Unit 1 Worksheet 1: Graphing Practice - Determine mathematical expressions for each data set by analyzing the relationship between variables.

Let's solve each data set step by step. The goal is to:

1. Graph the original data (with the first column as independent variable, second as dependent).
2. Determine if the relationship is linear or not.
3. Modify the data (if needed) so that it plots as a straight line.
4. Find the slope and y-intercept of the best-fit line.
5. Write a mathematical expression with units.

We'll go through each data set one at a time.

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🔹 Data Set 1: V (m³) vs P (pa)

| V (m³) | P (Pa) |
|--------|--------|
| 0.1 | 40 |
| 0.5 | 8 |
| 1 | 4 |
| 2 | 2 |
| 4 | 1 |
| 5 | 0.8 |
| 8 | 0.5 |
| 10 | 0.4 |

#### Step 1: Graph V vs P
Plot V on x-axis, P on y-axis.

Observation:
- As V increases, P decreases.
- The product $ V \times P $ appears constant?

Check:
- $ 0.1 \times 40 = 4 $
- $ 0.5 \times 8 = 4 $
- $ 1 \times 4 = 4 $
- $ 2 \times 2 = 4 $
- $ 4 \times 1 = 4 $
- $ 5 \times 0.8 = 4 $
- $ 8 \times 0.5 = 4 $
- $ 10 \times 0.4 = 4 $

✔ So $ PV = 4 $ → $ P = \frac{4}{V} $

This is an inverse relationship, so plotting $ P $ vs $ V $ gives a hyperbola.

To make it linear, we can plot $ P $ vs $ \frac{1}{V} $.

Let’s create a new column: $ \frac{1}{V} $

| V (m³) | P (Pa) | 1/V (1/m³) |
|--------|--------|------------|
| 0.1 | 40 | 10 |
| 0.5 | 8 | 2 |
| 1 | 4 | 1 |
| 2 | 2 | 0.5 |
| 4 | 1 | 0.25 |
| 5 | 0.8 | 0.2 |
| 8 | 0.5 | 0.125 |
| 10 | 0.4 | 0.1 |

Now plot $ P $ vs $ \frac{1}{V} $. This should be a straight line.

Let’s find the slope and intercept.

Use two points: (10, 40) and (0.1, 0.4)

Wait — better to use (1, 4) and (10, 40):

Actually, from above:
- When $ \frac{1}{V} = 10 $, $ P = 40 $
- When $ \frac{1}{V} = 1 $, $ P = 4 $
- When $ \frac{1}{V} = 0.1 $, $ P = 0.4 $

So $ P = 4 \times \frac{1}{V} $

Thus: $ P = 4 \cdot \left(\frac{1}{V}\right) $

So slope = 4 Pa·m³, y-intercept = 0

✔ Mathematical Expression #1:
$$
P = \frac{4\ \text{Pa·m}^3}{V}
$$
or
$$
P = \frac{4}{V} \quad \text{(with units: Pa = (Pa·m³)/m³)}
$$

> Note: This is Boyle’s Law behavior: $ PV = \text{constant} $

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🔹 Data Set 2: t (s) vs x (m)

| t (s) | x (m) |
|-------|-------|
| 0.1 | 0.03 |
| 0.2 | 0.12 |
| 0.5 | 0.75 |
| 1 | 3 |
| 2 | 12 |
| 3 | 27 |
| 4 | 48 |
| 5 | 75 |

Plot $ x $ vs $ t $

Try to see pattern:
- $ x $ vs $ t $: not linear? Let’s check ratios.

Try $ x / t $:
- 0.03 / 0.1 = 0.3
- 0.12 / 0.2 = 0.6
- 0.75 / 0.5 = 1.5
- 3 / 1 = 3
- 12 / 2 = 6
- 27 / 3 = 9
- 48 / 4 = 12
- 75 / 5 = 15

→ increasing → not constant velocity

Try $ x / t^2 $:
- 0.03 / (0.1)^2 = 0.03 / 0.01 = 3
- 0.12 / 0.04 = 3
- 0.75 / 0.25 = 3
- 3 / 1 = 3
- 12 / 4 = 3
- 27 / 9 = 3
- 48 / 16 = 3
- 75 / 25 = 3

✔ All give 3!

So $ x = 3 t^2 $

This suggests motion under constant acceleration starting from rest: $ x = \frac{1}{2} a t^2 $

Here: $ x = 3 t^2 $ → $ \frac{1}{2} a = 3 $ → $ a = 6\ \text{m/s}^2 $

To make it linear, plot $ x $ vs $ t^2 $

Let’s do that.

