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

2023
Views:

407

Pages:

3
Author:

Shahbaz A.

Q 27 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 154 shahbaz ahmed August 2024 Find slope of the curve at the given points y 2 + x2 = y 4 − 2x at (-2,1) and (-2,-1). Solution y 2 + x2 = y 4 − 2x d 4 d 2 [y + x2 ] = [y − 2x] dx dx d 2 d 2 d 4 d y + (x ) = (y ) − (2x) dx dx dx dx d 2 dy d 2 d 4 dy dx y + (x ) = (y ) −2 dy dx dx dy dx dx 2y dy dy + 2x = 4y 3 −2 dx dx 1 2y 2y dy dy = 4y 3 − 2 − 2x dx dx dy dy − 4y 3 = −2 − 2x dx dx [2y − 4y 3 ] dy = −2x − 2 dx 2y[1 − 2y 2 ] dy = −2[x + 1] dx y[1 − 2y 2 ] dy = −[x + 1] dx dy x+1 =− dx y(1 − 2y 2 ) Putting x=-2,y=1 in equation (1) −2 + 1 dy =− dx 1[1 − 2(12 )] −1 dy = −( ) = −1 dx −1 If, m denotes the slope, then slope at the point (-2,1), m=-1, at the point (-2,1) Similarly putting x=-2,y=-1 in equation (1) dy −2 + 1 =− dx (−1)[1 − 2(−1)2 ] dy −1 =− dx (−1)[1 − 2(1)] −1 dy =− dx (−1)(−1) dy −1 = −[ ] dx 1 2 (1) dy = −(−1) = 1 dx Similarly, if m denotes the slope, then slope at the point (-2,-1) m=1, at the point (-2,-1) The figure below shows the tangents at the points (-2,1) and (-2,-1) 3