Changes in direct proportion

Changes in direct proportion

Secondary 2

When \(y=kx^n\), multiplying \(x\) by a factor \(r\) multiplies \(y\) by \(r^n\). The constant \(k\) cancels, so the change in \(y\) can be found without knowing \(k\).

Why: new \(y=k(rx)^n=r^n\left(kx^n\right)=r^n\times\text{original}\hspace{0.5em}y\).

Change in \(x\)Multiplier for \(x\)\(y=kx\)\(y=kx^2\)\(y=kx^3\)
doubled\(2\)\(\times2\)\(\times4\)\(\times8\)
halved\(0.5\)\(\times0.5\)\(\times0.25\)\(\times0.125\)
increased by \(50\%\)\(1.5\)\(\times1.5\)\(\times2.25\)\(\times3.375\)
decreased by \(20\%\)\(0.8\)\(\times0.8\)\(\times0.64\)\(\times0.512\)

Method

  1. Convert the change in \(x\) to a multiplier: an increase of \(30\%\) gives \(1.3\); a decrease of \(30\%\) gives \(0.7\).
  2. Raise the multiplier to the power in the equation (use a square root for \(\sqrt{x}\)).
  3. Convert the result back: percentage change in \(y=(\text{multiplier}\hspace{0.5em}-1)\times100\%\). A negative answer is a decrease.

Example

The power, \(P\) watts, produced by a wind turbine is directly proportional to the cube of the wind speed, \(v\) m/s. The wind speed increases by \(20\%\). Find the percentage increase in \(P\).

Multiplier for \(v\): \(1.2\)

Multiplier for \(P\): \(1.2^3=1.728\)

Increase: \((1.728-1)\times100\%=72.8\%\)

Check: a \(20\%\) increase in \(v\) does not give a \(60\%\) increase in \(P\); the multiplier is cubed, not the percentage. Test with numbers: \(v=10\) gives \(P=1000k\); \(v=12\) gives \(P=1728k\), which is \(72.8\%\) more.

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