\( y \propto 1/x^4. If y=625 when x=2, find y when x=4. \)
Explanation
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Statement
A quantity is in inverse proportion to a power if it decreases as another variable increases, with the relationship involving a power of the variable. Mathematically:
\[
y \propto \tfrac{1}{x^k} \quad \Rightarrow \quad y = \tfrac{k}{x^n}
\]
Here, \(n\) is the power of \(x\), and \(k\) is a constant of proportionality.
Why it’s true
When one variable decreases as a power of another variable increases, their product involving that power remains constant.
For example, in physics, gravitational force \(F\) is inversely proportional to the square of distance: \(F \propto \tfrac{1}{r^2}\).
This reflects the fact that the influence spreads over an area (or volume), which grows as a power of distance.
Recipe (how to use it)
Write the formula as \(y = \tfrac{k}{x^n}\).
Use known values of \(x\) and \(y\) to find the constant \(k\).
Substitute into the equation to find unknown values of \(y\) or \(x\).
Remember: \(y \times x^n = k\) is constant.
Spotting it
Problems often say “\(y\) varies inversely with the square (or cube, etc.) of \(x\).” This signals inverse proportion to a power.
Engineering: Sound intensity and radiation spread.
Mini examples
Given: \(y\) varies inversely with \(x^2\). If \(y=20\) when \(x=2\), find \(y\) when \(x=5\).
Solution: \(k=yx^2=20×2^2=80.\) For \(x=5\), \(y=80/25=3.2.\)
Given: Intensity of light \(I \propto 1/d^2\). If intensity is 100 at distance 2 m, what is it at 10 m?
Solution: \(k=100×2^2=400.\) At 10 m, \(I=400/100=4.\)
Pitfalls
Forgetting the power: Always check whether it’s inverse square, cube, etc.
Confusing with simple inverse proportion: That’s only \(y=k/x\) (power of 1).
Arithmetic errors: Be careful with squaring/cubing before dividing.
Exam strategy
Write \(y \times x^n = k\) at the start.
Use the given pair of values to calculate \(k\).
Substitute into the formula to find the unknown.
Double-check powers and units if it’s a physics context.
Summary
Inverse proportion to a power means one quantity decreases as the power of another increases. Formula: \(y=k/x^n\). Typical examples include inverse-square laws in physics. Remember: always multiply \(y\) by \(x^n\) to get the constant.
Worked examples
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\( y ∝ 1/x². If y=20 when x=2, find y when x=5. \)
\( k=yx²=20×2²=80. \)
\( For x=5, y=80/25=3.2. \)
Answer:
3.2
\( y varies inversely with x². If y=18 when x=3, find y when x=6. \)
\( k=18×3²=162. \)
\( For x=6, y=162/36=4.5. \)
Answer:
4.5
\( y ∝ 1/x³. If y=40 when x=2, find y when x=4. \)
\( k=40×2³=40×8=320. \)
\( For x=4, y=320/64=5. \)
Answer:
5
\( y varies inversely with x². If y=10 when x=1, find y when x=10. \)
\( k=10×1²=10. \)
\( For x=10, y=10/100=0.1. \)
Answer:
0.1
\( y ∝ 1/x². If y=50 when x=2, find y when x=8. \)
\( k=50×2²=200. \)
\( For x=8, y=200/64=3.125. \)
Answer:
3.125
\( y ∝ 1/x³. If y=81 when x=1, find y when x=3. \)
\( k=81×1³=81. \)
\( For x=3, y=81/27=3. \)
Answer:
3
\( y ∝ 1/x². If y=72 when x=2, find y when x=12. \)
\( k=72×2²=288. \)
\( For x=12, y=288/144=2. \)
Answer:
2
\( In physics, F ∝ 1/r². If F=100 when r=5, find F when r=20. \)
\( k=100×5²=2500. \)
\( For r=20, F=2500/400=6.25. \)
Answer:
6.25
\( y ∝ 1/x⁴. If y=256 when x=1, find y when x=2. \)