Graphics Programs Reference
In-Depth Information
81 m/s 2 rep-
resents the gravitational acceleration. If the projectile istohit the target at the 45
angle shown in the figure, determine v ,
where t is the time measured from the instantoflaunch, and g
=
9
.
θ
and the timeofflight.
29.
y
2
200 mm
1
3
x
The three angles shown in the figure of the four-bar linkage are relatedby
150 cos
θ 1 +
180 cos
θ 2
200 cos
θ 3 =
200
150sin
θ 1 +
180sin
θ 2
200 sin
θ 3 =
0
75 .
Determine
θ 1 and
θ 2 when
θ 3 =
Note that there aretwo solutions.
4.7 Zeroes of Polynomials
Introduction
Apolynomialofdegree n has the form
a 1 x n
a 2 x n 1
P n ( x )
=
+
+···+
a n x
+
a n + 1
(4.9)
where the coefficients a i may be realor complex. We will concentrate onpolynomi-
als with real coefficients, but the algorithms presentedinthis article also work with
complex coefficients.
The polynomialequation P n ( x )
=
0hasexactly n roots, which may be realor
complex. If the coefficients are real, the complexroots always occur in conjugate pairs
( x r +
where x r and x i are the real and imaginary parts, respectively. For
real coefficients, the number of real roots can be estimated from the rule of Descartes :
ix i ,
x r
ix i )
,
The number of positive, real roots equals the number of sign changes in the
expression for P n ( x ), or less by an even number.
The number of negative, real roots isequal to the number of sign changes in
P n (
x ), or less by an even number.
=
x 3
2 x 2
+
.
As an example, consider P 3 ( x )
8 x
27
Since the sign changes
twice, P 3 ( x )
=
0haseither twoor zero positive real roots. On the other hand,
P 3 (
x )
=−
x 3
2 x 2
+
8 x
+
27 contains a single sign change; hence P 3 ( x ) possesses
one negative realzero.
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