#$&*
course mth 173
7/18 7
Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:If a sand pile 2.1 meters high has a mass of 5000.939 kg, then what would we expect to be the mass of a geometrically similar sand pile 6.1 meters high?
Q(x) = kx^3
5000.939 = k(2.1^3) = k(9.261)
k = 539.898
Q(x) = 539.898(x^3)
Q(6.1) = 539.898(6.1^3) = 122546.59 kg
If there are 1.323 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?
A(x) = kx^2
1.323 = k (2.1)^2 = k (4.41)
k = .3 billion
A(x) = .3(x^2)
A(6.1) = .3 (6.1^2) = 11.163 billion
Could you have done this using a relationship between mass and grains of sand exposed? if so how????
@&
Good.
*@
@&
Mass is proportional to the cube of the height, and number of exposed grains to the square of the height.
So
M = k1 x^3
A = k2 x^2
We can eliminate x to get the relationship between M and A.
For example:
From the first equation x = (M / k1)^(1/3).
Substituting this into the second equation we get
A = k2 * ( (M / k1 ) ^(1/3) ) ^ 2
so that
A = k2 / k1^(2/3) * M^(2/3).
k2 and k1 are just constants, so k2 / k1^(2/3) is a constant, so we might as well just call it k.
Thus
A = k M^(2/3).
*@
#$&*
course mth 173
7/18 7
Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:If a sand pile 2.1 meters high has a mass of 5000.939 kg, then what would we expect to be the mass of a geometrically similar sand pile 6.1 meters high?
Q(x) = kx^3
5000.939 = k(2.1^3) = k(9.261)
k = 539.898
Q(x) = 539.898(x^3)
Q(6.1) = 539.898(6.1^3) = 122546.59 kg
If there are 1.323 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?
A(x) = kx^2
1.323 = k (2.1)^2 = k (4.41)
k = .3 billion
A(x) = .3(x^2)
A(6.1) = .3 (6.1^2) = 11.163 billion
Could you have done this using a relationship between mass and grains of sand exposed? if so how????
@&
Good.
*@
@&
Mass is proportional to the cube of the height, and number of exposed grains to the square of the height.
So
M = k1 x^3
A = k2 x^2
We can eliminate x to get the relationship between M and A.
For example:
From the first equation x = (M / k1)^(1/3).
Substituting this into the second equation we get
A = k2 * ( (M / k1 ) ^(1/3) ) ^ 2
so that
A = k2 / k1^(2/3) * M^(2/3).
k2 and k1 are just constants, so k2 / k1^(2/3) is a constant, so we might as well just call it k.
Thus
A = k M^(2/3).
*@