The Central Idea
Scale, conversion, and variation problems all depend on preserving a relationship.
A scale problem preserves proportional shape or size:
\[ \boxed{\frac{\text{new measurement}}{\text{original measurement}}=k}. \]A conversion problem preserves physical quantity while changing units:
\[ \boxed{\text{quantity}\times\text{conversion factor} =\text{equivalent quantity}}. \]A variation problem preserves an algebraic dependency such as
\[ \boxed{y=kx}, \qquad \boxed{y=\frac{k}{x}}, \]or a related model.
In every case, the algebra should preserve what the problem says must remain unchanged.
What Is a Scale Factor?
A scale factor is the multiplier that changes one size into a corresponding size.
If an original length \(L\) becomes a new length \(L'\), then
\[ \boxed{L'=kL}. \]The number \(k\) is the scale factor.
- If \(k>1\), the new figure is an enlargement.
- If \(0<k<1\), the new figure is a reduction.
- If \(k=1\), the size is unchanged.
Finding a Scale Factor
Suppose a segment of length \(8\) is enlarged to length \(14\).
The scale factor is
\[ k=\frac{14}{8}. \]Simplify:
\[ k=\frac{7}{4}=1.75. \]Therefore,
\[ \boxed{k=1.75}. \]Using a Scale Factor
If every length in a drawing is multiplied by \(1.5\), then a length of \(12\) becomes
\[ 1.5(12)=18. \]Thus,
\[ \boxed{12\longrightarrow18}. \]The same multiplier must be applied to every corresponding linear measurement if the figure is to remain geometrically similar.
Scale Drawings
A scale drawing represents an object using a fixed ratio between drawing measurements and actual measurements.
A scale may be written
\[ 1\text{ cm}:5\text{ m}. \]This means every \(1\) centimeter on the drawing corresponds to \(5\) meters in the real object.
A drawing length of \(7.2\) centimeters therefore represents
\[ 7.2(5)=36 \]meters.
\[ \boxed{36\text{ m}}. \]Writing Scale as a Unit Ratio
The scale
\[ 1\text{ cm}:5\text{ m} \]can be written as
\[ \frac{5\text{ m}}{1\text{ cm}}. \]Multiplying a drawing measurement by this ratio converts drawing length to actual length:
\[ 7.2\text{ cm} \left( \frac{5\text{ m}}{1\text{ cm}} \right) = 36\text{ m}. \]The centimeter units cancel, leaving meters.
Map Scale
Suppose a map uses the scale
\[ 1\text{ in}:24\text{ mi}. \]Two locations are \(3.75\) inches apart on the map.
The actual distance is
\[ 3.75(24)=90. \]Therefore,
\[ \boxed{90\text{ miles}}. \]Finding a Drawing Measurement
Suppose a blueprint uses
\[ 1\text{ cm}:2.5\text{ m}. \]An actual wall is \(11\) meters long. Let \(x\) be its blueprint length in centimeters.
\[ \frac{1\text{ cm}}{2.5\text{ m}} = \frac{x\text{ cm}}{11\text{ m}}. \]Solve:
\[ 2.5x=11, \] \[ x=4.4. \]Therefore,
\[ \boxed{4.4\text{ cm}}. \]Scale and Similar Figures
In similar figures, corresponding lengths are proportional.
If a triangle with side lengths
\[ 3,\qquad4,\qquad5 \]is enlarged by scale factor \(2.5\), the new side lengths are
\[ 7.5,\qquad10,\qquad12.5. \]Every linear measurement is multiplied by the same factor.
Linear Scale Does Not Apply Directly to Area
If all lengths are multiplied by \(k\), areas are multiplied by
\[ \boxed{k^2}. \]For example, if a rectangle is enlarged by a linear scale factor of \(3\), its area becomes
\[ 3^2=9 \]times as large.
This happens because both length and width are multiplied by \(3\).
Volume Scale
If all linear dimensions of a three-dimensional object are multiplied by \(k\), its volume is multiplied by
\[ \boxed{k^3}. \]A model enlarged by scale factor \(2\) has volume
\[ 2^3=8 \]times the original volume.
