Using Proportions
If you change a variable we can predict whether other variables in the same equation will increase or decrease. The first
step is to decide if the relationship between the variables is directly proportional or inversely proportional.
$$ {\color{#09e}{\Uparrow} \atop y} {\atop = k } {\color{#09e}{\Uparrow} \atop x} \quad\quad\quad {\color{#f00}{\Downarrow}
\atop y} {\atop = k } {\color{#f00}{\Downarrow} \atop x} \quad\quad\quad \frac{ y \color{#f00}{\Downarrow}}{x \color{#f00}{\Downarrow}}
= k$$
directly proportional = a constant ratio, k, between two variables. Both variables increase and decrease together.
$$ {\color{#09e}{\Uparrow} \atop y} {\color{#f00}{\Downarrow} \atop x} {\atop = k } \quad\quad\quad {\color{#f00}{\Downarrow}
\atop y} {\color{#09e}{\Uparrow} \atop x} {\atop = k } \quad\quad\quad y {\color{#f00}{\Downarrow}} = \frac{k}{x \color{#09e}{\Uparrow}}
$$
inversely proportional = a constant ratio, k, between a variable and the inverse of a variable. Variables increase
and decrease
opposite to each other.
You can use proportions when everything else in an equation is constant except two variables. Then you predict how one
will change when the other changes.
$$F = ma$$
Example: A force of 100 N produces an acceleration of 10 m/s² for a 10 kg mass. How should you change the force
to reduce the acceleration?
solution
$$ \text{m is constant}$$ $$F = ma$$ $$ \text{F and a are directly proportional}$$ $${\color{#f00}{\Downarrow} \atop F} {\atop
=} { \atop m} {\color{#f00}{\Downarrow} \atop a} $$ $$ \text{Decreasing F will decrease a}$$
$$v = \frac{\Delta x}{\Delta t}$$
Example: You can run 100 m in about 15 s. If you want to decrease your time, should you increase or decrease
your velocity?
solution
$$ \text{Δx is constant}$$ $$v = \frac{\Delta x}{\Delta t}$$ $$v \Delta t = \Delta x$$ $$ \text{v and Δt are inversely proportional}$$
$${\color{#09e}{\Uparrow} \atop v} {\color{#f00}{\Downarrow} \atop \Delta t} {\atop =} { \atop \Delta x}$$ $$ \text{increasing
v will decrease Δt}$$
Actual vs. Goal
-
Is 10 N enough force?
Is 10 N too much force?
Is this almost the same as 100 J?
Is this different from 100 J?
Am I over the 10 kg limit?
Am I below the 10 kg limit?
-
Is 100 W enough power for 2 hours?
Is 100 W too much power for 2 hours?
Will it happen in time?
Will it be too late?
Is my speed above the speed limit?
Is my speed below the speed limit?
You might be able to answer these goal-based questions with this technique:
- Identify the goal value, but don't use it.
- Calculate the actual value, without using the goal value.
- Compare the actual to the goal, and see which is bigger.
$$v = \frac{\Delta x}{\Delta t}$$
Example: We are driving to a friend's house in the country. I know the house is on this road in about 67 miles.
If we drove at an average speed of 35 mph for 2 hours, will we pass the house?
1. Identify the goal value, but don't use it.
$$\text{The goal is 67 miles.}$$
2. Calculate the actual value, without using the goal value.
$$v=\frac{\Delta x}{\Delta t}$$ $$v\Delta t=\Delta x$$ $$(35 \, \mathrm{\tfrac{mile}{hour}} )(2 \,\mathrm{hour})=\Delta x$$
$$70 \,\mathrm{mile}=\Delta x$$
3. Compare the actual to the goal, and see which is bigger.
$$70 \,\mathrm{mile} > 67 \,\mathrm{mile}$$ $$\text{actual > goal}$$ $$\text{We will pass the house.}$$
$$v = \frac{\Delta x}{\Delta t}$$
Example: I claimed that I could run above a speed of 10 meters per second for 100 meters. When I actually ran
100 meters, it took me 15 seconds.
Did I exaggerate?
strategy
The goal is a velocity of 10 m/s. Don't use it.
Solve for the actual velocity using the 100 m and 15 s.
Compare the goal value to the actual value.
solution
$$v=\frac{\Delta x}{\Delta t}$$ $$v=\frac{100\,\mathrm{m}}{15 \,\mathrm{s}}$$ $$v=6.\overline{66} \,\mathrm{ \tfrac{m}{s}}$$
$$6.\overline{66} \,\mathrm{ \tfrac{m}{s}}
< 10 \,\mathrm{ \tfrac{m}{s}}$$ $$\text{actual < goal}$$ $$\text{I exaggerated :( }$$
More Practice
Problem solving is easier when you name the actual value, goal value, and relationship. These examples focus on proportions and multipliers.
Example: Solve the proportion 2 over 5 equals x over 20.
solution
$$\frac{2}{5}=\frac{x}{20}$$
$$5x=(2)(20)$$
$$x=\frac{40}{5}$$
$$x=8$$
Example: Solve the proportion 3 over 7 equals x over 42.
solution
$$\frac{3}{7}=\frac{x}{42}$$
$$7x=(3)(42)$$
$$x=\frac{126}{7}$$
$$x=18$$
Example: Solve the proportion 4 over 9 equals x over 36.
solution
$$\frac{4}{9}=\frac{x}{36}$$
$$9x=(4)(36)$$
$$x=\frac{144}{9}$$
$$x=16$$
Example: Solve the proportion 5 over 12 equals x over 60.
solution
$$\frac{5}{12}=\frac{x}{60}$$
$$12x=(5)(60)$$
$$x=\frac{300}{12}$$
$$x=25$$
Example: The actual value is 4 and the goal value is 20. What multiplier gets from actual to goal?
