XXIV Variation and Laws

Direct Variation

Direct variation refers to a proportional relationship of two quantities.

e.g. the relationship between wage and hours worked

W1W2=T1T2\frac {W1} {W2} = \frac {T1} {T2}

32006400=120240\frac {3200} {6400} = \frac {120} {240}

If one quanitity is altered in a certain ratio, the other quantity is affected the same way.

Notation for Proportional Variation

yxy \varpropto x, yy varies directly with xx, describes:

y=kxy = kx

Example

s=Distance travelleds = \text{Distance travelled}
t=Timet = \text{Time}

sts \varpropto t
s=kts = kt therefore k=stk = \frac s t

If s=40s = 40 and t=2.5t=2.5

40=2.5k40=2.5k
k=402.5k = \frac {40}{2.5}
k=16k = 16

Meaning the law connecting ss and tt for this particular velocity is

s=16ts = 16t

Graphs

Since the equation y=kxy = kx is equal to that of a straight line passing through the origin y=mxy = mx, any direct variation can be graphed accordingly.

Constant and Partly Variation

In practical work a variable might be dependent on a variable and a constant.

y=axby = ax - b

Where the values of aa and bb have to be figured out before the law can be stated.

Squared Direct Variation

yx2y \varpropto x^2, y=kx2y = kx^2

Implying a quadratic function drawing a parabola when graphed.

For any kind of exponential variation the index is part of the law.

Cubed Direct Variation

yx3y \varpropto x^3, y=kx3y = kx^3

The graph of x3x^3 is called a cubic curve.

Inverse Squared Variation

yx2y \varpropto \sqrt[2]{x}, y=x12y = x^{\frac 1 2}

The graph only draws y=x+y = \sqrt[+]{x} because a function may only yield one result.

Inverse Variation

y1xy \varpropto \frac 1 x, y=kxy = \frac k x

As one value increases the other decreases or vice-versa.

Inverse variants when plotted appear as hyperbolas.

Joint Variation

When a quantity directly depends on two or more other quantities.

yxzy \varpropto xz, y=kxzy = kxz

yxzy \varpropto \frac x z, y=kxzy = \frac {kx}{z}

Determination of Non-Linear Laws

Laws can not be derived from exponential equations in a straightforward manner because their curves not not be derived in their entirety from their parts.

Example

y=axn+by = ax^n + b

Since plotting for xx would result in a parabola, plotting against xnx^n and yy instead yields a straight line.

Logarithms can help finding unknown exponents:
y=cxny = cx^n
logy=nlogx+logc\log y = n \log x + \log c

So given two points (x1,y1)(x1, y1) and (x2,y2)(x2, y2):

3.236=1.398n+logc3.236 = 1.398n + \log c
2.795=1.255n+logc2.795 = 1.255n + \log c

Subtracting the equations cancels logc\log c:

0.441=0.143n0.441 = 0.143n

Thus

n=0.4410.1433.1n = \frac {0.441}{0.143} \approx 3.1

Substituting nn in one of the equations now leads to the value of cc.