# 5.8 Modeling using variation  (Page 2/14)

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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 $\text{\hspace{0.17em}}y.$

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

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
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Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1