solution
$$\mathrm{multiplier}=\frac{\mathrm{goal}}{\mathrm{actual}}$$
$$\mathrm{multiplier}=\frac{20}{4}$$
$$\mathrm{multiplier}=5$$
Example: The actual value is 7 and the goal value is 35. What multiplier gets from actual to goal?
solution
$$\mathrm{multiplier}=\frac{\mathrm{goal}}{\mathrm{actual}}$$
$$\mathrm{multiplier}=\frac{35}{7}$$
$$\mathrm{multiplier}=5$$
Example: The actual value is 12 and the goal value is 3. What multiplier gets from actual to goal?
solution
$$\mathrm{multiplier}=\frac{\mathrm{goal}}{\mathrm{actual}}$$
$$\mathrm{multiplier}=\frac{3}{12}$$
$$\mathrm{multiplier}=0.25$$
Example: The actual value is 5 and the goal value is 50. What multiplier gets from actual to goal?
solution
$$\mathrm{multiplier}=\frac{\mathrm{goal}}{\mathrm{actual}}$$
$$\mathrm{multiplier}=\frac{50}{5}$$
$$\mathrm{multiplier}=10$$
Example: The actual value is 80 and the goal value is 20. What multiplier gets from actual to goal?
solution
$$\mathrm{multiplier}=\frac{\mathrm{goal}}{\mathrm{actual}}$$
$$\mathrm{multiplier}=\frac{20}{80}$$
$$\mathrm{multiplier}=0.25$$
Example: A value changes at 3 units per second for 10 s. What is the total change?
solution
$$\mathrm{change}=\mathrm{rate}\times\mathrm{time}$$
$$\mathrm{change}=(3)(10)$$
$$\mathrm{change}=30$$
Example: A value changes at 5 units per second for 12 s. What is the total change?
solution
$$\mathrm{change}=\mathrm{rate}\times\mathrm{time}$$
$$\mathrm{change}=(5)(12)$$
$$\mathrm{change}=60$$
Example: A value changes at 0.5 units per second for 40 s. What is the total change?
solution
$$\mathrm{change}=\mathrm{rate}\times\mathrm{time}$$
$$\mathrm{change}=(0.5)(40)$$
$$\mathrm{change}=20$$
Example: A value changes at 8 units per second for 6 s. What is the total change?
solution
$$\mathrm{change}=\mathrm{rate}\times\mathrm{time}$$
$$\mathrm{change}=(8)(6)$$
$$\mathrm{change}=48$$
Example: Solve the proportion 6 over 6 equals x over 22.
solution
$$\frac{6}{6}=\frac{x}{22}$$
$$6x=(6)(22)$$
$$x=22$$
Example: Solve the proportion 7 over 7 equals x over 23.
solution
$$\frac{7}{7}=\frac{x}{23}$$
$$7x=(7)(23)$$
$$x=23$$
Example: Solve the proportion 8 over 8 equals x over 24.
solution
$$\frac{8}{8}=\frac{x}{24}$$
$$8x=(8)(24)$$
$$x=24$$
Example: Solve the proportion 9 over 9 equals x over 25.
solution
$$\frac{9}{9}=\frac{x}{25}$$
$$9x=(9)(25)$$
$$x=25$$
Example: Solve the proportion 2 over 10 equals x over 26.
solution
$$\frac{2}{10}=\frac{x}{26}$$
$$10x=(2)(26)$$
$$x=5.2$$
Example: Solve the proportion 3 over 11 equals x over 27.
solution
$$\frac{3}{11}=\frac{x}{27}$$
$$11x=(3)(27)$$
$$x=7.36$$
Example: Solve the proportion 4 over 3 equals x over 28.
solution
$$\frac{4}{3}=\frac{x}{28}$$
$$3x=(4)(28)$$
$$x=37.33$$
Example: Solve the proportion 5 over 4 equals x over 29.
solution
$$\frac{5}{4}=\frac{x}{29}$$
$$4x=(5)(29)$$
$$x=36.25$$
Example: Solve the proportion 6 over 5 equals x over 30.
solution
$$\frac{6}{5}=\frac{x}{30}$$
$$5x=(6)(30)$$
$$x=36$$
Example: Solve the proportion 7 over 6 equals x over 31.
solution
$$\frac{7}{6}=\frac{x}{31}$$
$$6x=(7)(31)$$
$$x=36.17$$
Example: Solve the proportion 8 over 7 equals x over 32.
solution
$$\frac{8}{7}=\frac{x}{32}$$
$$7x=(8)(32)$$
$$x=36.57$$
Example: Solve the proportion 9 over 8 equals x over 33.
solution
$$\frac{9}{8}=\frac{x}{33}$$
$$8x=(9)(33)$$
$$x=37.13$$
Example: Solve the proportion 2 over 9 equals x over 34.
solution
$$\frac{2}{9}=\frac{x}{34}$$
$$9x=(2)(34)$$
$$x=7.56$$
Example: Solve the proportion 3 over 10 equals x over 35.
solution
$$\frac{3}{10}=\frac{x}{35}$$
$$10x=(3)(35)$$
$$x=10.5$$
Example: Solve the proportion 4 over 11 equals x over 36.
solution
$$\frac{4}{11}=\frac{x}{36}$$
$$11x=(4)(36)$$
$$x=13.09$$
Example: Solve the proportion 5 over 3 equals x over 37.
solution
$$\frac{5}{3}=\frac{x}{37}$$
$$3x=(5)(37)$$
$$x=61.67$$
Example: Solve the proportion 6 over 4 equals x over 38.
solution
$$\frac{6}{4}=\frac{x}{38}$$
$$4x=(6)(38)$$
$$x=57$$