Question 1149570: The weight W of a cylindrical metal varies jointly as its legth l and the square of the diameter d of its base.
If W=6kg when l=6cm and d=3 cm,find the equation of variation
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39630)   (Show Source):
Answer by ikleyn(52903)  (Show Source):
You can put this solution on YOUR website! .
The weight W of a cylindrical metal varies jointly as its legth l and the square of the diameter d of its base.
If W=6kg when l=6cm and d=3 cm,find the equation of variation
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From the problem's description
W =  , (1)
where L is the length and d is the diameter, and "k" is the constant value coefficient.
From equation (1),
k =  =  = 0.111111...
Then the equation (1) takes the form
w =  .
It is the equation you are asking for.
Be aware : The equation and the value of "k", found by @josgatithmetic in his post,
are W R O N G, so you better simply IGNORE it/them for your safety.
There is another circumstance I want to aware you.
The coefficient "k" divided by  , i.e.  =  = 0.1414 kilograms per cm^3 = 141.4 grams per cm^3.
This ratio  in this problem, is nothing else as the specific weight (or density) of the metal.
The fact is that THERE IS NO such a heavy metal on the Earth. The heaviest is osmium with the specific weight (density) of 22.6 g/cm^3
So, the entire your problem is a F A K E, and you better throw it into the GARBAGE BOX --- the only right place for it . . .
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