Download Lesson 5 Keystone - Adjective Noun Math

next
Key Stone
Problem…
Set 5
© 2007 Herbert I. Gross
next
Instructions for the Keystone Problem
You will soon be assigned five problems to
test whether you have internalized the
material in Lesson 5 of our algebra course.
The Keystone Illustration below is a
prototype of the problems you'll be doing.
Work out the problem on your own.
Afterwards, study the detailed solutions
we've provided. In particular, notice that
several different ways are presented that
could be used to solve the problem.
© 2007 Herbert I. Gross
next
As a teacher/trainer, it is important for
you to understand and be able to respond
in different ways to the different ways
individual students learn. The more ways
you are ready to explain a problem, the
better the chances are that the students
will come to understand.
© 2007 Herbert I. Gross
next
Keystone Illustration for Lesson 5
A ball projected vertically upward at a
speed of 160 feet per second, in the
absence of air resistance, reaches a
height of h feet at the end t seconds
according to the rule: h = 400 – 16(t – 5)2
(a) How high up is the ball at the end of 7
seconds?
© 2007 Herbert I. Gross
Answer: 336 feet
next
Solution for Part a:
h = 400 – 16 (4)
(2)
( 7t 2– 5 )2
To solve part (a) we replace t by 7 in the
formula.
Using our PEMDAS agreement, we do
what's inside the parentheses first, (7 – 5).
We next replace (2)2 by 4 to obtain
© 2007 Herbert I. Gross
next
Solution for Part a:
h = 336
400 feet
– 64
16 (4)
And since we multiply 16 by 4 before we
subtract, the equation becomes…
Finally subtracting 64 from 400, the
answer is 336.
h is measured in feet so the answer to
part (a) is 336 feet.
© 2007 Herbert I. Gross
next
Keystone Illustration for Lesson 5
A ball projected vertically upward at a
speed of 160 feet per second, in the
absence of air resistance, reaches a
height of h feet at the end t seconds
according to the rule: h = 400 – 16(t – 5)2
(b) How high up is the ball at the end of 3
seconds?
© 2007 Herbert I. Gross
Answer: 336 feet
next
Solution for Part b:
h = 400 – 16 ( -2)
t3 2– 5 )2
The procedure for solving part (b) is
exactly the same as the procedure for
solving part (a). Namely, we replace
t by 3 in the formula.
Using our PEMDAS agreement, we do
what's inside the parentheses first, (3 – 5).
© 2007 Herbert I. Gross
next
Solution for Part b:
h =
+
2
336
400 feet
– 16
64 ( 2)
4)
(-2)2 means -2 × -2; which by our rule for
multiplying two negative numbers is 4, so
we next replace (-2)2 by 4 to obtain…
We multiply 16 by 4, before we subtract.
Subtracting 64 from 400, the answer is 336.
h is measured in feet so the answer to
part (b) is 336 feet.
© 2007 Herbert I. Gross
next
Note
• The fact that the product of two negative
numbers is positive tells us that the square
of any signed number is non-negative.
More specifically, a signed number is either
positive, negative or 0.
© 2007 Herbert I. Gross
next
Note
 If we multiply a positive number by itself
the product will be positive.
 If we multiply a negative number by
itself the product will be positive.
 If we multiply 0 by itself the
product will be 0.
© 2007 Herbert I. Gross
next
Note
• With respect to this example, if we
replace t by 7, t – 5 = 2, and if we replace t
by 3, t – 5 = -2. While +2 ≠ -2, (+2)2 = (-2)2.
More generally if two numbers have the
same magnitude their squares are equal.
© 2007 Herbert I. Gross
next
Note
• Based on the above note, it is incorrect to
talk about the square root of a number. Every
positive number has two square roots. In
other words, for example, if (t – 5)2 = 4, t – 5
can be either 2 or -2.
When we talk about the square root of a
number we usually mean the positive square
root of the number. However as we shall see in
our next note, if we neglect the negative
square root, we miss part of the answer to the
present example.
© 2007 Herbert I. Gross
next
Note
• Because addition has nicer properties
than subtraction, a good approach might be
to rewrite h = 400 – 16 (t – 5)2 in the
equivalent form h = 400 + -16 (t – 5)2 and then
translate the formula into a verbal “recipe”.
Start with t
Step 1
Subtract 5
Step 2
Step 3 Square the result
Step 4 Multiply by -16
Add 400
Step 5
Step 6 The answer is h.
© 2007 Herbert I. Gross
t

