If x- acost- y- asint- then d2 y-d x2 at t-4 is

The correct option is D 2√(2)/a Clearly x^2 +y^2 =a^2 and y (π/4)=a/√(2),(π/4)= 1 2x+2yy_1=0 ⇒ y_1 = x/y, so y_1 (π/4)= 1. Now x+yy_1 =0 ⇒ 1 +y^2_1 +yy_2 =0 ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If x= acost, ...

Question

If x = acost, y = asint, then $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4}$ is

$\frac{a}{2 \sqrt{2}}$

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$- \frac{a}{2 \sqrt{2}}$

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$\frac{2 \sqrt{2}}{a}$

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$- \frac{2 \sqrt{2}}{a}$

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Solution

The correct option is D
$- \frac{2 \sqrt{2}}{a}$

Clearly $x^{2} + y^{2} = a^{2}$ and $y (\frac{π}{4}) = \frac{a}{\sqrt{2}}, (\frac{π}{4}) = - 1$
$2 x + 2 y y_{1} = 0 \Rightarrow y_{1} = - \frac{x}{y}, s o y_{1} (\frac{π}{4}) = - 1. N o w x + y y_{1} = 0 \Rightarrow 1 + y_{1}^{2} + y y_{2} = 0$
$\Rightarrow y_{2} (\frac{π}{4}) = - \frac{1 + (y_{1} (\frac{π}{4}))^{2}}{y (\frac{π}{4})} = \frac{- 2 \sqrt{2}}{a}$

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If x = acost, y = asint, then d2ydx2 at t=π4 is

The correct option is D −2√2a Clearly x2+y2=a2 and y(π4)=a√2,(π4)=−1 2x+2yy1=0⇒y1=−xy,soy1(π4)=−1.Nowx+yy1=0⇒1+y21+yy2=0 ⇒y2(π4)=−1+(y1(π4))2y(π4)=−2√2a

If x = acost, y = asint, then $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4}$ is