Q19 Ex 3.7. Thomas Calculus Twelfth Edition Page 153 | University of Cambridge - Edubirdie

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

2023
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305

Pages:

4
Author:

Shahbaz A.

Q19 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 153 shahbaz ahmed August 2024 Q19 Use implicit differentiation to find x2 + y 2 = 1 Solution x2 + y 2 = 1 d 2 d [x + y 2 ] = [1] dx dx d 2 [x + y 2 ] = 0 dx d 2 d (x ) + (y 2 ) = 0 dx dx 1 dy dx and then d2 y dx2 d 2 d (x ) + (y 2 ) = 0 dx dx 2x + d 2 dy (y ) =0 dy dx 2y dy = −2x dx dy x =− dx y Or y dy = −x dx dx d dy [y ] = − dx dx dx y d dy dy dy [ ]+ = −1 dx dx dx dx 2 d2 y dy y 2 + [ ]2 = −1 dx dx Putting dy dx = − xy d2 y x y 2 + [− ]2 = −1 dx y Putting dy dx = − xy d2 y x2 y 2 + [ 2 ] = −1 dx y y x2 d2 y = −1 − dx2 y2 d2 y −y 2 − x2 y 2= dx y2 y 2 + x2 d2 y y 2 =− dx y2 Since x2 + y 2 = 1 d2 y 1 y 2 =− 2 dx y 3 d2 y 1 = − dx2 y3 4

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Q19 Ex 3.7. Thomas Calculus Twelfth Edition Page 153

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