QuestionA fungus grows exponentially then linearly. Given $R(t)$ , find $R(t_c)$ for continuity and graph $R(t)$ if $t_c=2$ .

Studdy Solution

STEP 1

Assumptions1. The function $R(t)$ models the growth of a fungus over time, where $t$ is time. . The fungus grows exponentially when $0 \leq t \leq t_{c}$ , and grows linearly when $t>t_{c}$ .
3. The time at which the fungus switches from exponential to linear growth is $t_{c}$ .
4. The constant $a$ is the rate of growth when the fungus is growing linearly.
5. We are given that $a=e^{t_c}$ .

STEP 2

(a) We are asked to write an expression using $t_{c}$ as the variable. Since $a=2e^{t_c}$ , we can substitute $a$ into the function $R(t)$ to get an expression in terms of $t_{c}$ .
$R(t)=\left\{\begin{array}{ll} 2 e^{t} & \text { if }0 \leq t \leq t_{c} \\ 2e^{t_c} & \text { if } t>t_{c} \end{array}\right.$

STEP 3

This is the expression for $R(t)$ in terms of $t_{c}$ . The function is continuous at $t=t_c$ because the value of $R(t)$ is the same whether $t$ is just less than or just greater than $t_{c}$ .
$R(t_c)=2e^{t_c}$

STEP 4

(b) We are asked to draw the graph of $R(t)$ as a function of $t$ , assuming that $t_{c}=2$ . First, we need to substitute $t_{c}=2$ into the function $R(t)$ .
$R(t)=\left\{\begin{array}{ll} 2 e^{t} & \text { if }0 \leq t \leq2 \\ 2e^{2} & \text { if } t>2\end{array}\right.$

STEP 5

The function $R(t)$ is defined piecewise, so we draw two separate graphs for $0 \leq t \leq2$ and $t>2$ .
For $0 \leq t \leq2$ , the function $R(t)=2e^{t}$ is an exponential function with base $e$ . It starts at $R(0)=2$ and grows rapidly as $t$ increases.
For $t>2$ , the function $R(t)=2e^{2}$ is a constant function. Its value is the same no matter what $t$ is, and it equals to $R(2)$ , which ensures the continuity of the function at $t=2$ .

STEP 6

Therefore, the graph of $R(t)$ starts at $R(0)=2$ and grows rapidly until $t=2$ . After $t=2$ , the graph becomes a horizontal line at the height of $R(2)=2e^{2}$ .

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