New column: $ t^2 $

| t (s) | t² (s²) | x (m) |
|-------|---------|-------|
| 0.1 | 0.01 | 0.03 |
| 0.2 | 0.04 | 0.12 |
| 0.5 | 0.25 | 0.75 |
| 1 | 1 | 3 |
| 2 | 4 | 12 |
| 3 | 9 | 27 |
| 4 | 16 | 48 |
| 5 | 25 | 75 |

Now plot $ x $ vs $ t^2 $ → straight line with slope = 3 m/s²

Y-intercept = 0

✔ Mathematical Expression #2:
$$
x = 3\ \frac{\text{m}}{\text{s}^2} \cdot t^2
$$
or
$$
x = 3t^2 \quad \text{(units: meters)}
$$

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🔹 Data Set 3: A (months) vs W (lbs)

| A (months) | W (lbs) |
|------------|---------|
| 1 | 7.3 |
| 2 | 9.4 |
| 3 | 10.5 |
| 4 | 12.0 |
| 5 | 13.0 |
| 6 | 14.3 |
| 7 | 15.2 |
| 8 | 16.7 |

Plot $ W $ vs $ A $

Check if linear:

Compute differences in W per month:
- 9.4 - 7.3 = 2.1
- 10.5 - 9.4 = 1.1
- 12.0 - 10.5 = 1.5
- 13.0 - 12.0 = 1.0
- 14.3 - 13.0 = 1.3
- 15.2 - 14.3 = 0.9
- 16.7 - 15.2 = 1.5

Not constant → not perfectly linear, but maybe approximately?

Try plotting and finding best-fit line.

Assume linear: $ W = m A + b $

Use first and last point:
- (1, 7.3), (8, 16.7)
- Slope $ m = \frac{16.7 - 7.3}{8 - 1} = \frac{9.4}{7} ≈ 1.343 $ lbs/month

Try $ W = 1.343 A + b $

At A=1: $ 1.343(1) + b = 7.3 $ → $ b = 7.3 - 1.343 = 5.957 $

Check at A=8: $ 1.343×8 + 5.957 ≈ 10.744 + 5.957 = 16.701 $ → matches

But let’s check middle values:

A=4: predicted = $ 1.343×4 + 5.957 = 5.372 + 5.957 = 11.329 $, actual = 12.0 → too low

A=5: pred = $ 1.343×5 + 5.957 = 6.715 + 5.957 = 12.672 $, actual = 13.0 → close

A=6: pred = $ 1.343×6 + 5.957 = 8.058 + 5.957 = 14.015 $, actual = 14.3 → close

A=7: pred = $ 1.343×7 + 5.957 = 9.391 + 5.957 = 15.348 $, actual = 15.2 → slightly high

So maybe acceptable approximation.

But wait — perhaps it's quadratic?

Check $ \Delta W $:
- ΔW: 2.1, 1.1, 1.5, 1.0, 1.3, 0.9, 1.5 → fluctuating

No clear trend.

Alternatively, try plotting $ W $ vs $ A $ directly.

It looks roughly linear, but with some scatter.

Let’s compute average slope over intervals.

From A=1 to A=8: total ΔW = 16.7 - 7.3 = 9.4 lbs over 7 months → avg rate = ~1.343 lbs/month

But growth slows down after A=3?

Wait — look again:

- Month 1–2: +2.1
- Month 2–3: +1.1 → drop
- Month 3–4: +1.5 → up
- Month 4–5: +1.0 → down
- Month 5–6: +1.3 → up
- Month 6–7: +0.9 → down
- Month 7–8: +1.5 → up

Fluctuates — possibly due to measurement error or biological variation.

But overall trend is approximately linear.

Let’s assume it's linear and find best-fit line using least squares or just pick two central points.

Take (3, 10.5) and (7, 15.2):

Slope $ m = \frac{15.2 - 10.5}{7 - 3} = \frac{4.7}{4} = 1.175 $ lbs/month

Then $ W = 1.175 A + b $

At A=3: $ 1.175×3 = 3.525 $, so $ b = 10.5 - 3.525 = 6.975 $

Check at A=1: $ 1.175×1 + 6.975 = 8.15 $, actual = 7.3 → too high

At A=8: $ 1.175×8 + 6.975 = 9.4 + 6.975 = 16.375 $, actual = 16.7 → close

Better than before?

Try another pair: (4,12.0) and (6,14.3)

$ m = (14.3 - 12.0)/(6 - 4) = 2.3 / 2 = 1.15 $ lbs/month

$ W = 1.15 A + b $

At A=4: $ 1.15×4 = 4.6 $, so $ b = 12.0 - 4.6 = 7.4 $

Then at A=1: $ 1.15 + 7.4 = 8.55 $, actual = 7.3 → still high

Wait — maybe there’s a non-linear component.

But given the fluctuations, and since this is likely a growth curve, it might be quadratic.

Try assuming $ W = a A^2 + b A + c $

But for simplicity, and since the problem says “modify data so that it plots as a straight line”, we need to see what transformation makes it linear.

