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

2023
Views:

459

Pages:

4
Author:

Shahbaz A.

Q 28 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 154 shahbaz ahmed August 2024 Find slope of the curve at the given points (x2 + y 2 )2 = (x − y)2 at(1,0) and (1,-1) Solution (x2 + y 2 )2 = (x − y)2 d d [(x2 + y 2 )2 ] = [(x − y)2 ] dx dx d d 2(x2 + y 2 ) [(x2 + y 2 )] = 2(x − y) [(x − y)] dx dx d dx dy d (x2 + y 2 )[ (x2 ) + (y 2 )] = (x − y)[ − ] dx dx dx dx d dy dy (x2 + y 2 )[2x + (y 2 ) ] = (x − y)[1 − ] dy dx dx (x2 + y 2 )[2x + 2y dy dy ] = (x − y)[1 − ] dx dx 1 dy dy = x − y − (x − y) dx dx dy dy = x − y − 2x(x2 + y 2 ) 2y(x2 + y 2 ) + (x − y) dx dx dy [2y(x2 + y 2 ) + (x − y)] = x − y − 2x(x2 + y 2 ) dx 2x(x2 + y 2 ) + 2y(x2 + y 2 ) dy x − y − 2x(x2 + y 2 ) = dx [2y(x2 + y 2 ) + (x − y)] (1) Putting x=1,y=0 in equation (1) 1 − 0 − 2(1)(12 + 02 ) dy = dx [2(0)(12 + 02 ) + (1 − 0)] dy 1 − 2(1 + 0) = dx [2(0)(1 + 0) + (1)] 1 − 2(1) dy = dx [0(1) + (1)] dy 1−2 = dx [0 + 1] dy −1 = dx 1 dy = −1 dx If m denote the slope, then slope of the curve at the point 2 (1,0) m=-1 Similarly, putting x=1,y=-1 in equation (1) dy 1 − (−1) − 2(1)[12 + (−1)2 ] = dx 2(−1)[12 + (−1)2 ] + [1 − (−1)] dy 1 − (−1) − 2(1)[12 + (−1)2 ] = dx 2(−1)[12 + (−1)2 ] + [1 − (−1)] dy 1 + 1 − 2([1 + 1] = dx −2[1 + 1] + [1 + 1] dy 2 − 2(2) = dx −2(2) + 2 dy 2−4 = dx −4 + 2 −2 dy = dx −2 dy =1 dx Hence if m denotes the slope ,then, slope of the curve at the point (1,-1) will be m=1 3 Graph below shows the tangent lines at the points (1,0) and (1,-1) 4