Math Problem Statement

what exponential decay function fits the following measurement points (time, weight): (0, 5960) (1, 2440) (70, 1390)

Solution

An exponential decay function is generally of the form:

y(t)=y0ekty(t) = y_0 \cdot e^{-kt}

where:

  • y(t)y(t) is the quantity at time tt,
  • y0y_0 is the initial quantity at t=0t = 0,
  • kk is the decay constant (rate),
  • tt is the time.

Given the points:

  • (0,5960)(0, 5960),
  • (1,2440)(1, 2440),
  • (70,1390)(70, 1390),

we can solve for y0y_0 and kk.

Step 1: Find y0y_0

From (0,5960)(0, 5960), when t=0t = 0: y(0)=y0=5960y(0) = y_0 = 5960

So, the equation becomes: y(t)=5960ekty(t) = 5960 \cdot e^{-kt}

Step 2: Solve for kk

Using the second point (1,2440)(1, 2440), substitute into the equation:

2440=5960ek12440 = 5960 \cdot e^{-k \cdot 1}

Solving for kk: 24405960=ek\frac{2440}{5960} = e^{-k}

0.409395=ek0.409395 = e^{-k}

Taking the natural logarithm of both sides: ln(0.409395)=k\ln(0.409395) = -k

k=ln(0.409395)0.892k = -\ln(0.409395) \approx 0.892

Step 3: Verify with the third point

Now that we have kk, we can verify the fit with the third point (70,1390)(70, 1390).

The function is: y(t)=5960e0.892ty(t) = 5960 \cdot e^{-0.892t}

Substituting t=70t = 70: y(70)=5960e0.89270y(70) = 5960 \cdot e^{-0.892 \cdot 70}

Evaluating this will confirm the validity of the fit. Let's compute that next.When using k=0.892k = 0.892, the calculation yields an extremely small value for y(70)y(70), approximately 4.55×10244.55 \times 10^{-24}, which is clearly much smaller than the observed value of 1390. This indicates that k=0.892k = 0.892 is too large, suggesting a slower decay rate than initially calculated.

Let me try recalculating kk using both the first two points while considering the third as a constraint. I'll compute it more carefully.After recalculating using all three points, the fitted values are:

  • y0=5960y_0 = 5960 (which matches the initial condition),
  • k0.893k \approx 0.893, similar to the earlier estimate.

The decay rate kk is slightly more refined, but this still results in an extremely fast decay when extrapolated to t=70t = 70. Given this, it's possible that the model might not perfectly follow a single exponential decay across the whole range, or some data points might reflect non-exponential behavior.

Would you like further clarification, or to adjust the model for a better fit?

Relative Questions:

  1. How does the decay constant kk affect the rate of decrease in an exponential decay function?
  2. What assumptions are involved in modeling with an exponential decay function?
  3. Could a different mathematical model better fit the data points?
  4. How can the exponential decay equation be used in real-world scenarios (e.g., radioactive decay)?
  5. What happens when kk becomes very small or very large?

Tip: If the model doesn't fit well, consider a multi-phase decay model or using logarithmic scaling for better visualization.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Logarithms

Formulas

y(t) = y_0 * e^(-kt)
Natural Logarithm: ln(a) = b if e^b = a

Theorems

Exponential Decay Law

Suitable Grade Level

Grades 11-12