But here, no obvious power law.

Alternatively, maybe it's linear with noise.

Given the context (weight vs age), often weight increases roughly linearly in early childhood.

But let’s accept that it's approximately linear.

Best fit: use all points.

Let’s do a quick linear regression.

Let me estimate:

Sum of A: $ 1+2+...+8 = 36 $

Sum of W: $ 7.3+9.4+10.5+12.0+13.0+14.3+15.2+16.7 = $

Calculate:
- 7.3 + 9.4 = 16.7
- +10.5 = 27.2
- +12.0 = 39.2
- +13.0 = 52.2
- +14.3 = 66.5
- +15.2 = 81.7
- +16.7 = 98.4

Sum W = 98.4 lbs

Sum A = 36

Sum A×W:
- 1×7.3 = 7.3
- 2×9.4 = 18.8
- 3×10.5 = 31.5
- 4×12.0 = 48.0
- 5×13.0 = 65.0
- 6×14.3 = 85.8
- 7×15.2 = 106.4
- 8×16.7 = 133.6

Sum A×W = 7.3 + 18.8 = 26.1; +31.5 = 57.6; +48 = 105.6; +65 = 170.6; +85.8 = 256.4; +106.4 = 362.8; +133.6 = 496.4

Sum A² = $ 1^2 + 2^2 + ... + 8^2 = \frac{8×9×17}{6} = 204 $

Formula for slope:
$$
m = \frac{n \sum(AW) - \sum A \sum W}{n \sum A^2 - (\sum A)^2}
$$

n = 8

Numerator: $ 8×496.4 - 36×98.4 = 3971.2 - 3542.4 = 428.8 $

Denominator: $ 8×204 - 36^2 = 1632 - 1296 = 336 $

So $ m = 428.8 / 336 ≈ 1.276 $ lbs/month

Then $ b = \bar{W} - m \bar{A} $

$ \bar{A} = 36/8 = 4.5 $, $ \bar{W} = 98.4 / 8 = 12.3 $

$ b = 12.3 - 1.276×4.5 = 12.3 - 5.742 = 6.558 $

So $ W = 1.276 A + 6.558 $

Check at A=1: $ 1.276 + 6.558 = 7.834 $, actual 7.3 → off

At A=8: $ 1.276×8 = 10.208 + 6.558 = 16.766 $, actual 16.7 → good

But at A=1: overestimate.

But maybe acceptable.

However, looking at the data, it may not be perfectly linear, but the instruction says "modify the data so that it will plot as a straight line".

Since no clear non-linear pattern (like quadratic or exponential), and it's approximately linear, we’ll proceed with linear fit.

So graph $ W $ vs $ A $ → straight line.

Slope ≈ 1.28 lbs/month, y-intercept ≈ 6.6 lbs

✔ Mathematical Expression #3:
$$
W = (1.28\ \text{lbs/month}) \cdot A + 6.6\ \text{lbs}
$$

Or rounded:
$$
W = 1.3 A + 6.6 \quad \text{(lbs)}
$$

Note: This implies at birth (A=0), weight is ~6.6 lbs — reasonable.

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🔹 Data Set 4: t (s) vs v (m/s)

| t (s) | v (m/s) |
|-------|---------|
| 0.3 | 10 |
| 1.2 | 20 |
| 2.7 | 30 |
| 4.8 | 40 |
| 7.5 | 50 |
| 10.8 | 60 |
| 14.7 | 70 |
| 19.2 | 80 |

Plot $ v $ vs $ t $

Check if linear:

v increases by 10 m/s every time, but t increases by:
- 0.3 → 1.2: +0.9
- 1.2 → 2.7: +1.5
- 2.7 → 4.8: +2.1
- 4.8 → 7.5: +2.7
- 7.5 → 10.8: +3.3
- 10.8 → 14.7: +3.9
- 14.7 → 19.2: +4.5

Δt increases by 0.6 s each time.

But Δv = 10 m/s each time.

So acceleration is changing?

Wait — but v increases by 10 m/s every time, but time between increases is increasing → speed is increasing at decreasing rate?

Wait — no: velocity is increasing linearly with time?

Let’s check: is $ v $ proportional to $ t $?

Try $ v / t $:
- 10 / 0.3 ≈ 33.3
- 20 / 1.2 ≈ 16.7
- 30 / 2.7 ≈ 11.1
- 40 / 4.8 ≈ 8.33
- 50 / 7.5 ≈ 6.67
- 60 / 10.8 ≈ 5.56
- 70 / 14.7 ≈ 4.76
- 80 / 19.2 ≈ 4.17

Decreasing → not linear

Try $ v $ vs $ t $: seems curved — concave down.

But look at the time intervals: the time between each 10 m/s increase is increasing.

That suggests velocity is increasing linearly with time?

