By Brooke Norman
Day 5
Understanding Direct Variation
Objectives:
1- Learn what direction variation means
2- Learn how to calculate the constant
of variation, write a direct variation equation, and use this to solve for
values of x or y.
3- Apply this knowledge in writing a
direct variation model.
1-
Remind
the students what a constant of variation, k, is and how it is applied. Remind them that as a formula changes,
a constant stays the same. Show
them that in the algebraic form of y=kx is a model of a direct variation where
x and y vary directly according to the constant of variation, k. In other words, use k=3. Y would be 3 times what x would be. If x was 5 and k was 3, then y would be
15.
Y=kx
Y=3x
Y=3(5)
Y=15
2-
Now,
if you were given x and y and were told that they were in direct variation of
each other and you had to find the constant of variation, we would need to do a
little bit of manipulation. LetŐs
use x=3 and y= 12. Now we need to
solve for k. First, you plug in
what you know into the formula.
Y=kx
12=k(3)
12/3=k
k=4.
The constant of variation for this example is 4.
Now that we have our k, what would y be if x=6?
Y=4x
Y=4*6
Y=24
3-
How
can we apply this knowledge to write an equation? LetŐs use an example of: The distance a frog can jump varies directly with the length
of its legs. From the following
data, find the constant of variation and write a direct variation equation.
|
Length of
Frog's Legs (in.) |
4 |
3.5 |
3 |
2.5 |
4.5 |
2 |
|
Distance
of hop (in.) |
12 |
10.5 |
9 |
7.5 |
13.5 |
6 |
To
find the constant of variation, use the general equation and solve for k. I
will do one below.
Hop
distance = k(Frog legs)
Hop
distance / Frog legs = k
Choose one
of the data sets
I will use
the second set.
10.5/3.5=k
k= 3.
In this
example, the constant of variation is 3.
The new
equation of direct variation is
Hop
distance = 3(Frog legs)
Now, set up
some questions for the students to answer such as:
a-
What
would be the distance of a hop of a frog whose legs are 6.5 inches long?
b-
How
long would a frogŐs legs be if he jumped a distance of 14 inches?
To solve:
a-
Hop
distance = 3 (frog legs)
Hop distance = 3(6.5)
Hop distance = 19.5 inches
b-
Hop
distance = 3 (frog legs)
14= 3 (frog legs)
14/3= frog legs
4.66667 inches= length of frog legs
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