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## Algebra (all content)

### Unit 13: Lesson 7

Direct and inverse variation- Intro to direct & inverse variation
- Recognizing direct & inverse variation
- Recognize direct & inverse variation
- Recognizing direct & inverse variation: table
- Direct variation word problem: filling gas
- Direct variation word problem: space travel
- Inverse variation word problem: string vibration
- Proportionality constant for direct variation

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# Inverse variation word problem: string vibration

Sal models a context about lengths of strings and the frequency of their vibrations! with an inverse variation equation. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- If we graph this variation we get a negative slope. I worked out the equation of the line and got y=-40x+840. Do we always get a negative slope with inverse variation and a positive slope with direct variation?(8 votes)
- We don't always get a positive slope with direct variation because the slope is the k in the equation Y=kx and that could be anything. If k is negitive than the slope is negitive, if k is positive the slope is positive.(11 votes)

- What is inverse variation?(2 votes)
- The simplest way I think of direct vs inverse variation is this way: in an equation, ask yourself what would the other variable do if one variable went up or down. If the other variable goes in the same direction, then the two variables vary directly. If they go in opposite directions, they vary inversely. 'Neither' applies when there is an addition or subtraction function in the equation, because the word 'proportion' no longer applies.(3 votes)

- given that x varies as y and inversely as zsquared and that x=12 when y=3 and z=3,find

{a}the eguation connecting x,y and z.

{b}the value of x when y=3 andz=4(3 votes) - 5 machines, when operated for 9 hours each day, can harvest a farm in 16 days. How many days would 8 machines take to harvest the same farm, if each machine is now operated for 10 hours each day?(1 vote)
- 5 machines working for 9 hours each do 45 machine-hours of work total. If the farm is harvested in 16 increments of 45 machine-hours, that means the farm has 16โข45=720 machine-hours of work to be done.

720 machine-hours divided by 8 machines is 90 hours of work. If the work is done for 10 hours a day, that gives 90/10=9 days.(5 votes)

- y=6x I don't understand this at all!(2 votes)
- ayo try resolving the problem it always helps(2 votes)

- Doesn't "constant" begin with a c, so why do we use K? (0:17) I really do not get it.(2 votes)
- It's arbitrary. There's nothing more to read into it. I've seen c,C,K,constant all used.(2 votes)

- what is 1 to the millionth [pwer(1 vote)
- 1 to almost any power besides 0 is still equal to 1(3 votes)

- I don't understand how this fits in the with function lessons. Is this an example of how a function can be used? Is it possible to fill in the constant K as the function..or something like that? Any help is appreciated!(1 vote)
- No, you can't fill in K as the function. Think of K as the slope of a line. In this problem think of l as a function of f, or f as a function of l. it works either way.(2 votes)

- How in the world does this apply to strings? #stuck(2 votes)
- I had read somewhere that there is a suitable formula including x2 and y2, and this really confuses me. Could someone tell me if there is such a thing ?(2 votes)

## Video transcript

We're told in this question
that on a string instrument, the length of a
string-- so let's call that l-- the
length of a string l varies inversely
as the frequency, so varies inversely
as the frequency. So l is going to be equal
to some constant times the inverse of the frequency. And I'll use f for
frequency-- the frequency of its vibrations. And then they tell
us the vibrations are what give the string
instruments their sound. That's nice. An 11-inch-- and it's
actually the vibrations of the string affecting
the air, and then the air compressions
eventually get to our eardrum. And that's actually what gives
us the perception of the sound. But we don't want to delve too
much into the physics of it. An 11-inch inches
string has a frequency. So 11-inch, so
this is its length. So the 11-inch string
has a frequency of 400 cycles per second. So this right here
is the frequency. A cycle per second is
also called a Hertz. Find the constant
of proportionality, and then find the frequency
of a 10-inch string. So they say an 11-inch string. So 11-inch string is
equal to some constant of proportionality times
1/400 cycles per second. So one over 400
cycles per second. I'll write second as sec. So to solve for the
constant of proportionality, we need to multiply both sides
by 400 cycles per second. Multiply the left hand side
by 400 cycles per second, and the left hand side
becomes 400 times 11. Well, 4 times 11 is 44. So 400 times 11 is 4,400. And then we have in our units,
just for out of interest, our units are cycles
times inches per second. So it's cycles times
inches in the numerator of our units divided by seconds. And that is equal to our
constant of proportionality. So we can say that the length is
equal to 4,400 times-- or 4,400 cycles times inches per second--
I want to get the units right-- per second, times 1
over the frequency. So we solved for our
constant of proportionality. And then we can use this
to find the frequency of a 10-inch string. So now we're talking
about a situation where our length is 10
inches, so 10 inches. So we get 10 inches
are equal to 4,400-- as you can imagine, these
units are a little cumbersome-- but 4,400 cycles times
inches per second times 1 over the frequency. And so we could do
a couple of things. We could just multiply
both sides of this equation by the frequency so that it
gets out of the denominator. So let's do that. Let's multiply both
sides by the frequency. And then we could also divide
both sides by 10 inches to get rid of this. And then we'll just have
frequency on the left hand side. So we divide both
sides by 10 inches. And then we divide
it by 10 inches. The left hand side, we're
just left with the frequency. On the right hand side, we
have 4,400 divided by 10. 4,400 divided by 10 is 440. And then you have cycles, inches
over second divided by inches. Well, the inches
cancel out with inches. And you're just left
with cycles per second. And then obviously, 1/f times
f cancels out to just being 1. So we get our frequency
when our string is when our string
is 10 inches long, our frequency has increased
to 440 cycles per second. From when it was 11 inches,
when it was slightly longer, our frequency was 400
cycles per second. When our string got little
bit shorter, one inch shorter, now our frequency increased
by 40 cycles per second.