SOLUTION: It is believed that two quantities, z and d are Connected by the relationship of the form z=kd^n where k and n are provided that d doesn't exceed some fixed (but unknown)values D.

Question 1163696: It is believed that two quantities, z and d are Connected by the relationship of the form z=kd^n where k and n are provided that d doesn't exceed some fixed (but unknown)values D.An experiment produced the following data
D 750 810 870 930 990 1050 1110 1170
Z 2.1 2.6 3.2 4.0 4.8 5.6 5.9 6.1
a)Plot the values of log10Z against log10 d.Use these points to suggest a
value for D.
b)It is known tht for d < D,n is a whole number.Use your graph to find the value of n.Show also that k=5×10^-9
c) Use your value of n and the estimate k=5×10^-9 to find the value of d for which z=30.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import numpy as np
import matplotlib.pyplot as plt
d = np.array([750, 810, 870, 930, 990, 1050, 1110, 1170])
z = np.array([2.1, 2.6, 3.2, 4.0, 4.8, 5.6, 5.9, 6.1])
log_d = np.log10(d)
log_z = np.log10(z)
plt.scatter(log_d, log_z, color='blue', label='Data points')
plt.plot(log_d, log_z, 'b--', alpha=0.3)
plt.xlabel('log10(d)')
plt.ylabel('log10(z)')
plt.title('log10(z) vs log10(d)')
plt.grid(True)
plt.savefig('log_plot.png')
# Calculate slopes between consecutive points
slopes = np.diff(log_z) / np.diff(log_d)
print(f"Slopes: {slopes}")
# Let's perform linear regression for the first few points to check n
# Trying first 5 points (d = 750 to 990)
for i in range(3, 8):
subset_log_d = log_d[:i]
subset_log_z = log_z[:i]
m, c = np.polyfit(subset_log_d, subset_log_z, 1)
print(f"Points 1 to {i}: slope (n) = {m:.4f}, intercept (log10 k) = {c:.4f}, k = {10**c:.2e}")

```
```text?code_stdout&code_event_index=2
Slopes: [2.7750937 2.90571474 3.34591322 2.91619507 2.61980574 0.93910191
0.63324575]
Points 1 to 3: slope (n) = 2.8372, intercept (log10 k) = -7.8355, k = 1.46e-08
Points 1 to 4: slope (n) = 2.9825, intercept (log10 k) = -8.2565, k = 5.54e-09
Points 1 to 5: slope (n) = 3.0028, intercept (log10 k) = -8.3155, k = 4.84e-09
Points 1 to 6: slope (n) = 2.9634, intercept (log10 k) = -8.2005, k = 6.30e-09
Points 1 to 7: slope (n) = 2.7732, intercept (log10 k) = -7.6442, k = 2.27e-08
```
[image-tag: code-generated-image-0-1776447487869153342]
To find the values of $k$, $n$, and $D$ for the relationship $z = k d^n$, we transform the equation by taking the logarithm (base 10) of both sides:
$$\log_{10} z = n \log_{10} d + \log_{10} k$$
This is a linear equation of the form $y = mx + c$, where the gradient is $n$ and the y-intercept is $\log_{10} k$.
### a) Plotting and suggesting a value for $D$
First, we calculate the values for $\log_{10} d$ and $\log_{10} z$:
| $d$ | $750$ | $810$ | $870$ | $930$ | $990$ | $1050$ | $1110$ | $1170$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $z$ | $2.1$ | $2.6$ | $3.2$ | $4.0$ | $4.8$ | $5.6$ | $5.9$ | $6.1$ |
| $\log_{10} d$ | $2.875$ | $2.908$ | $2.939$ | $2.968$ | $2.996$ | $3.021$ | $3.045$ | $3.068$ |
| $\log_{10} z$ | $0.322$ | $0.415$ | $0.505$ | $0.602$ | $0.681$ | $0.748$ | $0.771$ | $0.785$ |
**Analysis of the Plot:**
When plotting these points, we look for where the linear trend stops. Calculating the gradient (slope) between successive points shows:
* From $d=750$ to $d=1050$, the slope remains consistently near **$3.0$**.
* After $d=1050$, the slope drops drastically to $\approx 0.96$ (between 1050-1110) and $\approx 0.63$ (between 1110-1170).
**Suggested Value for $D$:**
Based on the graph, the linear relationship holds up to **$D \approx 1050$**. Beyond this value, the data points clearly deviate from the straight line.
---
### b) Finding $n$ and showing $k = 5 \times 10^{-9}$
**Determining $n$:**
Since $n$ is known to be a whole number for $d < D$, we look at the slopes of the linear portion:
* Slope ($870$ to $930$) $\approx \frac{0.602 - 0.505}{2.968 - 2.939} = \frac{0.097}{0.029} \approx 3.3$
* Slope ($930$ to $990$) $\approx \frac{0.681 - 0.602}{2.996 - 2.968} = \frac{0.079}{0.028} \approx 2.8$
The average slope is approximately 3. Thus, **$n = 3$**.
**Showing $k = 5 \times 10^{-9}$:**
Using the formula $k = \frac{z}{d^3}$ for values where $d < D$:
* For $d=750$: $k = \frac{2.1}{750^3} \approx 4.98 \times 10^{-9}$
* For $d=930$: $k = \frac{4.0}{930^3} \approx 4.97 \times 10^{-9}$
* For $d=990$: $k = \frac{4.8}{990^3} \approx 4.95 \times 10^{-9}$
Rounding to one significant figure, we find **$k = 5 \times 10^{-9}$**.
---
### c) Find the value of $d$ for which $z=30$
Using the model $z = (5 \times 10^{-9})d^3$, we substitute $z = 30$:
$$30 = (5 \times 10^{-9}) d^3$$
$$d^3 = \frac{30}{5 \times 10^{-9}}$$
$$d^3 = 6 \times 10^9$$
$$d = \sqrt[3]{6 \times 10^9}$$
$$d = 1000 \times \sqrt[3]{6}$$
$$d \approx 1000 \times 1.817$$
**$d \approx 1817$**
*(Note: While $1817 > D$, the calculation based strictly on the suggested constants yields this result.)*
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