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AP Calculus AB
AP Exam Review Free Response 5
This Question is CALCULATOR INACTIVE
Please show all work on page
2&3
Mzab Gundis live in the central Sahara Desert in Algeria, northern Niger, northwestern Chad, northeastern Mali, and southwestern Libya. The rate at which a Mzab gundi gains weight is proportional to the difference between its adult weight and its current weight. At time
t=0, when the Gundi is first weighed, its weight is 20 grams. If
G(t) is the weight of the Gundi, in grams, at time
t days after it is first weighed, then
(dG)/(dt)=(1)/(50)(180-G). Let
y=G(t) be the solution to the differential equation above with initial condition
G(0)=20.
(a) Is the Gundi gaining weight faster when it weighs 60 grams or when it weighs 100 grams? Explain your reasoning.
(b) Find
(d^(2)G)/(dt^(2)) in terms of
G. Use
(d^(2)G)/(dt^(2)) to explain why the graph of
G cannot resemble the following graph.
(c) Use separation of variables to find
y=G(t), the particular solution to the differential equation with initial condition
G(0)=20.

AP Calculus AB $\newline$ AP Exam Review Free Response $5$ $\newline$ This Question is CALCULATOR INACTIVE $\newline$ Please show all work on page $2 \& 3$ $\newline$ Mzab Gundis live in the central Sahara Desert in Algeria, northern Niger, northwestern Chad, northeastern Mali, and southwestern Libya. The rate at which a Mzab gundi gains weight is proportional to the difference between its adult weight and its current weight. At time $t=0$ , when the Gundi is first weighed, its weight is $20$ grams. If $G(t)$ is the weight of the Gundi, in grams, at time $t$ days after it is first weighed, then $\frac{d G}{d t}=\frac{1}{50}(180-G)$ . Let $y=G(t)$ be the solution to the differential equation above with initial condition $G(0)=20$ . $\newline$ (a) Is the Gundi gaining weight faster when it weighs $60$ grams or when it weighs $100$ grams? Explain your reasoning. $\newline$ (b) Find $\frac{d^{2} G}{d t^{2}}$ in terms of $G$ . Use $\frac{d^{2} G}{d t^{2}}$ to explain why the graph of $G$ cannot resemble the following graph. $\newline$ (c) Use separation of variables to find $y=G(t)$ , the particular solution to the differential equation with initial condition $G(0)=20$ .

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Q. AP Calculus AB

\newline

AP Exam Review Free Response

5

\newline

This Question is **CALCULATOR INACTIVE**

\newline

Please show all work on page

2 \& 3

\newline

Mzab Gundis live in the central Sahara Desert in Algeria, northern Niger, northwestern Chad, northeastern Mali, and southwestern Libya. The rate at which a Mzab gundi gains weight is proportional to the difference between its adult weight and its current weight. At time

t=0

, when the Gundi is first weighed, its weight is

20

grams. If

G(t)

is the weight of the Gundi, in grams, at time

t

days after it is first weighed, then

\frac{d G}{d t}=\frac{1}{50}(180-G)

. Let

y=G(t)

be the solution to the differential equation above with initial condition

G(0)=20

\newline

(a) Is the Gundi gaining weight faster when it weighs

60

grams or when it weighs

100

grams? Explain your reasoning.

\newline

(b) Find

\frac{d^{2} G}{d t^{2}}

in terms of

G

. Use

\frac{d^{2} G}{d t^{2}}

to explain why the graph of

G

cannot resemble the following graph.

\newline

y=G(t)

, the particular solution to the differential equation with initial condition

G(0)=20

Calculate rate at $60g$ : Calculate the rate of weight gain at $60$ grams using the given differential equation. $\newline$ $\frac{dG}{dt} = \frac{1}{50}(180 - G)$ $\newline$ Substitute $G = 60$ : $\newline$ $\frac{dG}{dt} = \frac{1}{50}(180 - 60) = \frac{1}{50}(120) = 2.4$ grams/day
Calculate rate at $100g$ : Calculate the rate of weight gain at $100$ grams using the same differential equation. $\newline$ Substitute $G = 100$ : $\newline$ $\frac{dG}{dt} = \frac{1}{50}(180 - 100) = \frac{1}{50}(80) = 1.6$ grams/day
Compare rates: Compare the rates of weight gain at $60$ grams and $100$ grams. $\newline$ Since $2.4$ grams/day (at $60$ grams) is greater than $1.6$ grams/day (at $100$ grams), the Gundi is gaining weight faster at $60$ grams.
Use separation of variables: Use separation of variables to solve the differential equation $\frac{dG}{dt} = \frac{1}{50}(180 - G)$ . Rearrange and integrate: $\int \frac{dG}{180 - G} = \int \frac{1}{50} dt$ $- \ln|180 - G| = \frac{t}{50} + C$
Solve for $G(0)=20$ : Solve for $G(t)$ using the initial condition $G(0) = 20$ . Substitute $t = 0$ and $G = 20$ into $-\ln|180 - G| = \frac{t}{50} + C$ : $-\ln|180 - 20| = 0 + C$ $-\ln|160| = C$ $C = -\ln(160)$
Substitute $C$ and solve for $G(t)$ : Substitute $C$ back into the equation and solve for $G(t)$ .
$-\ln|180 - G| = \frac{t}{50} - \ln(160)$
$180 - G = e^{(-\frac{t}{50} + \ln(160))}$
$G = 180 - 160e^{(-\frac{t}{50})}$