NTA Abhyas 2020
Exam NTA Abhyas
Year 2020
Subject Mathematics
Chapter Differentiation NTA Abhyas 2020 · Mathematics · Differentiation
If u = x^ 2 + y^ 2 and x = s + 3t, y = 2s - t, then& d^ 2 u ds^ 2...
If
u = x 2 + y 2 u = x^{2} + y^{2} u = x 2 + y 2 and
x = s + 3 t , y = 2 s − t , x = s + 3t, y = 2s - t, x = s + 3 t , y = 2 s − t , then&
d 2 u d s 2 \dfrac{d^{2}u}{ds^{2}} d s 2 d 2 u is equal to (where
t t t & is a constant)
Options A 12 12 12 B 32 32 32 C 36 36 36 D 10 10 10
Step-by-step solution
Given,
u = x 2 + y 2 , x = s + 3 t , y = 2 s − t u = x^{2} + y^{2}, x = s + 3t, y = 2s - t u = x 2 + y 2 , x = s + 3 t , y = 2 s − t Now ,
d x d s = 1 , d y d s = 2 … ( i ) d 2 x d s 2 = 0 , d 2 y d s 2 = 0 … ( i i ) \dfrac{dx}{ds} = 1, \dfrac{dy}{ds} = 2 \ldots ( i ) \dfrac{d^{2}x}{ds^{2}} = 0, \dfrac{d^{2}y}{ds^{2}} = 0 \ldots ( ii ) d s d x = 1 , d s d y = 2 … ( i ) d s 2 d 2 x = 0 , d s 2 d 2 y = 0 … ( ii ) Now,
u = x 2 + y 2 d u d s = 2 x d x d s + 2 y d y d s d 2 u d s 2 = 2 ( d x d s ) 2 + 2 x d 2 x d s 2 + 2 ( d y d s ) 2 + 2 y ( d 2 y d s 2 ) u = x^{2} + y^{2} \dfrac{du}{ds} = 2x\dfrac{dx}{ds} + 2y\dfrac{dy}{ds} \dfrac{d^{2}u}{ds^{2}} = 2{ \left(\dfrac{dx}{ds}\right)}^{2} + 2x\dfrac{d^{2}x}{ds^{2}} +2{ \left(\dfrac{dy}{ds}\right)}^{2} + 2y \left(\dfrac{d^{2}y}{ds^{2}}\right) u = x 2 + y 2 d s d u = 2 x d s d x + 2 y d s d y d s 2 d 2 u = 2 ( d s d x ) 2 + 2 x d s 2 d 2 x + 2 ( d s d y ) 2 + 2 y ( d s 2 d 2 y ) ⇒
d 2 u d s 2 = = 2 ( 1 ) 2 + 2 x ( 0 ) + 2 ( 2 ) 2 + 2 y ( 0 ) ( \dfrac{d^{2}u}{ds^{2}} ==2\left(1\right)2 + 2x\left(0\right) + 2\left(2\right)2 + 2y\left(0\right) ( d s 2 d 2 u == 2 ( 1 ) 2 + 2 x ( 0 ) + 2 ( 2 ) 2 + 2 y ( 0 ) ( from
( i ) a n d ( i i ) ) = 2 + 8 = 10 and ( i ) and ( ii ) ) = 2 + 8 = 10 \text{\text{ and } } ( i ) an d ( ii )) = 2 + 8 = 10 and
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