Direct Variation

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#13872

Direct Variation

 $$y\,\propto\,x$$

$$y=kx$$

where k is the constant of variation

The statement		The variation statement	The mathematical equation
y varies as x		y ∝ x	y = kx
the velocity of a body falling from rest varies with the time fallen		v ∝ t	v = kt
the circumference (C cm) of a circle varies directly as its radius (r cm)		C ∝ r	C = kr
the area of a circle varies with the square of the radius		A ∝ r2	A = kr2
the surface area of a cube varies directly as the square of the length of a side		S ∝ l2	S = kl2
b varies directly as the cube of a		b ∝ a3	b = ka3
the length of the rectangle varies with the square root of the extension of the spring		l ∝ √e	l = k√e

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#13874

The energy e produced by a falling object travelling at speed s is directly proportional to the square of its speed. The energy produced by a certain object falling at a speed of 4m/s is 136 joules. Find the energy produced when the speed of the object is 6.5m/s.

e = ks²
136 = k × 4²
k = ¹³⁶/₁₆
k = 8.5

e = 8.5 × 6.5²
e = 359.125

359.125 joules

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#13875

If c varies with the cube of v, and when v = 60, c = 12, find the value of c when v = 90.

c  $$\alpha$$ v³
c = kv³
12 = k × 60³
12 = k × 216000
k = ¹²/₂₁₆₀₀₀
k = ¹/₁₈₀₀₀
c = ¹/₁₈₀₀₀ × 90³
c = 40.5

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#13876

The weight of a sphere is directly proportional to the cube of its radius. If a sphere with radius 4cm has a weight of 5 kilograms find the weight of a sphere of radius 8cm.

w = $$\alpha r^3$$
w = kr³
5 = k × 4³
$$k=\frac{5}{64}$$
w = ⁵/₆₄ × 8³
w = 40kg

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#18819

The rate at which water comes out of a house is directly proportional to the square of the hose’s diameter.
a.   If a hose of diameter 3cm gives out water at the rate of 3.6 litres/ second find the rate at which water comes out of a fireman’s hose which is 8cm across.
b.   What diameter hose do I use if I want water to come out at 1 litre/ second?

a. w ∝ d²
w = kd²
3.6 = k × 3²
k = 0.4

w = 0.4d²

if d = 8
w = 0.4 × 8²
w = 25.6 L/s

b. 1 = 0.4d²
2.5 = d²
d = 1.58 cm

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#18820

The area of a circle varies directly as the square of its diameter. If a circle of diameter 10 cm has an area 80 cm2, then the area of a circle of diameter 20 cm is:
$$a\propto d^2$$	First we form the variation statement
a = kd2	Then we form the mathematical equation
80 = k × 102 80 = 100k k = 0.8	Now we substitute to solve for k. We know that a = 80 when d = 10
a = 0.8d2	Now we rewrite and replace our k with our solution this is the equation we must now use for any other part of the question. (since k is constant, it will not change)
a = 0.8 × 202 320 cm2	The last part of the question requires us to find the area when d = 20, so we simply substitute 20 into d

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