Let P vary inversely as the square of x. What is the effect on P if x is increased by %?
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P=k/x2 where k is a constant defines the relationship.

If x is increased by p% then its value becomes x+px/100. Let r=p/100, so x becomes x(1+r).

P becomes k/[x2(1+r)2] so if the original value of P is P0 and Pr is its new value after the increase, then:

Pr-P0 is the change in P=k/[x2(1+r)2]-k/x2,

Pr-P0=(k/x2)(1/(1+r)2-1)=P0(1-(1+r)2)/(1+r)2,

Pr-P0=P0(-r(2+r))/(1+r)2=-rP0(2+r)/(1+r)2, which is a decrease.

We can also write this in terms of p (the percentage increase in x):

Change in P=-pP0(2+p/100)/(1+p/100)2 so P decreases by a factor p(2+p/100)/(1+p/100)2% of its original value.

If p=100% then Pr=P1 and P1-P0=k/(4x2)-k/x2=-¾P0, a decrease of 75%.

If p=10%, r=0.1 and the change is -10(2.1)/1.21=-21/1.21=-17.355% or a decrease of about 17.355%.

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