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Section 5.4 Direct Variation

Subsection 1. Solve a variation equation

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

Subsubsection Examples

Example 5.25.

Solve \(~231.90=18.85r^2~\)

Solution

The equation is quadratic. We solve by extraction of roots.

\begin{align*} 231.90 \amp = 18.85 r^2 \amp\amp \blert{\text{Isolate the squared expression.}}\\ 12.302 \amp = r^2 \amp\amp \blert{\text{Take square roots.}}\\ r \amp = 35 \end{align*}
Example 5.26.

Solve \(~62x^3 = 4860.8~\)

Solution

This equation is cubic. We isolate the variable, then take cube roots.

\begin{align*} 62x^3 \amp = 4860.8 \amp\amp \blert{\text{Divide both sides by 62.}}\\ x^3 \amp = 78.4 \amp\amp \blert{\text{Take cube roots.}}\\ x \amp = 16.9 \end{align*}

Subsubsection Exercises

Solve \(~1371.8=25R^3\)

Answer
\(3.8\)

Solve \(~6.3t^2 = 18.4\)

Answer
\(1.7\)

Subsection 2. Sketch a variation graph

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

Subsubsection Example

Example 5.29.

Sketch a graph of \(V=0.2s^3\text{.}\)

Solution

We know that the graph has the shape of the basic function \(y=x^3\text{,}\) so all we need are a few points to "anchor" the graph.

\begin{align*} \text{If } s=1, \amp V = 0.2(1)^3 = 0.2\\ \text{If } s=2, \amp V = 0.2(2)^3 = 1.6\\ \text{If } s=3, \amp V = 0.2(3)^3 = 5.4 \end{align*}

The graph is shown below.

cubic

Subsubsection Exercise

Plot three points and sketch a graph of \(d=\dfrac{3}{8}t^2\text{.}\)

grid
Answer
\(\left(1,\dfrac{3}{8}\right)\text{,}\) \(\left(2,\dfrac{3}{2}\right)\text{,}\) \((4,6)\)
parabola

Subsection 3. Find the constant of variation

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

Subsubsection Example

Example 5.31.

Find the constant of variation and the variation equation:

\(~~~y\) varies directly with the square of \(x\text{,}\) and \(y=100\) when \(x=2.5\text{.}\)

Solution

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

\begin{align*} 100 \amp = k(2.5)^2 \amp\amp \blert{\text{Solve for}~k.}\\ k \amp = \dfrac{100}{2.5^2} = 16 \end{align*}

The constant of variation is 16, and the variation equation is \(~y=16x^2\text{.}\)

Subsubsection Exercise

Find the constant of variation and the variation equation:

\(y\) varies directly with the cube of \(x\text{,}\) and \(y=119,164\) when \(x=6.2\text{.}\)

Answer
\(k=500\) and \(y=500x^3\)