Example: Area under Scaling
A square has side length \(6\). It is enlarged by scale factor \(1.5\).
The original area is
\[ 6^2=36. \]The new side length is
\[ 1.5(6)=9. \]The new area is
\[ 9^2=81. \]Equivalently,
\[ 36(1.5)^2=36(2.25)=81. \]Therefore,
\[ \boxed{81}. \]Example: Volume under Scaling
A solid has volume \(50\text{ cm}^3\). Every linear dimension is doubled.
The volume scale factor is
\[ 2^3=8. \]Therefore, the new volume is
\[ 50(8)=400. \]Thus,
\[ \boxed{400\text{ cm}^3}. \]What Is a Conversion Factor?
A conversion factor is a ratio equal to \(1\) that expresses the same quantity in two different units.
Since
\[ 1\text{ ft}=12\text{ in}, \]both
\[ \frac{12\text{ in}}{1\text{ ft}} \]and
\[ \frac{1\text{ ft}}{12\text{ in}} \]represent the number \(1\).
Choosing the correct orientation makes unwanted units cancel.
Dimensional Analysis
Dimensional analysis uses conversion factors so that units cancel algebraically.
Convert \(7\) feet to inches:
\[ 7\text{ ft} \left( \frac{12\text{ in}}{1\text{ ft}} \right). \]The feet cancel:
\[ 7(12)\text{ in}=84\text{ in}. \]Therefore,
\[ \boxed{7\text{ ft}=84\text{ in}}. \]Choose the Conversion Factor by Unit Cancellation
Suppose we want to convert \(30\) inches to feet.
Because inches must cancel, use
\[ \frac{1\text{ ft}}{12\text{ in}}. \]Then
\[ 30\text{ in} \left( \frac{1\text{ ft}}{12\text{ in}} \right) = 2.5\text{ ft}. \]Therefore,
\[ \boxed{30\text{ in}=2.5\text{ ft}}. \]Multi-Step Unit Conversions
Convert \(3.5\) hours to seconds.
Use two conversion factors:
\[ 3.5\text{ hr} \left( \frac{60\text{ min}}{1\text{ hr}} \right) \left( \frac{60\text{ s}}{1\text{ min}} \right). \]Hours and minutes cancel, leaving seconds:
\[ 3.5(60)(60)=12{,}600. \]Therefore,
\[ \boxed{3.5\text{ hr}=12{,}600\text{ s}}. \]Conversion Chains
A long conversion can be handled as one multiplication chain.
Convert \(2.4\) kilometers to centimeters using
\[ 1\text{ km}=1000\text{ m}, \]and
\[ 1\text{ m}=100\text{ cm}. \]Then
\[ 2.4\text{ km} \left( \frac{1000\text{ m}}{1\text{ km}} \right) \left( \frac{100\text{ cm}}{1\text{ m}} \right) = 240{,}000\text{ cm}. \]Therefore,
\[ \boxed{240{,}000\text{ cm}}. \]Converting Rates
Units may appear in both numerator and denominator.
Convert \(72\) kilometers per hour to meters per second.
\[ 72\frac{\text{km}}{\text{hr}} \left( \frac{1000\text{ m}}{1\text{ km}} \right) \left( \frac{1\text{ hr}}{3600\text{ s}} \right). \]Units cancel to give
\[ 72\left(\frac{1000}{3600}\right) \frac{\text{m}}{\text{s}} = 20\frac{\text{m}}{\text{s}}. \]Thus,
\[ \boxed{72\text{ km/h}=20\text{ m/s}}. \]Squared Units Require Squared Conversion Factors
Suppose
\[ 1\text{ m}=100\text{ cm}. \]Then
\[ 1\text{ m}^2 = (100\text{ cm})^2 = 10{,}000\text{ cm}^2. \]Therefore,
\[ \boxed{1\text{ m}^2=10{,}000\text{ cm}^2}. \]The conversion factor must be squared because area has two linear dimensions.
