Page 22 - (1-1)eng

(1 - 1) The First Derivative of a Function

S.B. Example (14)
3
2
Find the value of the parameter z at which the curve x = 2 z – 5 z – 4 z + 12
2
and y = 2 z + z – 4 has
(a) a horizontal tangent

(b) a vertical tangent
Solution:
3
2
2
 x = 2 z – 5 z – 4 z + 12, y = 2 z + z – 4
dx 2 dy
 = 6 z – 10 z – 4 = 4 z + 1
dz dz

dy dy dz
 = 
dx dz dx

1
= ( 4 z + 1 ) ×
6 z 2 10 z 4

4 z 1
=
6 z 2 10 z 4

(a) If the tangent is horizontal , then

dy 1
 0  4 z + 1= 0  z = –
dx 4

(b) If the tangent is vertical , then

dy
= undefined
dx

2
6 z – 10 z – 4 = 0  ( 2 )

2
 3 z – 5 z – 2 = 0

 ( 3 z + 1 ) ( z – 2 ) = 0

 3 z + 1 = 0 or z – 2 = 0

1
 z = – z = 2
3

Calculus 21 Unit (1)