Problem Solving

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


You might be able to answer these goal-based questions with this technique:

  1. Identify the goal value, but don't use it.
  2. Calculate the actual value, without using the goal value.
  3. 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 $$v=\frac{\Delta x}{\Delta t}$$ $$v=\frac{100\,\mathrm{m}}{15 \,\mathrm{s}}$$ $$v=6.\overline{66} \,\mathrm{ \tfrac{m}{s}}$$ $$6.\overline{66} \,\mathrm{ \tfrac{m}{s}} < 10 \,\mathrm{ \tfrac{m}{s}}$$ $$\text{actual < goal}$$ $$\text{I exaggerated :( }$$