Cubic Units Require Cubed Conversion Factors
Since
\[ 1\text{ m}=100\text{ cm}, \]it follows that
\[ 1\text{ m}^3 = (100\text{ cm})^3 = 1{,}000{,}000\text{ cm}^3. \]Therefore,
\[ \boxed{1\text{ m}^3=1{,}000{,}000\text{ cm}^3}. \]Exact and Approximate Conversions
Some unit relationships are exact by definition, while others are commonly used in approximate form.
For example,
\[ 1\text{ ft}=12\text{ in} \]is exact.
A conversion such as
\[ 1\text{ mi}\approx1.609\text{ km} \]may be used as a rounded approximation when the exact defined conversion is unnecessary for the problem.
The precision of the conversion factor can affect the precision of the final answer.
Units Are Part of the Algebra
Units are not decoration added after the calculation. They help determine whether a setup is structurally correct.
For example,
\[ 5\text{ hr} \left( \frac{60\text{ min}}{1\text{ hr}} \right) \]correctly leaves minutes.
But
\[ 5\text{ hr} \left( \frac{1\text{ hr}}{60\text{ min}} \right) \]leaves
\[ \frac{\text{hr}^2}{\text{min}}, \]signaling that the conversion factor has been oriented incorrectly.
What Is Variation?
Variation describes a systematic way one quantity changes in relation to another.
The most common elementary forms include:
- direct variation;
- inverse variation;
- joint variation;
- combined variation;
- power variation.
Each model contains a constant that determines the strength of the relationship.
Direct Variation
A variable \(y\) varies directly with \(x\) if
\[ \boxed{y=kx}, \]where \(k\) is a nonzero constant called the constant of variation or constant of proportionality.
The ratio
\[ \frac{y}{x}=k \]remains constant whenever \(x\neq0\).
Recognizing Direct Variation
If
\[ y=4x, \]then doubling \(x\) doubles \(y\), tripling \(x\) triples \(y\), and so on.
For example,
\[ x=3\Longrightarrow y=12, \]while
\[ x=6\Longrightarrow y=24. \]The ratio remains
\[ \frac{y}{x}=4. \]Finding the Constant of Direct Variation
Suppose \(y\) varies directly with \(x\), and
\[ y=18 \]when
\[ x=6. \]Use
\[ y=kx. \]Substitute:
\[ 18=6k. \]Therefore,
\[ k=3. \]The variation equation is
\[ \boxed{y=3x}. \]Using a Direct-Variation Model
If
\[ y=3x, \]find \(y\) when \(x=14\).
Substitute:
\[ y=3(14)=42. \]Therefore,
\[ \boxed{y=42}. \]Direct Variation and Proportions
If \(y=kx\), then for two corresponding pairs
\[ (x_1,y_1) \]and
\[ (x_2,y_2), \]we have
\[ \frac{y_1}{x_1} = \frac{y_2}{x_2} = k. \]Therefore,
\[ \boxed{\frac{y_1}{x_1}=\frac{y_2}{x_2}}. \]Direct variation is therefore the algebraic generalization of equivalent ratios.
Direct Variation Graphs
The graph of
\[ y=kx \]is a straight line through the origin.
The constant \(k\) is the slope:
\[ \boxed{k=\frac{y}{x}}. \]A linear relationship of the form
\[ y=mx+b \]is a direct variation only when
\[ b=0. \]Inverse Variation
A variable \(y\) varies inversely with \(x\) if
\[ \boxed{y=\frac{k}{x}}, \qquad x\neq0. \]Multiplying both sides by \(x\) gives
\[ \boxed{xy=k}. \]Thus the product of corresponding values remains constant.
Recognizing Inverse Variation
Suppose
\[ y=\frac{24}{x}. \]If \(x=3\), then
\[ y=8. \]If \(x=6\), then
\[ y=4. \]Doubling \(x\) halves \(y\), and the product stays constant:
\[ 3(8)=6(4)=24. \]Finding the Constant of Inverse Variation
Suppose \(y\) varies inversely with \(x\), and \(y=10\) when \(x=6\).
