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Game Math Chapter 4 Solutions
Game Math Chapter 4 Solutions
Frank Luna
www.gameinstitute.com
Monday, January 3, 2004
1. Given the following right triangle information, find x and y .
θ = , where y is the length of the
opposite edge, x is the length of the adjacent edge, and r is the length of the hypotenuse.
Using these definition with our given data we can solve for x and y like so:
sin
θ = and ( )
yr
cos
x r
sin

=⇔= ≈
y
y
10 sin

π
3.82683
80

8
cos

=⇔= ≈
x
x
10 cos

9.23879
80

2. Given the following right triangle information, find x and r .
Frank Luna
Page 1
www.gameinstitute.com
From definition, recall that ()
π



π
π



8
736239434.008.png 736239434.009.png 736239434.010.png
Game Math Chapter 4 Solutions
θ = , where y is the length of the
opposite edge, x is the length of the adjacent edge, and r is the length of the hypotenuse.
Using these definition with our given data we can solve for x and r like so:
sin
θ = and ( )
yr
cos
x r
sin
 =⇔= ≈
3
r
3
7.83937
( )
8
r
sin
π
8
cos

=
x
⇔=
x
7.83937 cos

π
≈
7.24263

7.83937

8
3. Given the following right triangle information, find y and r .
θ = , where y is the length of the
opposite edge, x is the length of the adjacent edge, and r is the length of the hypotenuse.
Using these definition with our given data we can solve for y and r like so:
sin
θ = and ( )
yr
cos
x r
cos
 =⇔= ≈
7
r
7
7.57674
( )
8
r
cos 8
π
sin

=
y
⇔=
y
7.57674 sin

π
≈
2.89949

7.57674

8
4. Suppose you have n functions, labeled f 1 ( r , t ), f 2 ( r , t ), ..., f n ( r , t ), and that the i th
function models the height of a water ripple at time t at a distance r away from the impact
point ( x i , y i ). Assuming all of the ripples occur in a single pond, write down an expression
describing the height of an arbitrary point ( x , y ) at time t . (The distance between two
points ( a , b ) and ( c , d ) is
(
a
−
c
)
2
+
(
b
−
d
)
2
.)
Frank Luna
Page 2
www.gameinstitute.com
From definition, recall that ()
π

π


8
From definition, recall that ()
π

π


8
736239434.011.png 736239434.001.png
Game Math Chapter 4 Solutions
The key ideas here are that there are n ripples in the pond; the i th ripple originates from a
point ( )
i
,
i
f , with just this one
ripple, the height of an arbitrary point ( )
x y in the pond is given by
f
( ) ( )
xx yy t
−+−
2
2
,
.
1
1
1
x y is a function of the distance of the point from the
origin of the ripple. This makes sense intuitively because we expect the ripple to be
larger near the source and then diminish in magnitude with distance.
Now, our pond does not have one ripple, but n ripples. So how to handle n
ripples? Well you can think of a ripple as a wave, and there is a physics principle called
the superposition of waves , which states that when you have multiple waves, you can sum
the wave functions to yield a net wave function. So, in order to get the height of an
arbitrary point with n ripples (waves), we sum the heights given from all n ripples to
obtain a net height:
( ) ( ) ( )
=
f
xx yyt
− + − +
2
2
,
 
f
( ) ( )
− + − +
2
2
,
1
1
1
 
1
2
2
...
+
f
( ) ( )
x x
− + −
2
y y t
2
,
n
n
n
Observe that since the ripples originate from different point sources, the distance from
( )
5. Polynomials can be used to represent can be used to approximate trigonometric
functions (or even represent them perfectly, for polynomials of infinite degree).
Determine which of the trig functions is approximated by the following polynomial:
x
2
x
4
x
6
f
(
x
)
=
1
−
+
−
1
×
2
1
×
2
×
3
×
4
1
×
2
×
3
×
4
×
5
×
6
Graphing the function yields the following:
Frank Luna
Page 3
www.gameinstitute.com
x y ; and the i th function f models the height of the i th water ripple as a
function of time and distance. So for example, take the first ripple
Notice the height of the point ( )
f xyt
,,
xx yyt
 
x y to each ripple source point is different.
736239434.002.png
Game Math Chapter 4 Solutions
From the graph, we see that the function looks a lot like the cosine function for the
interval [-2, 2].
Students of calculus will recognize the function as the first four terms of the
Maclaurin polynomial for the cosine function. (Students who have not studied Calculus
can skip to the next exercise solution.) The nth Maclaurin polynomial for f is defined as
follows:
n
f
k
() ( ) ( ) ( )
f
0
f
(1)
0
f
( 2)
0
f
( )
()
0
()
∑
Px
=
x
= +
x
+
x
+ +
x
n
k
!
0!
1!
2!
n
!
k
=
0
For cosine we have the following derivatives:
Frank Luna
Page 4
www.gameinstitute.com
n
0
k
2
...
736239434.003.png 736239434.004.png
Game Math Chapter 4 Solutions
f
() ()
() ()
() ()
() ()
() ()
() ()
() ()
= =
f
(1)
0
=− =
sin 0 0
f
(2)
0
=− =−
cos 0
1
f
(3)
0 in00
= =
f
(4)
0 s01
= =
f
(5)
0
=− =
sin 0 0
f
(6)
0
=− =−
cos 0
1
Plugging these values into the Maclaurin polynomial gives:
1
0! 2! 4! 6!
x
2
x x
6
Px =−+−,
which was the function given in the exercise.
Incidentally, we noticed that the given function only approximately agreed with
the cosine function in the interval [-2, 2]. By adding more terms to the Maclaurin
polynomial, say 15, you get a wider interval of agreement.
6. Given Pythagorean's theorem and what you know of trigonometry, determine the area
of the triangle shown in the figure (the area of a triangle is used to compute its mass in
some physics simulations).
From definition we have ()
sin
θ
=⇔= . Also by definition, we
hx h x
sin
( )
have ()
cos
θ
=⇔= . From the theorem of Pythagoras, we have
ax a x
cos
()
θ
Frank Luna
Page 5
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0 s01
4
()
θ
736239434.005.png 736239434.006.png 736239434.007.png

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