essential question how can you recognize when two quantities vary directly or inversely? 1 exploration: recognizing direct variation go to bigideasmath.com for an interactive tool to investigate this exploration. work with a partner. you hang different weights from the same spring. a. describe the relationship between the weight x and the distance d the spring stretches from equilibrium. explain why the distance is said to vary directly with the weight. b. estimate the values of d from the figure. then draw a scatter plot of the data. what are the characteristics of the graph? c. write an equation that represents d as a function of x. d. in physics, the relationship between d and x is described by hooke’s law. how would you describe hooke’s law?

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Analyze the Relationship
As the weight $ x $ (in kg) increases (from 0 kg to 0.7 kg), the distance $ d $ (in cm) the spring stretches from equilibrium also increases. This is because the force exerted by the weight (gravity) causes the spring to stretch, and a larger weight means a larger force.
## Step2: Direct Variation Reason
For direct variation, the relationship is of the form $ d = kx $ (where $ k $ is a constant). As $ x $ increases by a constant factor (e.g., from 0.1 to 0.2 kg, it doubles), $ d $ should also increase by the same factor (if $ k $ is constant). From the figure, as weight increases, the stretch distance increases proportionally, so $ d $ varies directly with $ x $ because $ \frac{d}{x} $ should be constant (the constant of proportionality $ k $).

# Answer:
As the weight $ x $ increases, the distance $ d $ the spring stretches also increases. The distance varies directly with the weight because the ratio $ \frac{d}{x} $ is constant (proportional relationship, $ d = kx $ for some constant $ k $).

### Part (b)
# Explanation:
## Step1: Estimate $ d $ Values
- For $ x = 0 $ kg, $ d = 0 $ cm (equilibrium, no stretch).
- For $ x = 0.1 $ kg, estimate $ d \approx 1 $ cm (visually, the spring stretches a small amount).
- For $ x = 0.2 $ kg, $ d \approx 2 $ cm.
- For $ x = 0.3 $ kg, $ d \approx 3 $ cm.
- For $ x = 0.4 $ kg, $ d \approx 4 $ cm.
- For $ x = 0.5 $ kg, $ d \approx 5 $ cm.
- For $ x = 0.6 $ kg, $ d \approx 6 $ cm.
- For $ x = 0.7 $ kg, $ d \approx 7 $ cm.
## Step2: Scatter Plot Characteristics
When we plot the points $(0,0)$, $(0.1,1)$, $(0.2,2)$, ..., $(0.7,7)$ on the scatter plot, the points lie on a straight line passing through the origin $(0,0)$. This is a characteristic of a direct variation relationship (linear, proportional).

# Answer:
Estimated $ d $ values: $ (0,0) $, $ (0.1,1) $, $ (0.2,2) $, $ (0.3,3) $, $ (0.4,4) $, $ (0.5,5) $, $ (0.6,6) $, $ (0.7,7) $. The scatter plot is a straight line passing through the origin, showing a linear (proportional) relationship.

### Part (c)
# Explanation:
## Step1: Determine the Constant $ k $
From the estimated values (e.g., $ x = 0.1 $ kg, $ d = 1 $ cm), use $ d = kx $. Solve for $ k $: $ k=\frac{d}{x}=\frac{1}{0.1} = 10 $ cm/kg. Check with another point: $ x = 0.2 $ kg, $ d = 2 $ cm, $ k=\frac{2}{0.2}=10 $. So $ k = 10 $.
## Step2: Write the Equation
Using $ d = kx $ with $ k = 10 $, the equation is $ d = 10x $.

# Answer:
The equation is $ \boldsymbol{d = 10x} $ (where $ d $ is in centimeters and $ x $ is in kilograms).

### Part (d)
# Explanation:
## Step1: Recall Hooke's Law
From the exploration, the stretch distance $ d $ of a spring is directly proportional to the applied force (which is proportional to the weight $ x $ here, since $ F = mg $, and $ m $ is $ x $ in kg). So Hooke's Law states that the force $ F $ (or the stretch distance $ d $, since $ d \propto F $) exerted by a spring is directly proportional to the displacement (stretch) from its equilibrium position, as long as the elastic limit is not exceeded. In this context, the stretch distance $ d $ is directly proportional to the weight (force) applied, so $ d = kx $ (or $ F = kd $ where $ F $ is force, $ k $ is spring constant).

# Answer:
Hooke’s Law states that the distance a spring stretches (from equilibrium) is directly proportional to the weight (force) applied to it, as long as the spring is within its elastic limit. Mathematically, $ d = kx $ (or $ F = kd $ where $ F $ is the force, $ d $ is the stretch distance, and $ k $ is the spring constant).

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Analyze the Relationship

Step2: Direct Variation Reason

Step1: Estimate \( d \) Values

Step2: Scatter Plot Characteristics

Step1: Determine the Constant \( k \)

Step2: Write the Equation

Answer:

Part (b)