Use
\[ y=\frac{k}{x}. \]Substitute:
\[ 10=\frac{k}{6}. \]Therefore,
\[ k=60. \]The variation equation is
\[ \boxed{y=\frac{60}{x}}. \]Using an Inverse-Variation Model
If
\[ y=\frac{60}{x}, \]find \(y\) when \(x=15\).
\[ y=\frac{60}{15}=4. \]Therefore,
\[ \boxed{y=4}. \]Inverse Variation from Two Data Pairs
If \(xy=k\), then
\[ x_1y_1=x_2y_2. \]Suppose \(y=12\) when \(x=5\), and find \(y\) when \(x=8\).
\[ 5(12)=8y. \] \[ 60=8y. \] \[ y=7.5. \]Therefore,
\[ \boxed{y=7.5}. \]Direct and Inverse Variation Behave Differently
In direct variation,
\[ y=kx, \]so multiplying \(x\) by a factor \(c\) multiplies \(y\) by the same factor.
In inverse variation,
\[ y=\frac{k}{x}, \]multiplying \(x\) by \(c\) divides \(y\) by \(c\).
Recognizing this directional behavior is often enough to identify which model is appropriate.
Joint Variation
A variable varies jointly with two or more variables when it is proportional to their product.
If \(z\) varies jointly with \(x\) and \(y\), then
\[ \boxed{z=kxy}. \]The constant \(k\) is found from known corresponding values.
Example: Joint Variation
Suppose \(z\) varies jointly with \(x\) and \(y\), and
\[ z=48 \]when
\[ x=4,\qquad y=3. \]Use
\[ z=kxy. \]Substitute:
\[ 48=k(4)(3). \]Therefore,
\[ k=4. \]The model is
\[ \boxed{z=4xy}. \]Using a Joint-Variation Model
If
\[ z=4xy, \]find \(z\) when \(x=5\) and \(y=7\).
\[ z=4(5)(7)=140. \]Therefore,
\[ \boxed{z=140}. \]Combined Variation
A relationship may be direct with some variables and inverse with others.
If \(z\) varies directly with \(x\) and inversely with \(y\), then
\[ \boxed{z=\frac{kx}{y}}, \qquad y\neq0. \]This is often called combined variation.
Example: Combined Variation
Suppose
\[ z=\frac{kx}{y}, \]and \(z=12\) when \(x=8\) and \(y=4\).
Substitute:
\[ 12=\frac{8k}{4}. \]Simplify:
\[ 12=2k. \]Therefore,
\[ k=6. \]The model is
\[ \boxed{z=\frac{6x}{y}}. \]Power Variation
A variable may vary directly with a power of another variable.
If \(y\) varies directly with the square of \(x\), then
\[ \boxed{y=kx^2}. \]If \(y\) varies directly with the cube of \(x\), then
\[ \boxed{y=kx^3}. \]More generally,
\[ \boxed{y=kx^n}. \]Example: Square Variation
Suppose \(y\) varies directly with \(x^2\), and \(y=45\) when \(x=3\).
Write
\[ y=kx^2. \]Substitute:
\[ 45=k(3^2)=9k. \]Therefore,
\[ k=5. \]The model is
\[ \boxed{y=5x^2}. \]Inverse Power Variation
A quantity may vary inversely with a power.
If \(y\) varies inversely with the square of \(x\), then
\[ \boxed{y=\frac{k}{x^2}}, \qquad x\neq0. \]If \(x\) is doubled, then \(x^2\) becomes four times as large, so \(y\) becomes one-fourth as large.
Scaling Laws Are Variation Laws
The earlier area and volume scaling rules are examples of power variation.
For similar figures, area varies as the square of a characteristic length:
\[ A=kL^2. \]Volume varies as the cube:
\[ V=kL^3. \]Thus scaling and variation are not separate ideas. Scaling laws are structured variation relationships.
