# 5.8 Modeling using variation  (Page 2/14)

 Page 2 / 14

Do the graphs of all direct variation equations look like [link] ?

No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through $\text{\hspace{0.17em}}\left(0,0\right).$

The quantity $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly with the square of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}y=24\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 4.

$\frac{128}{3}$

## Solving inverse variation problems

Water temperature in an ocean varies inversely to the water’s depth. The formula $\text{\hspace{0.17em}}T=\frac{14,000}{d}\text{\hspace{0.17em}}$ gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

If we create [link] , we observe that, as the depth increases, the water temperature decreases.

$d,\text{\hspace{0.17em}}$ depth $T=\frac{\text{14,000}}{d}$ Interpretation
500 ft $\frac{14,000}{500}=28$ At a depth of 500 ft, the water temperature is 28° F.
1000 ft $\frac{14,000}{1000}=14$ At a depth of 1,000 ft, the water temperature is 14° F.
2000 ft $\frac{14,000}{2000}=7$ At a depth of 2,000 ft, the water temperature is 7° F.

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations .

For our example, [link] depicts the inverse variation    . We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula $\text{\hspace{0.17em}}y=\frac{k}{x}\text{\hspace{0.17em}}$ for inverse variation in this case uses $\text{\hspace{0.17em}}k=14,000.\text{\hspace{0.17em}}$

## Inverse variation

If $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are related by an equation of the form

$y=\frac{k}{{x}^{n}}$

where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a nonzero constant, then we say that $\text{\hspace{0.17em}}y$ varies inversely    with the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ power of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ In inversely proportional    relationships, or inverse variations , there is a constant multiple $\text{\hspace{0.17em}}k={x}^{n}y.\text{\hspace{0.17em}}$

## Writing a formula for an inversely proportional relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

Recall that multiplying speed by time gives distance. If we let $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ represent the drive time in hours, and $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ represent the velocity (speed or rate) at which the tourist drives, then $\text{\hspace{0.17em}}vt=\text{distance}\text{.}\text{\hspace{0.17em}}$ Because the distance is fixed at 100 miles, $\text{\hspace{0.17em}}vt=100\text{\hspace{0.17em}}$ so $t=100/v.\text{\hspace{0.17em}}$ Because time is a function of velocity, we can write $\text{\hspace{0.17em}}t\left(v\right).$

$\begin{array}{ccc}\hfill t\left(v\right)& =& \frac{100}{v}\hfill \\ & =& 100{v}^{-1}\hfill \end{array}$

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that time varies inversely with velocity.

Given a description of an indirect variation problem, solve for an unknown.

1. Identify the input, $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ and the output, $\text{\hspace{0.17em}}y.$
2. Determine the constant of variation. You may need to multiply $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the specified power of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.

## Solving an inverse variation problem

A quantity $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies inversely with the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}y=25\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 6.

The general formula for inverse variation with a cube is $\text{\hspace{0.17em}}y=\frac{k}{{x}^{3}}.\text{\hspace{0.17em}}$ The constant can be found by multiplying $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$

$\begin{array}{ccc}\hfill k& =& {x}^{3}y\hfill \\ & =& {2}^{3}\cdot 25\hfill \\ & =& 200\hfill \end{array}$

Now we use the constant to write an equation that represents this relationship.

$\begin{array}{ccc}\hfill y& =& \frac{k}{{x}^{3}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}k=200\hfill \\ y\hfill & =& \frac{200}{{x}^{3}}\hfill \end{array}$

Substitute $\text{\hspace{0.17em}}x=6\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}y.$

$\begin{array}{ccc}\hfill y& =& \frac{200}{{6}^{3}}\hfill \\ & =& \frac{25}{27}\hfill \end{array}$

#### Questions & Answers

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5