Q25 Ex 3.7. Thomas Calculus Twelfth Edition Page 153 | University of Cambridge - Edubirdie

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

2023
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216

Pages:

3
Author:

Shahbaz A.

Q 25 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 153 shahbaz ahmed August 2024 If x2 + y 2 = 16, find the value of d2 y dx2 at the point (2,2) Solution x2 + y 2 = 16 d 2 d [x + y 2 ] = [16] dx dx d d 2 [x ] + y 2 ] = 0 dx dy 2x + d 2 dy [y ] =0 dy dx 2x + 2y dy =0 dx dy = −2x dx dy y = −x dx 2y 1 dy x =− dx y dy dx = − xy Again using the result dy = −x dx d dy dx [y ] = − dx dx dx dy dy d dy = −1 y [ ]+ dx dx dx dx y d2 y dy y 2 + [ ]2 = −1 dx dx Putting dy x =− dx y x d2 y y 2 + [− ]2 = −1 dx y y d2 y x2 + = −1 dx2 y 2 d2 y x2 y 2 = −1 − 2 dx y 2 x2 d2 y y 2 = −[1 + 2 ] dx y d2 y y 2 + x2 y 2 = −[ ] dx y2 Putting x2 + y 2 = 16 d2 y 16 y 2 =− 2 dx y 16 d2 y =− 3 2 dx y Putting y=2 16 d2 y = − = −2 dx2 23 3

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Q25 Ex 3.7. Thomas Calculus Twelfth Edition Page 153

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