- Find the equations of direct variation, inverse variation, and combined variation.
- Solve applied problems involving variation.

[SLIDE 1]

Direct Variation.

y varies directly as x, or y is directly proportional to x.

This situation can be modeled with the following function:

`f(x) = kx` or `y = kx`. Where `k > 0`.

And k is called the constant of proportionality or the variation constant.

The graph of a direct variation is a diagonal line through the origin. And notice that as the x-values are increasing, the y-values are also increasing.


[SLIDE 2]

Find the equation of variation if y varies directly as x, and `y = 25` when `x = 3`. So since y varies directly as x, we write the formula `y = kx`. And we need to solve for the constant of proportionality by plugging in 25 for y and 3 in for x. So `25/3` is the constant of proportionality, k, and then the equation of variation is `y = 25/3x`.


[SLIDE 3]

Health experts believe that a dog's weight, W, varies directly with the number of grams of protein, P, that should be included in its diet each day. A 30 pound dog requires 25 grams of protein each day. How many grams of protein should a 60 pound dog eat each day?

So since W varies directly with P, we can setup a direct variation equation: `W = k*P`. And we're given two values, 30 pounds and 25 grams of protein. And we can solve for the constant of proportionality, k. It's `30/25`, which can be simplified to `6/5`.

So our direct variation equation is `W = 6/5*P`.

And now they want us to find out how much protein a 60 pound dog should eat. So we've got to plug 60 in for W. `60 = 6/5*P` or `60*5/6 = P`. 60 divides by 6, 10 times, and we're left with `50 = P`.

So a 60 pound dog needs 50 grams of protein each day.


[SLIDE 4]

Inverse Variation.

y varies inversely as x, or y is inversely proportional to x.

This situation can be modeled with the following function:

`f(x) = k/x` or `y = k/x`, where `k > 0`. And k is called the constant of proportionality or the variation constant.

This is the graph of an inverse variation. Notice that as the x-values are increasing, the y-values are decreasing.


[SLIDE 5]

Find the equation of variation if y varies inversely as x and `y = 2` when `x = 15`.

So since y varies inversely as x, we setup the equation `y = k/x`. And we need to solve for k. We're given y is 2, when x is 15.

So `30 = k`, the constant of proportionality. That means `y = 30/x`. This is our equation of variation.


[SLIDE 6]

The time needed to drive a certain distance varies inversely with the speed of the vehicle. If it takes 7 hours at a speed of 80 miles per hour to drive a fixed distance, how long will it take to travel the same distance at a speed of 35 miles per hour.

So let's let `T = "time"` and `S = "speed"`. And since the time needed to drive a certain distance varies inversely with the speed of the vehicle, our inverse variation equation is `T = k/S`.

We're given that it takes 7 hours at a speed of 80 miles per hour to travel a fixed distance. So `T = 7` and `S = 80`. Now let's solve for our constant of proportionality, k. `7 = k/80` or `560 = k`. So our formula is `T = 560/S`.

So what is T when S is 35? `T = 560/35`, therefore, `T = 16" hours"`. So it takes 16 hours to travel the same distance at 35 miles per hour.


[SLIDE 7]

Combined Variation.

y varies directly as the nth power of x if `y = kx^n`.

y varies inversely as the nth power of x if `y = k/(x^n)`.

And y varies jointly as x and z if `y = kxz`. 

k is a positive number called the constant of proportionality or the variation constant.


[SLIDE 8]

Find the equation of variation if y varies directly as the square of x, and `y = 0.25` when `x = 0.1`.

So y varies directly as the square of x, and y is 0.25, when x is 0.1. So the constant of proportionality, k, is equal to `0.25/(0.1)^2` or `k = 25`.

So `y = 25*x^2`.


[SLIDE 9]

Find the equation of variation if y varies jointly as x and z, and `y = 64` when `x = 2` and `z = 8`.

`y = k*x*z` and `y = 64` when x is 2 and z is 8. So `64/16 = k` and `4 = k`.

So `y = 4*x*z`.


[SLIDE 10]

Find the equation of variation if y varies jointly as x and z and inversely as the product of w and the square of p. Given that `y = 3/14` when `x = 2`, `z = 10`, `w = 7`, and `p = 8`.

So let's start with "y varies jointly as x and z." That means `y = k*x*z`. "And inversely as the product of w and the square of p." So that means that the other stuff which varies inversely, is on the bottom of our fraction. `w*p^2`.

Now, let's plug in the information, `3/14` for y, 2 for x, 10 for z, 7 for w, and 8 for p.

So, `3/14 = (20k)/448` or `k = 3/14*448/20`, which is `24/5`. So the equation of variation is `y = 24/5*(xz)/(wp^2)`.


[SLIDE 11]

The velocity, V, with which a fluid flows in a circular pipe varies directly with the volume of the liquid, q, and inversely as the square of the diameter, d, of the pipe. So let's setup our equation.

`V = (kq)/(d^2)`.

Suppose that V is 9.8 feet per second when q is 600 gallons per minute and d is 5 inches.

`9.8 = (k*600)/(5^2)`. So `k = 49/120`. And our equation becomes `V = 49/120*q/(d^2)`.

Now how much slower would the fluid flow if the volume of fluid stayed the same, but the diameter of the pipe was tripled?

So, if the diameter of the pipe was tripled, that means `5*3`, so our new diameter is 15. And then q stays at 600. So `V = 49/120 * 600/(15^2)`. So the velocity is equal to about 1.09 feet per second. So they want to know how much slower the fluid flows if the volume of the fluid stayed the same, but the diameter of the pipe was tripled. So since they want to know how much slower, we have to take `9.8 - 1.09` to get our difference. `9.8 - 1.09 = 8.71" feet per second"`.

So the fluid flows 8.71 feet per second slower when the diameter of the pipe increased 3 times.