University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

294

Pages:

4
Author:

Shahbaz A.

Q 26 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 154 shahbaz ahmed August 2024 If xy + y 2 = 1, find the value of d2 y dx2 at the point (0,-1) Solution d d [xy + y 2 ] = (1) dx dx d d (xy) + (y 2 ) = 0 dx dx dx dy d dy y + x + (y 2 ) =0 dx dx dy dx dy dy + 2y =0 dx dx dy dy x + 2y = −y dx dx dy [x + 2y] = −y dx dy y =− dx x + 2y y+x 1 dy dx y = − x+2y Again differentiating implicitly, the equation dy y =− dx x + 2y (x + 2y) dy = −y dx dy dy d [(x + 2y) ] = − dx dx dx dy d d dy dy [(x + 2y)] + (x + 2y) ( ) = − dx dx dx dx dx d d2 y dy dy dx [ + (2y)] + (x + 2y) 2 + =0 dx dx dx dx dx dy dy d2 y dy [1 + 2 ] + (x + 2y) 2 + =0 dx dx dx dx dy 2 d2 y dy dy + 2( ) + (x + 2y) 2 + =0 dx dx dx dx dy dy 2 d2 y 2 + 2( ) + (x + 2y) 2 = 0 dx dx dx 2 Putting dy y =− dx x + 2y y d2 y y 2 ] + 2(− ) + (x + 2y) 2 = 0 2[− x + 2y x + 2y dx 2y 4y 2 d2 y − + + (x + 2y) 2 = 0 x + 2y (x + 2y)2 dx 4y 2 d2 y 2y − ] + (x + 2y) 2 = 0 −[ x + 2y (x + 2y)2 dx −[ 2y(x + 2y)2 − 4y 2 (x + 2y) d2 y ] + (x + 2y) =0 (x + 2y)2 dx2 d2 y 2y(x2 + 4y 2 + 4xy) − 4y 2 (x + 2y) ] + (x + 2y) 2 = 0 −[ (x + 2y)2 dx −[ 2yx2 + 8y 3 + 8xy 2 − 4y 2 x − 8y 3 ) d2 y ] + (x + 2y) =0 (x + 2y)2 dx2 2x2 y + 8xy 2 − 4xy 2 ) d2 y ] + (x + 2y) 2 = 0 −[ (x + 2y)2 dx 2x2 y + 4xy 2 ) d2 y −[ ] + (x + 2y) 2 = 0 (x + 2y)2 dx 3 d2 y 2x2 y + 4xy 2 ) (x + 2y) 2 = [ ] dx (x + 2y)2 d2 y 2x2 y + 4xy 2 ) =[ ] dx2 (x + 2y)3 d2 y 2xy(x + 2y) = dx2 (x + 2y)3 2xy d2 y = [ ] dx2 (x + 2y)2 Putting x=0,y=-1 d2 y =0 dx2 4