Tables and Direct Variation
Consider the table
\[ \begin{array}{c|cccc} x & 2 & 4 & 6 & 8\\ \hline y & 7 & 14 & 21 & 28 \end{array} \]Compute
\[ \frac{y}{x}. \]Every pair gives
\[ \frac{7}{2}=3.5. \]Therefore,
\[ \boxed{y=3.5x}. \]Tables and Inverse Variation
Consider
\[ \begin{array}{c|cccc} x & 2 & 3 & 4 & 6\\ \hline y & 12 & 8 & 6 & 4 \end{array} \]Compute the products:
\[ 2(12)=3(8)=4(6)=6(4)=24. \]Since \(xy\) is constant,
\[ \boxed{y=\frac{24}{x}}. \]Determining Whether Data Show Direct Variation
A data set shows exact direct variation when the ratio
\[ \frac{y}{x} \]is constant for all nonzero \(x\).
For example,
\[ (2,6),\quad(5,15),\quad(9,27) \]all satisfy
\[ \frac{y}{x}=3. \]Therefore,
\[ y=3x. \]Determining Whether Data Show Inverse Variation
A data set shows exact inverse variation when the product
\[ xy \]is constant.
For example,
\[ (2,30),\quad(3,20),\quad(5,12) \]all satisfy
\[ xy=60. \]Therefore,
\[ y=\frac{60}{x}. \]Not Every Increasing Relationship Is Direct Variation
Consider
\[ y=3x+5. \]This function increases as \(x\) increases, but
\[ \frac{y}{x} \]is not constant.
It also does not pass through the origin.
Therefore,
\[ \boxed{y=3x+5} \]is linear, but it is not a direct variation.
Not Every Decreasing Relationship Is Inverse Variation
A decreasing function is not automatically an inverse variation.
For example,
\[ y=10-x \]decreases as \(x\) increases, but the product
\[ xy=x(10-x) \]is not constant.
Therefore, it is not of the form
\[ y=\frac{k}{x}. \]Variation Statements Translate Directly into Formulas
The language of a variation problem determines the algebraic model.
- “\(y\) varies directly with \(x\)” means \(y=kx\).
- “\(y\) varies inversely with \(x\)” means \(y=\frac{k}{x}\).
- “\(z\) varies jointly with \(x\) and \(y\)” means \(z=kxy\).
- “\(z\) varies directly with \(x\) and inversely with \(y\)” means \(z=\frac{kx}{y}\).
- “\(y\) varies directly with the square of \(x\)” means \(y=kx^2\).
Example: Translate and Solve
Suppose \(P\) varies directly with \(q\) and inversely with \(r\).
The model is
\[ P=\frac{kq}{r}. \]If \(P=18\) when \(q=12\) and \(r=4\), then
\[ 18=\frac{12k}{4}=3k. \]Thus
\[ k=6. \]Therefore,
\[ \boxed{P=\frac{6q}{r}}. \]Example: Use the Model with New Values
Continue with
\[ P=\frac{6q}{r}. \]Find \(P\) when \(q=15\) and \(r=9\).
\[ P=\frac{6(15)}{9}. \]Simplify:
\[ P=10. \]Therefore,
\[ \boxed{P=10}. \]Unit Conversion as Multiplication by One
Conversion factors do not change the underlying physical quantity because they represent \(1\).
For example,
\[ \frac{100\text{ cm}}{1\text{ m}}=1 \]in the sense that the numerator and denominator represent equal lengths.
Therefore,
\[ 2.3\text{ m} \left( \frac{100\text{ cm}}{1\text{ m}} \right) \]changes the unit representation but not the length itself.
Scale Conversion and Unit Conversion Can Be Combined
Suppose a map scale is
\[ 1\text{ cm}:8\text{ km}. \]A map distance is \(45\) millimeters.
