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Game Math Chapter 3 Solutions
Game Math Chapter 3 Solutions
Frank Luna
www.gameinstitute.com
Wednesday, December 15, 2004
1. Calculate the following:
a.
(
2
x
2
−
4
x
3
)
+
(
−
2
x
2
−
2
x
3
+
9
x
)
b.
(
x
+
x
2
+
x
3
)
×
(
−
2
+
3
x
+
5
x
5
)
c.
(
x
3
+
2
x
−
5
÷
(
x
+
1
a.
(2
x
2
−+−−+
4 ) ( 2
x
3
x
2
2
x
3
9 )
x
=−−−+
=− +
24229
69
x
2
x
3
x
2
x
3
x
x
3
x
b.
(
x
+
x
2
+
x
3
)
×
(
−
2
+
3
x
+
5
x
5
)
=−++−++−++
=+++++−
23 5 2 3 5 2 3 5
5553
x x
2
x
6
x
2
x
3
x
7
x
3
x
4
x
8
x
8
x
7
x
6
x x x
4
3
2
2
x
c.
(
x
3
+
2
x
−
5
÷
(
x
+
1
xx
xx x x
xx
x x
xx
x
x
−+
+++−
−−
−+−
++
−
−−
−
2
3
1
3
0
2
2 5
3
2
2
25
2
35
33
8
(
x
3
+−÷+=−++ +
2 ) ( )
x
x
x x
2
3
−
8
x
1
2. Approximate the square root function with a quadratic polynomial over the interval [0,
10].
Frank Luna
Page 1
www.gameinstitute.com
736239448.005.png
Game Math Chapter 3 Solutions
0, 0 , 5, 5 , 10, 10 . Given these three data points we can fit a
parabola through them. This parabola will be our quadratic polynomial approximation.
Note that it is important to select “good” points from the square root function that will
give us a good approximation. So we pick the start and end points so that the quadratic
polynomial agrees with the initial and end behavior; we also pick the middle point for our
third point so that our approximation agrees with the square root function in the middle.
So we want to find the three coefficients of the quadratic polynomial
=++. Our given data tells us three points that our parabola must pass
through, so we have the following conditions:
2
==⋅ +⋅+
(2) ()
P
00 0 0
a
2
b
c
==⋅ +⋅ +
(3) ()
P
5
5
a
5
2
b
5 0
P
10
==⋅ +⋅ +
10
a
10
2
b
10 0
Condition (1) tells us directly that c = 0. Conditions (2) and (3) for a system of two linear
equations, of which there are a myriad of ways to solve:
(4)
25 5
a +=
5
(5)
100 10
a += .
10
Multiplying equation (4) by –2 gives:
(6)
−−=− .
50 10
ab
2 5
Adding equations (5) and (6) yields:
50
a =−.
10 2 5
So,
(7)
a
=
10 2 5
50
−
.
Substituting (7) into (4) and solving for b we get:
25
10 2 5
−
+= ⇒= +
5
b
5
b
5
2 5
−
10
.
50
5
10
Thus our quadratic polynomial that fits the given data points is:
Frank Luna
Page 2
www.gameinstitute.com
To do this we’ll take three known points on the square root function over the
interval [0, 10]: ( ) ( ) ( )
()
P x ax bx c
(1) ()
736239448.006.png
Game Math Chapter 3 Solutions
()
=
10 2 5
−
 
+ +
5 2 5
−
10
x
 
50
5
10
 
If we graph this polynomial onto of ( )
f x
= we get the following:
So you can visually see how accurate our approximation is. Note that the polynomial
agrees exactly at the given data points ( ) ( ) ( )
0, 0 , 5, 5 , 10, 10 .
3. Find the general form of a monomial that passes through the point ( x , y ).
= . And the graph of a
monomial is a horizontal line at y = c . So the monomial that passes through the point ( x,
y ) is ()
A monomial is just a constant polynomial; that is, ( )
Px c
Px y
= .
4. Find the general form of a binomial that passes through the points ( x 0 , y 0 ) and ( x 1 , y 1 ).
= + , which is a line. So the problem wants
us to find the line that passes through the two given points. You equation of a line is
A binomial is of the form ( )
Px ax c
Frank Luna
Page 3
www.gameinstitute.com
2
Px
x
 
x
736239448.007.png 736239448.008.png
Game Math Chapter 3 Solutions
given by
yy mxx
−= − , where ( )
1
( )
1
x y is a fixed point on the line; m is the slope;
11
,
x y is any point on the line. The slope of the line that passes through the given
two points is:
,
m
=
yy
1
−
0
x x
−
1
0
Plugging that into our line equation yields:
yy
−= −⇔=
yy
1
−
0
( )
xx y
   
yy
−
0
x
−
yy
−
0
x y ymxb
+⇔=+
,
1
xx
−
1
xx
−
xx
−
1
1
   
1
0
1
0
1
0
where
b
=−
x y
xx
1
−
0
+
.
−
1
1
 
1
0
5. In order to map the texture onto the polygon, as you display each pixel of the polygon
you must find out which location in the bitmap corresponds to that pixel. Describe how
you might use binomials to solve this problem.
Use linear interpolation (see pg., 16) to interpolate across the texture map.
6. Suppose the angle between a surface and a light source is denoted by x , measured in
degrees, from 0 to 180. Develop a quadratic function to model the intensity of the surface
from 0 to 1 (assuming that at x = 90 degrees, the surface is at its brightest, and at x = 0 =
180 degrees, the surface is at its dimmest).
π π , and we want to fit a quadratic polynomial through them. So
from our data we have the following conditions:
2 , 1 , , 0
==⋅ +⋅+
(9) ( ) ( ) 2
P
00 0 0
a
2
b
c
P
π
21
==⋅ +⋅ +
a
π
2
b
π
20
(10) ()
P
π
==⋅ +⋅+
0
a
π π
2
b
0
Frank Luna
Page 4
www.gameinstitute.com
and ( )
1
1
   
yy
 
This problem is similar to problem #2. We are given the data points
( ) ( ) ( )
0, 0 ,
(8) ()
736239448.001.png 736239448.002.png
Game Math Chapter 3 Solutions
Condition (8) tells us that c = 0. Equations (9) and (10) give a system of two equations
and two unknowns. Solving (10) for a yields:
a
⋅ +⋅ =⇔⋅ =−⋅ ⇔=− .
2
b
0
a
π π
2
b
a
b
π
Substituting a into equation (9) and solving for b gives:

π π
2
−
b
π π
2
2
b
−
b
π π π
2
b
b
4
a
⋅ +⋅ =⇔ ⋅ + =⇔ + =⇔ =⇔ =
b
1
1
1
1
b
.
2
2
π
4
4
4
4
4
π
So our quadratic polynomial is,
()
Px
=− + .
4
x
2
π
π
2
To test our polynomial approximation, let us see if it agrees exactly with the given
data points.
P
()
=− + =
4
0
2
4
0 0
π
2
π
P π
( ) ( ) 2
2
2
=−
4
π π
2
+ =− + =
4
2
1 2 1
π
4
4
P π π π
()
=− + =− + =
2
44 0
π
2
The graph of this polynomial is as follows:
Frank Luna
Page 5
www.gameinstitute.com
π π

4
0
736239448.003.png 736239448.004.png

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