Question

suppose that you are given the task of learning 100% of a block of knowledge. human nature is such that we retain only a percentage p of knowledge t weeks after we have learned it. the ebbinghaus learning model asserts that p is given by p(t)=q+(100 - q)e^(-kt) where q is the percentage that we would never forget and k is a constant that depends on the knowledge learned. suppose that q = 45 and k = 0.7. complete parts (a) through (e) below.
b) find lim p(t) as t→∞. lim p(t)=45% (simplify your answer.)
c) sketch a graph of p. choose the correct graph below.
d) find the rate of change of p with respect to time t. p(t)=

Explanation:

Step1: Recall the given function

The Ebbinghaus - learning model is $P(t)=Q+(100 - Q)e^{-kt}$, where $Q = 45$ and $k = 0.7$.

Step2: Find $\lim_{t

ightarrow\infty}P(t)$
As $t
ightarrow\infty$, the term $e^{-kt}=e^{-0.7t}
ightarrow0$ since the exponent $- 0.7t
ightarrow-\infty$ as $t
ightarrow\infty$. Then $P(t)=45+(100 - 45)e^{-0.7t}$. Substituting the limit, we have $\lim_{t
ightarrow\infty}P(t)=45+(100 - 45)\times0=45$.

Step3: Analyze the function for graph - sketching

The function $P(t)=45 + 55e^{-0.7t}$. When $t = 0$, $P(0)=45+55e^{0}=45 + 55=100$. As $t$ increases, $e^{-0.7t}$ decreases, so $P(t)$ decreases towards $45$. The graph starts at $(0,100)$ and decays exponentially towards the horizontal asymptote $y = 45$.

Step4: Differentiate the function

Using the chain - rule, if $P(t)=45+55e^{-0.7t}$, then $P^\prime(t)=\frac{d}{dt}(45)+\frac{d}{dt}(55e^{-0.7t})$. Since $\frac{d}{dt}(45)=0$ and $\frac{d}{dt}(55e^{-0.7t})=55\times(-0.7)e^{-0.7t}=-38.5e^{-0.7t}$.

Answer:

b) $45$
c) The graph that starts at $(0,100)$ and decays exponentially towards the horizontal asymptote $y = 45$.
d) $P^\prime(t)=-38.5e^{-0.7t}$