**Unformatted text preview: **\5\\ To Find_ the Points where the_Curve has
a horizontal_ tangent line _
_dys
dyl do_
dx /do = 0
dy
Fle) Sin (A ) + F 10 ) copp)
d x
F ( A ) COs ( D ) _ F (@ ) sin ( b )
< =\ FQ ) = 5 CSC(DL = _
Sun( D )
FIA )
- 5 Cos(D )
sin = ( D )
5_ Cos(D )_ siAter_
Cos(D)
dy
since
+
Sin?(@)
O
5
-- Sin ( A )
X P
- 5 cos ( A )_ Cosle)_
sin? ( D)
Sin CAL
5 COS ( D ) +
5 COS ( Q) - O
sin A
sing
- 5 Cos (0 ) sing + 5 Cos ( 4 ) Sin ( D) = Q
Sins ( Q )
- 5 Cos ( Q ) Sin ( 6 ) + 5 Cos ( D ) = Q
- 5
toslot- (1 - 5ing ) = 0
_5 {es( 4) = 6
L_ since) _ Of
# = It
3 TT
T 2)
D = [
7
So the Points are
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- Spring '08
- KijuwaIthink
- Trigraph, #, 6 l, toslot