QuestionBertalanffy growth function:
1. Find $\lim_{t \rightarrow \infty} L(t)$ and interpret.
2. With $L_{0}=1 \mathrm{~cm}$ , $L_{T}=38 \mathrm{~cm}$ , graph $L(t)$ for various $k$ .

Studdy Solution

STEP 1

Assumptions1. The Bertalanffy growth function is given by $(t)=_{}-\left(L_{}-_{0}\right) e^{-k t}$ $is a number approaching$ \infty $<br />3.$ k $is a positive constant4.$ _{0} $is the length of the fish at birth5.$ _{} $is the length of the fish as$ t $approaches$ \infty$

STEP 2

We are asked to calculate the limit of $(t)$ as $t$ approaches $\infty$ . $\lim{t \rightarrow \infty} L(t)$

STEP 3

Substitute the given function into the limit.
$\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) e^{-k t}$

STEP 4

As $t$ approaches $\infty$ , the term $e^{-kt}$ approaches0 because $k$ is a positive constant. Thus, the expression becomes $\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) \times0$

STEP 5

implify the expression.
$\lim{t \rightarrow \infty} L_{}-0$

STEP 6

The limit of a constant as $t$ approaches $\infty$ is just the constant itself. Therefore, the limit of $(t)$ as $t$ approaches $\infty$ is $_{}$ .
$\lim{t \rightarrow \infty} L(t) = L_{}$ This means that as the age of the fish increases indefinitely, the length of the fish approaches $_{}$ .

STEP 7

Now, we are asked to graph $(t)$ for several values of $k$ given that $_{0}=1 \mathrm{~cm}$ and $_{}=38 \mathrm{~cm}$ .We can do this by substituting these values into the function and plotting the function for different values of $k$ .
$(t)=38-(38-1) e^{-k t}$ This step would be done using a graphing calculator or a graphing software, so it cannot be shown here.

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QuestionBertalanffy growth function:
1. Find $\lim_{t \rightarrow \infty} L(t)$ and interpret.
2. With $L_{0}=1 \mathrm{~cm}$ , $L_{T}=38 \mathrm{~cm}$ , graph $L(t)$ for various $k$ .

Studdy Solution

STEP 1

Assumptions1. The Bertalanffy growth function is given by $(t)=_{}-\left(L_{}-_{0}\right) e^{-k t}$ $is a number approaching$ \infty $<br />3.$ k $is a positive constant4.$ _{0} $is the length of the fish at birth5.$ _{} $is the length of the fish as$ t $approaches$ \infty$

STEP 2

We are asked to calculate the limit of $(t)$ as $t$ approaches $\infty$ . $\lim{t \rightarrow \infty} L(t)$

STEP 3

Substitute the given function into the limit.
$\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) e^{-k t}$

STEP 4

As $t$ approaches $\infty$ , the term $e^{-kt}$ approaches0 because $k$ is a positive constant. Thus, the expression becomes $\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) \times0$

STEP 5

implify the expression.
$\lim{t \rightarrow \infty} L_{}-0$

STEP 6

The limit of a constant as $t$ approaches $\infty$ is just the constant itself. Therefore, the limit of $(t)$ as $t$ approaches $\infty$ is $_{}$ .
$\lim{t \rightarrow \infty} L(t) = L_{}$ This means that as the age of the fish increases indefinitely, the length of the fish approaches $_{}$ .

STEP 7

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QuestionBertalanffy growth function: 1. Find lim⁡t→∞L(t)\lim_{t \rightarrow \infty} L(t)limt→∞​L(t) and interpret.2. With L0=1 cmL_{0}=1 \mathrm{~cm}L0​=1 cm, LT=38 cmL_{T}=38 \mathrm{~cm}LT​=38 cm, graph L(t)L(t)L(t) for various kkk.

Studdy Solution

STEP 1

STEP 2

We are asked to calculate the limit of (t)(t)(t) as ttt approaches ∞\infty∞.lim⁡t→∞L(t)\lim{t \rightarrow \infty} L(t)limt→∞L(t)

STEP 3

Substitute the given function into the limit.lim⁡t→∞L−(L−0)e−kt\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) e^{-k t}limt→∞L​−(L​−0​)e−kt

STEP 4

As ttt approaches ∞\infty∞, the term e−kte^{-kt}e−kt approaches0 because kkk is a positive constant. Thus, the expression becomeslim⁡t→∞L−(L−0)×0\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) \times0limt→∞L​−(L​−0​)×0

STEP 5

implify the expression.lim⁡t→∞L−0\lim{t \rightarrow \infty} L_{}-0limt→∞L​−0

STEP 6

STEP 7

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QuestionBertalanffy growth function:
1. Find $\lim_{t \rightarrow \infty} L(t)$ and interpret.
2. With $L_{0}=1 \mathrm{~cm}$ , $L_{T}=38 \mathrm{~cm}$ , graph $L(t)$ for various $k$ .

We are asked to calculate the limit of $(t)$ as $t$ approaches $\infty$ . $\lim{t \rightarrow \infty} L(t)$

Substitute the given function into the limit.
$\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) e^{-k t}$

As $t$ approaches $\infty$ , the term $e^{-kt}$ approaches0 because $k$ is a positive constant. Thus, the expression becomes $\lim{t \rightarrow \infty} L_{}-\left(L_{}-_{0}\right) \times0$

implify the expression.
$\lim{t \rightarrow \infty} L_{}-0$