HR11, HR14 - Constructing proportional models

Proportional to a function

The symbol $\propto$ means “is proportional to”. If $y\propto f(x)$, then $y=kf(x)$ for a constant $k$. For an inverse relationship, use the reciprocal of the stated function: $y\propto1/f(x)$.

Worked example 1: square and inverse-square scaling

Question. $y$ is directly proportional to $x^2$; $z$ is inversely proportional to $x^2$. Write their equation forms with separate constants. What happens when $x$ triples?

Plan. Apply the square to the scale factor, and take its reciprocal for the inverse model.

Replacing $x$ by $3x$ gives $a(3x)^2=9ax^2$ and $b/(3x)^2=b/(9x^2)$.

Answer. $y$ is multiplied by $9$; $z$ is divided by $9$.

Check. For $a=b=1$, changing $x$ from $2$ to $6$ changes $y$ from $4$ to $36$ and $z$ from $1/4$ to $1/36$, with the stated factors.

Worked example 2: cube and inverse-root scaling

Question. $p$ is directly proportional to the cube of $x$, and $q$ is inversely proportional to the square root of $x$. Write equation forms and describe the effect of multiplying positive $x$ by $4$.

Plan. Use $4^3$ for the cube model and $1/\sqrt4$ for the inverse-root model.

Since $(4x)^3=64x^3$ and $\sqrt{4x}=2\sqrt{x}$, the first output becomes $64$ times as large and the second halves.

Answer. $p$ is multiplied by $64$; $q$ is divided by $2$.

Check. With constants $1$, $x=1$ gives both outputs $1$; $x=4$ gives $p=64$, $q=1/2$.

Watch out

The inverse square model is $k/x^2$, not $k/x$, and the inverse-root model is $k/\sqrt{x}$, not $\sqrt{k/x}$.

Worked example 3: finding a root-model constant

Question. $y$ is directly proportional to $\sqrt{x}$. When $x=0.25$, $y=0.6$. Find the formula, calculate $y$ when $x=6.25$, and find $x$ when $y=4.8$.

Plan. Use $y=k\sqrt{x}$, then substitute the known pair to determine $k$.

The formula is $y=1.2\sqrt{x}$. At $x=6.25$, $y=1.2\times2.5=3$. If $y=4.8$, then $\sqrt{x}=4.8/1.2=4$, so $x=16$.

Answer. $y=1.2\sqrt{x}$; the requested values are $y=3$ and $x=16$.

Check. The formula gives $1.2\sqrt{0.25}=0.6$ for the original pair and $1.2\sqrt{16}=4.8$ for the reversed calculation.

Worked example 4: combining models

Question. $y$ is inversely proportional to $w^3$, and $w$ is directly proportional to $\sqrt{x}$. When $w=6$, $x=9$ and $y=1/4$. Find $y$ in terms of $x$, then find $y$ when $x=36$.

Plan. Determine the two constants separately, then replace $w$ by its expression in $x$.

Also, $w=c\sqrt{x}$, so $6=c\sqrt9=3c$, giving $c=2$. Substitute $w=2\sqrt{x}$:

Here $\sqrt{x}=x^{1/2}$, so $(\sqrt{x})^3=x^{3/2}$. At $x=36$, $w=12$, so $y=54/12^3=1/32$.

Answer. $y=(27/4)x^{-3/2}$, or $27/(4x^{3/2})$; at $x=36$, $y=1/32$.

Check. $x$ changed from $9$ to $36$, a factor of $4$. Therefore $w$ doubled and $y$ divided by $2^3=8$: $(1/4)/8=1/32$.