Question

The bacteria concentration in a reservoir varies as

e^(t )-t^3/6-e^(0.3 t)-t^2/2 -t

where t is the time in seconds. Use the Newton-Raphson method to estimate the
time required for the bacteria concentration to reach 1 (correct up to 2 decimal
places)

Answer 1

Answer on Question #51466 - Math - Algorithms | Quantitative Methods

The bacteria concentration in a reservoir varies as $e^{t} - \frac{t^{3}}{6} - e^{0.3t} - \frac{t^{2}}{2} - t$ where $t$ is the time in seconds. Use the Newton-Raphson method to estimate the time required for the bacteria concentration to reach 1 (correct up to 2 decimal places)

Solution.

We need to find the first positive root of the following equation:

$$f (t) = e ^ {t} - e ^ {0. 3 t} - \frac {t ^ {3}}{6} - \frac {t ^ {2}}{2} - t - 1 = 0;$$

Use the Newton-Raphson method. Firstly compute $f'(t)$ :

$$f ^ {\prime} (t) = e ^ {t} - 0. 3 e ^ {0. 3 t} - \frac {t ^ {2}}{2} - t - 1 = f (t) + 0. 7 e ^ {0. 3 t} + \frac {t ^ {3}}{6};$$

So:

$$t _ {0} = 2;$$ $$t _ {1} = t _ {0} - \frac {f (t _ {0})}{f ^ {\prime} (t _ {0})};$$ $$f (t _ {0}) = e ^ {2} - e ^ {0. 6} - \frac {1 9}{3} \approx - 0. 7 6 6;$$ $$f ^ {\prime} (t _ {0}) = f (t _ {0}) + 0. 7 e ^ {0. 6} + \frac {4}{3} \approx 1. 8 4 3;$$

So:

$$t _ {1} = 2 + \frac {0 . 7 6 6}{1 . 8 4 3} \approx 2. 4 1 6;$$ $$f (t _ {1}) \approx 0. 4 5 2;$$ $$f ^ {\prime} (t _ {1}) \approx 4. 2 4 7;$$

So:

$$t _ {2} = 2. 4 1 6 - \frac {0 . 4 5 2}{4 . 2 4 7} \approx 2. 3 0 9;$$ $$f (t _ {2}) \approx 0. 0 3 9;$$ $$f ^ {\prime} (t _ {2}) \approx 3. 4 9;$$

So:

$$t _ {3} = 2. 3 0 9 - \frac {0 . 0 3 9}{3 . 4 9} \approx 2. 2 9 8;$$ $$f(t_3) \approx 0.001;$$ $$f'(t_3) \approx 3.418;$$

So:

$$t_4 = 2.298 - \frac{0.001}{3.418} \approx t_3.$$

Answer: 2.298

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