If x-a cost-y-a sin t t- then d2 y-d x2 at t-4 is

The correct option is D 2 √(2)/aClearly x^2+y^2=a^2 and y ( π/4 )=a/√(2), x ( π/4 )=a/√(2). Differentiating we get, 2x+ 2yy_1=0 ⇒ y_1= x/y, so y_1 ( π/4 )= 1 ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

If x=a cost,y...

Question

If x = a cos t, y = a sin t, 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 $y (\frac{π}{4}) = \frac{a}{\sqrt{2}}, x (\frac{π}{4}) = \frac{a}{\sqrt{2}}$ . Differentiating we get,
$2 x + 2 y y_{1} = 0 \Rightarrow y_{1} = - \frac{x}{y}, s o y_{1} (\frac{π}{4}) = - 1$
Now $x + y y_{1} = 0 \Rightarrow 1 + y_{1}^{2} + y y_{2} = 0$
$\Rightarrow y_{2} (\frac{π}{4}) = - \frac{1 + {(y_{1} (\frac{π}{4}))}_{1}^{2}}{y (\frac{π}{4})} = \frac{- 2 \sqrt{2}}{a}$

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If x = a cos t, y = a sin t, then d2ydx2 at t=π4 is

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

If x = a cos t, y = a sin t, then $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4}$ is