College Algebra, Chapter 5, 5.4, Section 5.4, Problem 86

The function $P(t) = M - Ce^{-k}$ represents the learning curve that measures the $P(t)$ performance of someone that learns a skill as a function of the training time t. Initially, the rate of learning is rapid. Then, as performance increases and approached to its maximal value M, the rate of learning decreases.

a.) Find the expression of the learning time $t$ as a function of the performance level $P$.

b.) For a pole-vaulter in training, the learning curve is given by $P(t) = 20 - 14e^{0.024 t}$ where $P(t)$ is the height he is able to pole-vault after $t$ months. After how many months of training is he able to vault $12$ ft?

c.) Draw a graph of the learning curve in part (b).



a.)


$
\begin{equation}
\begin{aligned}

P(t) =& M - Ce^{-kt}
&& \text{Model}
\\
\\
Ce^{-kt} =& M - P
&& \text{Add $Ce^{-kt}$ and subtract $P$}
\\
\\
e^{-kt} =& \frac{M - P}{C}
&& \text{Divide by } C
\\
\\
-kt =& \ln \left( \frac{M - P}{C} \right)
&& \text{Take $\ln$ on both sides}
\\
\\
t =& \frac{\displaystyle - \ln \left( \frac{M - P}{C} \right)}{k}
&& \text{Divide by } -k

\end{aligned}
\end{equation}
$


b.) If $P = 12 $ ft, then


$
\begin{equation}
\begin{aligned}

12 =& 20 - 14 e^{-0.024 t}
&& \text{Add $14e^{-0.024t}$ and subtract } 12
\\
\\
14e^{-0.024 t} =& 8
&& \text{Divide by } 14
\\
\\
e^{-0.024 t} =& \frac{4}{7}
&& \text{Take $\ln$ of both sides}
\\
\\
-0.024 t =& \ln \left( \frac{4}{7} \right)
&& \text{Solve for } t

\end{aligned}
\end{equation}
$


$t = 23.3173 $ months or $24$ months. Therefore, it will take a little more than $23$ months for the pole-vaulter to vault $12$ ft.

c.)