Page 21 - (1-1)eng

(1 - 1) The First Derivative of a Function

S.B. Example (13)
dy
Find of each of the following curves at the indicated value of t
dx

2
(a) x = ( t + 7 ) ( t – 2 ) and y = ( t + 1 ) ( t – 2 ) at t = 1

(b) x = 3t  2 and y = 4t 1 at t = 2
Solution:
2
(a)  x = ( t + 7 ) ( t – 2 ) , y = ( t + 1 ) ( t – 2 )

2
2
3
= t + 5 t – 14 = t – 2 t + t – 2
dx dy
2
 = 2 t + 5 = 3 t – 4 t + 1
dt dt
dy dy dt
 = 
dx dt dx

2
= ( 3 t – 4 t + 1 ) × 1
2t 5

3t 2  4t 1
=
2t  5

 dy  3   1 2 4   1 1
   = = 0
 dx t 1 2   1 5


(b)  x = 3t  2 , y = 4t 1

dx 3 dy 4 2
 = = =
dt 2 3t  2 dt 2 4t 1 4t 1

dy dy dt
 = 
dx dt dx

2 2 3t  2
= ×
4t 1 3

4 3t  2
=
3 4t 1

 dy  4 3   2 2 8
   = =
 dx t 2 3 4   2 1 9

Calculus 20 Unit (1)