First convert millimeters to centimeters:
\[ 45\text{ mm} \left( \frac{1\text{ cm}}{10\text{ mm}} \right) = 4.5\text{ cm}. \]Then apply the map scale:
\[ 4.5\text{ cm} \left( \frac{8\text{ km}}{1\text{ cm}} \right) = 36\text{ km}. \]Therefore,
\[ \boxed{36\text{ km}}. \]Example: Scale with Area
Two similar floor plans have linear scale factor
\[ \frac{5}{4}. \]If the smaller plan represents an area of \(96\text{ m}^2\), the larger corresponding area is
\[ 96\left(\frac{5}{4}\right)^2. \]Compute:
\[ 96\left(\frac{25}{16}\right)=150. \]Therefore,
\[ \boxed{150\text{ m}^2}. \]Example: Scale with Volume
Two similar containers have linear scale factor \(1.2\).
If the smaller container has volume \(250\text{ cm}^3\), then the larger volume is
\[ 250(1.2)^3. \]Since
\[ 1.2^3=1.728, \]the new volume is
\[ 250(1.728)=432. \]Therefore,
\[ \boxed{432\text{ cm}^3}. \]Example: Conversion with Squared Units
Convert \(2.5\text{ m}^2\) to square centimeters.
Since
\[ 1\text{ m}=100\text{ cm}, \]use
\[ 2.5\text{ m}^2 \left( \frac{100\text{ cm}}{1\text{ m}} \right)^2. \]Then
\[ 2.5(10{,}000)=25{,}000. \]Therefore,
\[ \boxed{2.5\text{ m}^2=25{,}000\text{ cm}^2}. \]Example: Conversion with Cubic Units
Convert \(0.35\text{ m}^3\) to cubic centimeters.
\[ 0.35\text{ m}^3 \left( \frac{100\text{ cm}}{1\text{ m}} \right)^3. \]Since
\[ 100^3=1{,}000{,}000, \]we obtain
\[ 0.35(1{,}000{,}000)=350{,}000. \]Therefore,
\[ \boxed{350{,}000\text{ cm}^3}. \]Checking Scale Problems
Ask whether the direction of change matches the scale factor.
If \(k=3\), lengths should become larger.
If
\[ k=\frac{1}{4}, \]lengths should become smaller.
A result moving in the opposite direction is evidence that the ratio or scale factor may have been reversed.
Checking Unit Conversions
Use both unit cancellation and magnitude.
Converting \(5\) meters to centimeters should produce a number greater than \(5\) because centimeters are smaller units.
\[ 5\text{ m}=500\text{ cm}. \]Converting \(500\) centimeters to meters should produce a smaller number:
\[ 500\text{ cm}=5\text{ m}. \]Checking Variation Models
After finding \(k\), substitute the known data back into the model.
Suppose direct variation gives
\[ y=2.4x. \]If the original data were \(x=5\), \(y=12\), then
\[ 2.4(5)=12. \]The model reproduces the known pair, so the constant was calculated correctly.
Common Error: Reversing the Scale Factor
If an original length \(8\) becomes \(20\), the scale factor from original to new is
\[ \frac{20}{8}=2.5. \]The reciprocal
\[ \frac{8}{20}=0.4 \]would describe the scale factor from the new figure back to the original.
Always state the direction of the scale.
Common Error: Applying a Linear Scale Factor to Area
If lengths are doubled, area is not merely doubled.
The correct area multiplier is
\[ 2^2=4. \]Thus a region of area \(30\) becomes
\[ 30(4)=120, \]not \(60\).
Common Error: Applying a Linear Scale Factor to Volume
If all dimensions are tripled, volume is multiplied by
\[ 3^3=27, \]not by \(3\).
A volume of \(10\) becomes
\[ 10(27)=270. \]Common Error: Flipping a Conversion Factor Incorrectly
To convert feet to inches, the feet must cancel:
\[ 4\text{ ft} \left( \frac{12\text{ in}}{1\text{ ft}} \right). \]Using
\[ \frac{1\text{ ft}}{12\text{ in}} \]would not cancel the starting unit correctly.
Common Error: Ignoring Powers on Units
If
\[ 1\text{ m}=100\text{ cm}, \]then
\[ 1\text{ m}^2\neq100\text{ cm}^2. \]Instead,
\[ 1\text{ m}^2=(100\text{ cm})^2=10{,}000\text{ cm}^2. \]The exponent applies to the conversion factor as well as the unit.
