How do you find an equation of variation: y varies inversely as the square of x, and y = 0.15 when x= 0.1?

1 Answer
Apr 5, 2016

Answer:

#y= (0.0015)/x^2#

Or if you prefer #y=15/(10000x^2)#

Or as scientific notation: #y=15/(x^2)xx10^(-4)#

Explanation:

Splitting the question down into its component parts:

y varies inversely as: #->y=1/?#

the square of #->y=1/(?^2)#

#x" "-> y=1/(x^2)#

But we need a constant of variation
Let the constant of variation be #k# then we have:

#y=kxx1/(x^2)#
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#color(blue)("Determine the value of "k)#

Known condition: at #y=0.15"; "x=0.1#

So by substitution we have

#0.15=kxx1/((0.1)^2)#

#=>k=0.15xx(0.1)^2#

#k=0.15xx0.01#

To calculate this directly think of #0.01" as "1/100#

Then we have:

#k=0.15/100 = 0.0015#

So the equation becomes:

#y= (0.0015)/x^2#

Or if you prefer #y=15/(10000x^2)#
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