Direct Variation

Direct Variation exercise

• Draw the graph that represents the described direct variation.

  Hints

  The equation for direct variation is $y=kx$.

  $k$ is the constant of variation and is also the slope.

  $x$ represents the height reached by Schulzy, while $y$ represents the height reached by Schmiddy.

  Solution

  On the planet MPI, Schulzy can jump $1$ ft, and on the planet Beaumont, Schmiddy can jump $6$ ft high.

  The scores have a direct variation. So we can use the equation $y=kx$ to calculate the constant of variation k, or we can calculate the slope.

  The slope of a line is defined by $m=\frac{y_2-y_1}{x_2-x_1}$.

  • Schulzy jumps 1 ft and Schmiddy jumps 6 ft - this data is represented by the point $(1,6)$.

  • If $x=0$, the direct variation gives us $y=0$. This is the point $(0,0)$, or the origin of the coordinate system. Lines formed by data that has direct variation always pass through the origin.

  We can write the equation: $k=m=\frac{6-0}{1-0}=\frac61=6$.

• Determine the factor $k$.

  Hints

  Use the equation $y=kx$ and substitute the given values for $x$ and $y$.

  Solve this equation for $k$.

  You can reduce a fraction by dividing the numerator as well as the denominator by the same number.

  Solution

  In the equation $y=kx$, we can substitute each given pair of $x$ and $y$ and solve for $x$:

  $\mathbf{x=100,~y=2}$

  • This results in the equation: $2=k\times 100$
  • Divide by $100$: $k=\frac2{100}=\frac{2\div2}{100\div2}=\frac1{50}$

  $\mathbf{x=150,~y=3}$

  • This results in the equation: $3=k\times 150$
  • Divide by $150$: $k=\frac3{150}=\frac{3\div3}{150\div3}=\frac1{50}$

  $\mathbf{x=200,~y=4}$

  • This results in the equation: $4=k\times 200$
  • Divide by $200$: $k=\frac4{200}=\frac{4\div4}{200\div4}=\frac1{50}$

  As a result, we get the same value for $k=\frac1{50}$ for all the pairs and demonstrates there is a direct variation.

  This results in the equation $y=\frac1{50}x$.

• Determine body weight on different planets.

  Hints

  $y$ represents the body weight on the other planets.

  $x$ represents the body weight of Susan or Johnny on the planet Earth.

  The relationship between the weight on Earth and the weight on other planets varies directly, and the equation to calculate the constant of variation is: $y=kx$.

  $100~lbs$ on Earth equals $140~lbs$ on Neptune.

  Solution

  To calculate the constant of variation, use the equation: $y=kx$.

  On Earth, $x$ is the given body weight of Susan, $140~lbs$, and $x$ is also the given body weight of Johnny, $174~lbs$. $k$ is the constant of variation.

  We have to multiply $x$ by the constant of variation to get the body weight of Susan and Johnny on the different planets.

  Jupiter, $\mathbf{k=2.63}$

  • Susan: $y=2.63\times 140=368.2$
  • Johnny: $y=2.63\times 174=457.62$

  Venus, $\mathbf{k=\frac1{1.15}}$

  • Susan: $y=\frac1{1.15}\times 140 \approx 121.74$
  • Johnny: $y=\frac1{1.15}\times 174 \approx 151.3$

  Neptune, $\mathbf{k= 1.4}$

  • Susan: $y=1.4\times 140=196$
  • Johnny: $y=1.4\times 174=243.6$

  Saturn, $\mathbf{k= 1.13}$

  • Susan: $y=1.13\times 140=158.2$
  • Johnny: $y=1.13\times 174=196.62$

  Mars, $\mathbf{k=\frac1{2.63}}$

  • Susan: $y=2.63\times 140 \approx 53.23$
  • Johnny: $y=2.63\times 174 \approx 66.16$

  Mercury, $\mathbf{k=\frac1{2.9}}$

  • Susan: $y=\frac1{2.9}\times 140 \approx 48.28$
  • Johnny: $y=\frac1{2.9}\times 174=60$

• Identify the equations.

  Hints

  Each situation demonstrates a relationship that is directly variable.

  The equation to describe a direct variation is: $y=kx$.

  If you know to points on the line, you can use the slope formula to determine the constant of variation, $k$:

  $\large k=m=\frac{y_2-y_1}{x_2-x_1}$

  Solution

  What do we know?

  • Each situation is directly variable.
  • The equation to describe a direct variation is: $y=kx$.
  • We have to determine the constant of variation, $k$.

  Ice cream

  • We know the points $(2,2.4)$ and $(4,4.8)$.
  • Therefore, we can find $k$ by using the slope formula $k=m=\frac{4.8-2.4}{4-2}=\frac{2.4}{2}=1.2$.
  • $y=1.2x$.

  Lemonade

  • We know the two points $(3,96)$ and $(5,160)$.
  • Therefore, we can find $k$ by using the slope formula $k=m=\frac{160-96}{5-3}=\frac{64}{2}=32$.
  • $y=32x$.

  Susan and Johnny

  • We know the two points $(2,12.8)$ and $(1.5,9.6)$.
  • Therefore, we can find $k$ by using the slope formula $k=m=\frac{12.8-9.6}{2-1.5}=\frac{3.2}{0.5}=6.4$.
  • $y=6.4x$.

• Summarize the characteristics of direct variations.

  Hints

  The table at right shows a direct variation.

  A linear equation in slope-intercept form is given by $y=mx+b$.

  • $m$ is the slope
  • $b$ is the y-intercept

  For $y=mx+b$, if the variable $b$ is equal to $0$, there is a direct variation.

  Solution

  The equation to solve for the constant of variation is: $y=kx$. When there is a direct variation, $b=0$ for $y=mx+b$. The graph of a direct variation always passes through the origin of a coordinate system and has a y-intercept of 0.

  If you double the value for either variable, $x$ or $y$, the value of the other variable will double as well.

  In a situation such as where the height of a jump depends on gravity, if Schmiddy jumps 6 ft while Schulzy jumps just 1 ft, we get the following values for the doubled height:

  • $2\times 1= 2$ ft for Schmiddy
  • $2\times 6=12$ ft for Schulzy

• Find the equations that are represented by the graphs.

  Hints

  To solve for the constant of variation, k, you can use the formula: $y=kx$.

  $k$ is the slope of the line. Keep in mind that the slope is rise over run.

  The slope is given by the change in $y$ divided by the change in $x$.

  Solution

  Each line represents data that is directly variable and corresponds with the equation $y=kx$.

  The light purple line

  • passes through the point $(20,50)$
  • therefore, $k=m=\frac{50-0}{20-0}=\frac{50}{20}=\frac52$
  • the equation is $y=\frac52x$

  The green line

  • passes through the point $(30,50)$
  • therefore, $k=m=\frac{50-0}{30-0}=\frac{50}{30}=\frac53$
  • the equation is $y=\frac53x$

  The dark purple line

  • passes through the point $(70,50)$
  • therefore, $k=m=\frac{50-0}{70-0}=\frac{50}{70}=\frac57$
  • the equation is $y=\frac57x$

  The orange line

  • passes through the point $(50,50)$
  • therefore, $k=m=\frac{50-0}{50-0}=\frac{50}{50}=1$
  • the equation is $y=x$