Common Error: Treating Every Linear Function as Direct Variation
The function
\[ y=5x+2 \]is linear, but it is not a direct variation because it does not have the form
\[ y=kx. \]In particular,
\[ y(0)=2\neq0. \]Common Error: Treating Every Decreasing Pattern as Inverse Variation
To verify inverse variation, check whether
\[ xy \]is constant.
A relationship can decrease without satisfying
\[ y=\frac{k}{x}. \]The behavior alone is not enough; the algebraic invariant must be checked.
Common Error: Losing the Constant of Variation
The statement “\(y\) varies directly with \(x^2\)” does not imply
\[ y=x^2. \]The correct general model is
\[ \boxed{y=kx^2}. \]The constant \(k\) must be determined from data unless it is otherwise specified.
Common Error: Using Addition Instead of Multiplicative Variation
If \(y\) varies jointly with \(x\) and \(z\), the model is
\[ y=kxz, \]not
\[ y=k(x+z). \]The word jointly indicates proportionality to the product.
A Reliable Procedure for Scale Problems
- Identify the original and corresponding new measurements.
- Determine the direction of the scale.
- Compute or identify the linear scale factor \(k\).
- Multiply corresponding lengths by \(k\).
- Use \(k^2\) for corresponding areas.
- Use \(k^3\) for corresponding volumes.
- Check whether the result enlarged or reduced as expected.
A Reliable Procedure for Unit Conversions
- Write the starting quantity with its unit.
- Identify the desired final unit.
- Choose conversion factors equal to \(1\).
- Orient each factor so unwanted units cancel.
- Multiply the numerical factors.
- Verify that only the desired units remain.
- Check whether the numerical size is reasonable for the new unit.
A Reliable Procedure for Variation Problems
- Translate the verbal variation statement into an equation containing \(k\).
- Substitute the known corresponding values.
- Solve for the constant of variation \(k\).
- Write the complete variation equation.
- Substitute the new input values.
- Solve for the requested quantity.
- Check the result against the expected direct or inverse behavior.
Practice 1: Find a Scale Factor
A length of \(12\) is enlarged to \(21\). Find the scale factor.
Practice 2: Use a Map Scale
A map uses the scale
\[ 1\text{ cm}:18\text{ km}. \]Two points are \(6.5\) centimeters apart on the map. Find the actual distance.
Practice 3: Find a Drawing Length
A blueprint uses the scale
\[ 1\text{ in}:4\text{ ft}. \]How many inches long should a \(30\)-foot wall appear on the blueprint?
Practice 4: Area Scale
A figure is enlarged by linear scale factor \(2.5\). By what factor does its area change?
Practice 5: Volume Scale
A solid is reduced by linear scale factor
\[ \frac{1}{2}. \]By what factor does its volume change?
Practice 6: One-Step Conversion
Convert \(9.5\) feet to inches.
Practice 7: Multi-Step Conversion
Convert \(2.75\) hours to seconds.
Practice 8: Convert a Rate
Convert
\[ 90\text{ km/h} \]to meters per second.
Practice 9: Convert Area Units
Convert
\[ 1.8\text{ m}^2 \]to square centimeters.
Practice 10: Direct Variation
Suppose \(y\) varies directly with \(x\), and \(y=28\) when \(x=7\). Find the variation equation.
Practice 11: Use Direct Variation
Using the relationship from Practice 10, find \(y\) when \(x=13\).
Practice 12: Inverse Variation
Suppose \(y\) varies inversely with \(x\), and \(y=9\) when \(x=8\). Find the variation equation.
Practice 13: Use Inverse Variation
Using the relationship from Practice 12, find \(y\) when \(x=12\).
Practice 14: Joint Variation
Suppose \(z\) varies jointly with \(x\) and \(y\), and \(z=60\) when \(x=5\) and \(y=4\). Find the variation equation.
Practice 15: Combined Variation
Suppose \(P\) varies directly with \(q\) and inversely with \(r\). If \(P=15\) when \(q=10\) and \(r=4\), find \(P\) when \(q=18\) and \(r=6\).
