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Section 5.5 Inverse Variation

Subsection 1. Solve a variation equation

In these examples, we assume all variables are positive. We round answers to tenths.

Subsubsection Examples

Example 5.33.

Solve \(~2.8125 = \dfrac{36}{n}\)

Solution

We must first get the variable out of the denominator.

\begin{align*} n(2.8125) \amp = \dfrac{36}{n} n \amp\amp \blert{\text{Multiply boyh sides by}~n.}\\ 2.8125n \amp = 36 \amp\amp \blert{\text{Divide both sides by 2.8125.}}\\ n \amp = 12.8 \end{align*}
Example 5.34.

Solve \(~0.5547 = \dfrac{1500}{d^2}\)

Solution

We must first get the variable out of the denominator.

\begin{align*} d^2(0.5547) \amp = \dfrac{1500}{d^2} d^2 \amp\amp \blert{\text{Multiply boyh sides by}~d^2.}\\ 0.5547d^2 \amp = 1500 \amp\amp \blert{\text{Divide both sides by 0.5547.}}\\ d^2 \amp = 2704.16 \amp\amp \blert{\text{Take square roots.}}\\ d \amp = 52 \end{align*}

Subsubsection Exercises

Solve \(~13.03=\dfrac{380}{h^2}\)

Answer
\(5.4\)

Solve \(~0.065=\dfrac{12}{p}\)

Answer
\(184.6\)

Subsection 2. Sketch a variation graph

The graphs of inverse variations are transformations of the basic graphs \(y=\dfrac{1}{x^n}\text{.}\)

Subsubsection Example

Example 5.37.

Sketch a graph of \(H=\dfrac{48}{w}\text{.}\)

Solution

We know that the graph has the shape of the basic function \(y=\dfrac{1}{x}\text{,}\) so all we need are a few points to "anchor" the graph.

\begin{align*} \text{If } w=2, \amp H = \dfrac{48}{2} = 24\\ \text{If } w=6, \amp H = \dfrac{48}{6} = 8\\ \text{If } w=12, \amp H = \dfrac{48}{12} = 4 \end{align*}

The graph is shown below.

cubic

Subsubsection Exercise

Plot three points and sketch a graph of \(B=\dfrac{0.8}{d^2}\text{.}\)

reciprocal squared
Answer
\((1,0.8),~(2,0.2),~(4,0.05)\)
reciprocal squared

Subsection 3. Find the constant of variation

If we know the type of variation and the coordinates of one point on the graph, we can find the variation equation.

Subsubsection Example

Example 5.39.

Find the constant of variation and the variation equation:

\(~~~y\) varies inversely with the square of \(x\text{,}\) and \(y=4687.5\) when \(x=0.16\text{.}\)

Solution

Because \(y\) varies inversely with the square of \(x\text{,}\) we know that \(y=\dfrac{k}{x^2}\text{.}\) We substitute the given values to find

\begin{align*} 4687.5 \amp = \dfrac{k}{0.16^2} \amp\amp \blert{\text{Solve for}~k.}\\ k \amp = 4687.5(0.16)^2=120 \end{align*}

The constant of variation is 120, and the variation equation is \(~y=\dfrac{120}{x^2}\text{.}\)

Subsubsection Exercise

Find the constant of variation and the variation equation:

\(y\) varies inversely with \(x\text{,}\) and \(y=31.25\) when \(x=640\text{.}\)

Answer
\(k=20,000\) and \(y=\dfrac{20,000}{x}\)