Wait — no: if velocity increased linearly, then $ v = at $, so $ v/t $ should be constant.

But it's decreasing → so not linear.

Wait — maybe it's quadratic?

Try $ v $ vs $ t^2 $? Unlikely.

Wait — could it be acceleration is constant?

If acceleration is constant, then $ v = at + v_0 $

But here, if $ v $ increases by 10 m/s each time, but time between steps increases, then acceleration is decreasing.

But look: from t=0.3 to t=1.2: Δt=0.9 s, Δv=10 → a = 10/0.9 ≈ 11.1 m/s²

Next: t=1.2 to 2.7: Δt=1.5 s, Δv=10 → a = 10/1.5 ≈ 6.67 m/s²

Next: Δt=2.1 s → a = 10/2.1 ≈ 4.76 m/s²

So acceleration is decreasing

But that would mean jerk (rate of change of acceleration) is constant?

Wait — Δt increases by 0.6 s each time: 0.9, 1.5, 2.1, 2.7, 3.3, 3.9, 4.5 → difference = 0.6 s

And Δv = 10 each time

So $ a = \Delta v / \Delta t $, but $ \Delta t $ increasing → a decreasing

But notice: $ \Delta t $ increases linearly: by 0.6 s each step.

So the time between velocity increments is increasing linearly.

But velocity increases by fixed amount.

So $ v $ vs $ t $: piecewise constant Δv, but Δt increasing → slope decreasing.

But perhaps v is proportional to sqrt(t)?

Try $ v $ vs $ \sqrt{t} $

Compute $ \sqrt{t} $:

- √0.3 ≈ 0.547
- √1.2 ≈ 1.095
- √2.7 ≈ 1.643
- √4.8 ≈ 2.190
- √7.5 ≈ 2.738
- √10.8 ≈ 3.286
- √14.7 ≈ 3.834
- √19.2 ≈ 4.382

Now plot $ v $ vs $ \sqrt{t} $:

- 10 vs 0.547
- 20 vs 1.095
- 30 vs 1.643
- 40 vs 2.190
- 50 vs 2.738
- 60 vs 3.286
- 70 vs 3.834
- 80 vs 4.382

Now check ratio $ v / \sqrt{t} $:

- 10 / 0.547 ≈ 18.3
- 20 / 1.095 ≈ 18.25
- 30 / 1.643 ≈ 18.25
- 40 / 2.190 ≈ 18.26
- 50 / 2.738 ≈ 18.25
- 60 / 3.286 ≈ 18.26
- 70 / 3.834 ≈ 18.26
- 80 / 4.382 ≈ 18.25

✔ All ≈ 18.25 m/s / √s

So $ v = 18.25 \cdot \sqrt{t} $

Therefore, to make it linear, plot $ v $ vs $ \sqrt{t} $

Then slope = 18.25 m/s / √s, y-intercept = 0

So $ v = k \sqrt{t} $, where $ k = 18.25 $

But why?

This is typical of motion under constant force with air resistance, or free fall from rest with drag, but more likely here: object falling from rest with terminal velocity? No.

Actually, $ v \propto \sqrt{t} $ is not common.

Wait — recall: for free fall from rest, $ v = gt $ → linear

But here, $ v \propto \sqrt{t} $ suggests something else.

Wait — perhaps it's position vs time? But here it's velocity.

Another possibility: velocity as function of distance?

But we have $ v $ vs $ t $

But the data shows $ v = k \sqrt{t} $

So to make it linear, plot $ v $ vs $ \sqrt{t} $

Then it's a straight line.

✔ Mathematical Expression #4:
$$
v = 18.25\ \frac{\text{m/s}}{\sqrt{\text{s}}} \cdot \sqrt{t}
$$
or
$$
v = 18.25 \sqrt{t} \quad \text{(units: m/s)}
$$

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✔ Final Answers:

#### Mathematical Expression #1:
$$
P = \frac{4\ \text{Pa·m}^3}{V}
$$

#### Mathematical Expression #2:
$$
x = 3\ \frac{\text{m}}{\text{s}^2} \cdot t^2
$$

#### Mathematical Expression #3:
$$
W = (1.28\ \text{lbs/month}) \cdot A + 6.6\ \text{lbs}
$$

#### Mathematical Expression #4:
$$
v = 18.25\ \frac{\text{m/s}}{\sqrt{\text{s}}} \cdot \sqrt{t}
$$

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📌 Summary of Transformations:

| Data Set | Original Plot | Modified Plot | Relationship |
|---------|---------------|----------------|--------------|
| 1 | P vs V | P vs 1/V | Inverse proportion |
| 2 | x vs t | x vs t² | Quadratic |
| 3 | W vs A | W vs A | Linear |
| 4 | v vs t | v vs √t | Square root |

These are classic modeling strategies in physics and science!