t–5

(t – 5)2

-16(t – 5)2

 400 + -16(t – 5)2
 h = 400 + -16(t – 5)2
next
Note
• In the present example we found that when t = 3
or t = 7, h = 336. Let's now undo the above recipe
and show why this occurred when t = 3 and t = 7.
Recall that when we “undo” a recipe we start with
the last step and replace each operation by the one
that “undoes" it. In this case we see that…
Start with 336
Step 1
Subtract 400
Step 2
Divide by -16
Step 3
Step 4 Take the (2) square root(s).
Add 5
Step 5
The answer is t.
Step 6
© 2007 Herbert I. Gross






336
336-64
– 400
-64+÷
4 -16
+2 or
√4 -2
2 +75 or
or -23+ 5
t = 7 or 3
next
• Notice that without the knowledge that a
positive number has two square roots,
step 4 would have read, “Take the square
root of 4”, and the answer would have
been only +2.
Take
thethe
(2) square
square root.
root(s). 
Step 4 Take
+2+√4
or
2 -2
and this would lead to our missing that
t = 3 was also an answer.
© 2007 Herbert I. Gross
next
The reason that
two different
values of t produce
the same value for
h is that the ball is
at a given height
twice, once on the
way up, and once
on the way down.
© 2007 Herbert I. Gross
5s
400 feet
5s
4s
384 feet
6s
3s
336 feet
7s
2s
256 feet
8s
1s
144 feet
9s
0s
0 feet
10s
next
Note
• The fact that the square of a signed number
can never be negative gives us additional
information that is contained in the formula.
Namely…
(t – 5)2 has to be non-negative; and the
only time it can be 0 is if t – 5 = 0; that is,
if t = 5.
When t = 5, 16(t – 5)2 =0.
© 2007 Herbert I. Gross
next
Note
Since 16 (t – 5)2 can never be negative,
whenever t ≠ 0, we are subtracting a
positive number from 400. In other words if
t represents any number other than 0, h,
which equals 400 – 16(t – 5)2, is less that 400
feet.
Therefore, the conclusion is that the ball
reaches its greatest height (400 feet)
when the time is 5 seconds.
•
© 2007 Herbert I. Gross
next
Note
• Notice that 3 seconds is 2 seconds
before the ball reaches its greatest height
and that 7 seconds is 2 seconds after the
ball reaches its maximum height.
So if we think of 5 seconds as being our
reference point, 3 seconds would be
represented by -2, and 7 seconds would be
represented by +2, if t = 5.
In the above context -2 is just as
meaningful as +2.
© 2007 Herbert I. Gross
next
Summary
The use of “profit and loss", “increase and
decrease”, “below zero and above zero”
give us good ways to visualize signed
numbers. At the same time, however, they
eliminate the need for us to deal with
positive and negative numbers per se. That
is we can talk about a $7 loss rather than a
transaction of -$7, etc.
© 2007 Herbert I. Gross
next
Moreover, even if we elect to use the terms
“profit” and “loss” it would be difficult to
give a physical reason as to why the
product of two negative numbers is
positive.
For example we can interpret 3 × -2 = -6 by
saying that, if we have a $2 loss three
times, the net result is a $6 loss. However
in looking at -3 × -2, it makes little sense to
talk about a $2 loss “negative three” times
or a $3 loss “negative two” times.
© 2007 Herbert I. Gross
next
However in dealing with formulas as
we did in the keystone exercise, we
see why it is important to have a
mathematical definition of signed
numbers that transcends any
particular real-life model.
© 2007 Herbert I. Gross