QUESTION 2 [TOTAL MARKS: 10 MARKS] A control system engineer is modelling the temperature, in degree Celsius, $$ T(t) $$ of a device over time (in seconds). Near the time $$ t=0 $$, the temperature is approximated by $$ T(t)=e^{2 t} $$. i) Expand $$ T(t) $$ up to the $$ t^{3} $$ terms using Taylor's Series. [8 marks] ii) Using your answer from QUESTION 2 i), find the temperature at 0.1 seconds. [2 marks]

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**i) Expand $$ T(t) = e^{2t} $$ up to the $$ t^{3} $$ terms using Taylor's Series**

The Taylor series expansion of $$ e^{x} $$ around $$ x=0 $$ is given by

$$
e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
$$

For $$ T(t) = e^{2t} $$, substitute $$ x = 2t $$:

$$
e^{2t} = 1 + (2t) + \frac{(2t)^2}{2!} + \frac{(2t)^3}{3!} + \cdots
$$

Now, compute and simplify each term:

1. The constant term:  
   $$
   1
   $$

2. The linear term:  
   $$
   2t
   $$

3. The quadratic term:  
   $$
   \frac{(2t)^2}{2!} = \frac{4t^2}{2} = 2t^2
   $$

4. The cubic term:  
   $$
   \frac{(2t)^3}{3!} = \frac{8t^3}{6} = \frac{4t^3}{3}
   $$

Thus, the expansion up to the $$ t^3 $$ term is

$$
T(t) \approx 1 + 2t + 2t^2 + \frac{4t^3}{3}
$$

---

**ii) Find the temperature at $$ t=0.1 $$ seconds using the expansion**

Substitute $$ t = 0.1 $$ into the expansion:

$$
T(0.1) \approx 1 + 2(0.1) + 2(0.1)^2 + \frac{4(0.1)^3}{3}
$$

Now, calculate each term:

1. Constant term:  
   $$
   1
   $$

2. Linear term:  
   $$
   2(0.1) = 0.2
   $$

3. Quadratic term:  
   $$
   2(0.1)^2 = 2(0.01) = 0.02
   $$

4. Cubic term:  
   $$
   \frac{4(0.1)^3}{3} = \frac{4(0.001)}{3} = \frac{0.004}{3} \approx 0.00133
   $$

Adding these together:

$$
T(0.1) \approx 1 + 0.2 + 0.02 + 0.00133 \approx 1.22133
$$
**i) Taylor Series Expansion of $$ T(t) = e^{2t} $$ up to $$ t^3 $$:** $$ T(t) \approx 1 + 2t + 2t^2 + \frac{4t^3}{3} $$ **ii) Temperature at $$ t = 0.1 $$ seconds:** $$ T(0.1) \approx 1.22133 \text{ degrees Celsius} $$

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