Page 42 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
Solution Solution
When y = –2, From y = 3x – 9x + 4
2
y = 6 + 2x y = 6 + 2x dy = 6x – 9
x x dx
–1
–2 = 6 + 2x = 6x + 2 At the point (2, –2),
x 1 3 2 dy
–2x = 6 + 2x At the point – , –2 , Gradient of tangent, m = dx = 6(2) – 9
2
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1
4x = –6 \ dy = –6x –2 = 3
x = – 3 dx 6 Equation of tangent line,
2 = – 2 y – y = m (x – x )
1
1
1
1 – 3 2 2 y – (–2) = 3(x – 2)
y + 2 = 3x – 6
= – 24 y = 3x – 8
9
= – 8 Since m × m = –1
2
1
3 3 × m = –1
2
Try Question 1 in ‘Try This! 2.4’ Therefore, the gradient of normal, m = – 1
2 3
Equation of normal line,
B Determining the equation of y – y = m (x – x )
1
1
2
tangent and normal to a curve at a 1
point y – (–2) = – (x – 2)
3
1. The equation of the tangent to a curve y = f(x) y + 2 = – x + 2
1
at a given point (x , y ) is determined by using 3 3
1
1
the formula y – y = m (x – x ), such that y = – x + – 2
1
2
1
1
1
dy 3 3
m = = f'(x ).
1 dx 1 1 4
= – x –
3 3
y
y = f(x)
(x y ) C Solving problems involving tangent
1, 1
x and normal
O tangent
1. There are a few things involved in the problems
involving the tangent and normal, which are
2. The equation of the normal to a curve y = f(x) at (a) problems about the curve y = f(x) such that
a given point (x , y ) on the curve is determined the information about the point and the
1
1
by using the formula y – y = m (x – x ), such that equation of tangent or normal are given, Form 5
1
2
1
m × m = –1. (b) problems about finding the point (x , y )
1
2
1
1
such that the information about the curve
y y = f(x) and the equation of tangent or
m 2 m 1
normal are given,
(x y ) (c) problems which involve the equations of
1, 1
Normal tangent or normal or both such that the
x
O information about the curve y = f(x) and the
point (x , y ) are given.
1
1
13 2. Normally, the problems involving tangent and
Find the equations of the tangent and normal to the normal involve layers of systematic solving
curve y = 3x – 9x + 4 at the point (2, –2). steps.
2
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