Practice Answers
Practice 1
\[ k=\frac{21}{12}=\frac{7}{4}=1.75. \]Therefore,
\[ \boxed{k=1.75}. \]Practice 2
\[ 6.5(18)=117. \]Therefore,
\[ \boxed{117\text{ km}}. \]Practice 3
\[ \frac{1\text{ in}}{4\text{ ft}} = \frac{x\text{ in}}{30\text{ ft}}. \] \[ 4x=30. \] \[ x=7.5. \]Therefore,
\[ \boxed{7.5\text{ in}}. \]Practice 4
Area changes by the square of the linear factor:
\[ (2.5)^2=6.25. \]Therefore,
\[ \boxed{6.25}. \]Practice 5
Volume changes by the cube of the linear factor:
\[ \left(\frac{1}{2}\right)^3=\frac{1}{8}. \]Therefore, the new volume is
\[ \boxed{\frac{1}{8}} \]of the original volume.
Practice 6
\[ 9.5\text{ ft} \left( \frac{12\text{ in}}{1\text{ ft}} \right) = 114\text{ in}. \]Therefore,
\[ \boxed{114\text{ in}}. \]Practice 7
\[ 2.75(60)(60)=9900. \]Therefore,
\[ \boxed{9900\text{ s}}. \]Practice 8
\[ 90\frac{\text{km}}{\text{hr}} \left( \frac{1000\text{ m}}{1\text{ km}} \right) \left( \frac{1\text{ hr}}{3600\text{ s}} \right) = 25\frac{\text{m}}{\text{s}}. \]Therefore,
\[ \boxed{25\text{ m/s}}. \]Practice 9
\[ 1.8\text{ m}^2 \left( \frac{100\text{ cm}}{1\text{ m}} \right)^2 = 18{,}000\text{ cm}^2. \]Therefore,
\[ \boxed{18{,}000\text{ cm}^2}. \]Practice 10
Use
\[ y=kx. \]Substitute:
\[ 28=7k. \]Therefore,
\[ k=4, \]so
\[ \boxed{y=4x}. \]Practice 11
\[ y=4(13)=52. \]Therefore,
\[ \boxed{y=52}. \]Practice 12
Use
\[ y=\frac{k}{x}. \]Substitute:
\[ 9=\frac{k}{8}. \]Therefore,
\[ k=72. \]The model is
\[ \boxed{y=\frac{72}{x}}. \]Practice 13
\[ y=\frac{72}{12}=6. \]Therefore,
\[ \boxed{y=6}. \]Practice 14
Use
\[ z=kxy. \]Then
\[ 60=k(5)(4)=20k. \]Therefore,
\[ k=3. \]The model is
\[ \boxed{z=3xy}. \]Practice 15
Write
\[ P=\frac{kq}{r}. \]Use the known values:
\[ 15=\frac{10k}{4}. \] \[ 60=10k. \] \[ k=6. \]Therefore,
\[ P=\frac{6q}{r}. \]For \(q=18\) and \(r=6\),
\[ P=\frac{6(18)}{6}=18. \]Thus,
\[ \boxed{P=18}. \]What to Remember
A linear scale factor \(k\) changes lengths according to
\[ \boxed{L'=kL}. \]Corresponding areas change by
\[ \boxed{k^2}, \]and corresponding volumes change by
\[ \boxed{k^3}. \]Unit conversions use ratios equal to \(1\), oriented so unwanted units cancel.
Direct variation has the form
\[ \boxed{y=kx}, \]while inverse variation has the form
\[ \boxed{y=\frac{k}{x}}. \]Joint, combined, and power variation extend the same idea to more complex dependencies.
The most important habit is to identify what must remain invariant: proportional shape, physical quantity, unit equivalence, a constant ratio, or a constant product.
Where We Go Next
Rates are among the most important applications of proportional reasoning. Once distance, rate, and time are linked by a single equation, many travel, motion, and scheduling problems become systematic algebra problems.
The next lesson, Distance, Rate, and Time, develops that relationship and its most common problem families.