PA_M7_S5_T2_Inverse Variation Transcript
Key Phrases
Translation
“y varies inversely as x”
“y is inversely proportional to x”
I want to take you back to the direct variation statement where we had
distance was equal to the rate times the time, d = r*t. If I take this
direct variation and I divide both sides by t, I end up with this
variation; that the rate of speed is inversely proportional to the time
if I know what my distance is. So that I can calculate an average rate of
speed needed to go through a certain distance if I know how much time I
have to spend.
So here's an example; I need to cover 240 miles and I need to do it in
four hours. By setting up this direct variation I can calculate my
average speed, 240 divided by 4 means that I have to travel at an average
speed of 60 miles per hour. My proportionality constant, or my constant
of variation, is my distance, so my r is going to be equal to my D over t
gives me my relationship between the variable r and the variable t. If
240 is my proportionality constant. My formula becomes r = 240 (in
miles)/t. This way if I'm given time is 5 hours, I can plug that in and
find out that my average rate of speed to go 240 miles is just given by
this ratio; 240 mi/5 hr, which is 48 mi/hr.
Here is another example; if y varies inversely as x and y = 5
when x = 2, find the constant of proportionality and then
find the equation that relates the two variables. This is a
typical problem that you'll see. So y varies inversely with
x, or as x, tells me I have a constant of proportionality and
x is in the denominator. This is my general statement. So now
I take what I know; y = 5 when x = 2, I set that up, 5 = k/2.
I solve this for k by multiplying both sides by 2 and I
get that 10 is equal to k. Now I can go back and set up
my general formula, y = 10/x, and that's my equation that
relates y and x as my two variables.
Here is another example. The weight of an object varies inversely as the
square of its distance from the center of the earth. If an object weighs
100 pounds on the surface of the earth, which is about 4000 miles to the
center, then how much will it weigh at 1000 miles above the earth's
surface?
Well, I'm going to set it up
and say y = k/x. That's my
inverse variation form that my
formula will take on.
Now I'm going to use what I
know, that I have an object
that weighs 100 pounds at the surface. That's 100 pounds, that's my y,
and my k goes up on top and my x is the distance from the center of the
earth which is about 4000 miles. So I'm going to set up my
proportionality equation. I'm going to let y be my weight, and I'm going
to let x be the distance to the center of the earth. So now recall that I
said y was inversely proportional to the square of the distance to the
center of the earth. So I'm going to square the x in the bottom. This is
my generic statement. Now I'm going to look at my given information. I'm
told that my object that weighs 100 pounds at the surface of the earth
and the surface of the earth is 4000 and I'm going to square that.
I want to solve this for k and I find out that k is equal to
.
It's usually easier to write things like this in scientific notation, so
that I have it nice and neatly placed here. So now I want to find how
much this object is going to weigh when it's 1000 miles above the surface
of the earth, so I'm going to say y =
that I found out was my k,
over 1000 miles above the earth gives me
4000 miles to the
surface plus 1000
miles above the
surface squared which
is going to be
/
. I do that
arithmetic and I find
that my object will
weigh 64 pounds.
This is how I set up and solve inverse variation problems.