ \frac{d^{2} y}{d x^{2}} =$