Question
At time t = 0, a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91 degrees Celsius (°C) at time t = 0, and the internal temperature of the potato is greater than 27 C° for all times t > 0. The internal temperature of the potato at time t minutes can be modeled by the function H that satisfies the differential equation \(\frac{dH}{dt}=-\frac{1}{4}(H-27),\) where H(t) is measured in degrees Celsius and H(0) = 91 .
(a) Write an equation for the line tangent to the graph of H at t = 0. Use this equation to approximate the internal temperature of the potato at time t = 3.
(b) Use \(\frac{d^{2}H}{dt^{2}}\) to determine whether your answer in part (a) is an underestimate or an overestimate of the
internal temperature of the potato at time t = 3.
(c) For t < 10, an alternate model for the internal temperature of the potato at time t minutes is the function G that satisfies the differential equation \(\frac{dG}{dt}= -(G-27)^{2/3},\) where G (t) is measured in degrees Celsius and G(0) = 91. Find an expression for G(t). Based on this model, what is the internal temperature of the potato at time t = 3 ?
Answer/Explanation
Ans:
(a) \(H'(0)=-\frac{1}{4}(91-27)=-16\)
H(0) = 91
An equation for the tangent line is y = 91 – 16 . The internal temperature of the potato at time t = 3 minutes is approximately 91 – 16 ⋅ 3 = 43 degrees Celsius.
(b) \(\frac{d^{2}H}{dt^{2}}=-\frac{1}{4}\frac{dH}{dt}=\left ( -\frac{1}{4} \right )\left ( -\frac{1}{4} \right )(H-27)=\frac{1}{16}(H-27)\)
\(H > 27 for t>0\Rightarrow \frac{d^{2}H}{dt^{2}}=\frac{1}{16}(H-27)>0 for t>0\)
Therefore, the graph of H is concave up for t > 0. Thus, the answer in part (a) is an underestimate.
(c) \(\frac{dG}{(G-27)^{2/3}}=-dt\)
\(\int \frac{dG}{(G-27)^{2/3}}=\int (-1)dt\)
3(G-27)1/3 = -t + C
3(91-27)1/3 = 0 + C ⇒ C = 12
3(G-27)1/3 = 12 – t
\(G(t)=27 +\left ( \frac{12-t}{3} \right )^{3}for 0\leq t\leq 10\)
The internal temperature of the potato at time t = 3 minutes is
\(27 +\left ( \frac{12-t}{3} \right )^{3}=54\) degrees Celsius.
Question
A bacteria population is growing at a rate of \(200+9\sqrt{t}-5t\), where t is the time given in hours.
(A) The population size, P(t), is the antiderivative of the rate of growth. Find P(t).
(B) If it is given that the initial population, P(0), is 2,000, find the constant C from the integration in part (a).
(C) Find the population, P(t), after t = 4 h.
Answer/Explanation
(A)\(P(t)=\int (200+9t^{1/2}-5t)dt=200t+9\frac{t^{3/2}}{\frac{3}{2}}-5t^{2}+C=200t+6t^{3/2}-\frac{5}{2}t^{2}+C\)
(B)\(P(0)=200(0)+6(0)-\frac{5}{2}(0)+C\Rightarrow C=2,000\)
(C)\(P(4)=200(4)+6(4)^{3/2}-\frac{5}{2}(4)^{2}+2,000=800+48-40+2,000=2808\)
Question
The marginal revenue that a manufacturer receives for goods, q, is given by MR =100-0.5q
(A) Find the antiderivative of MR to get a function for the total revenue.
(B) How many goods must be produced to generate a total revenue of $10,000?
(C) At what point in production will you reach the point of diminishing returns (when revenue begins to decrease)?
Answer/Explanation
(A)\(MR_{total}=\int (100-.5q)dq=100q-\frac{1}{4}q^{2}+C\).Since the manufacturer receives nothing when nothing is produced,\(MR_{total}(0)=0\Rightarrow C=0\).Then \(MR_{total}=100q-\frac{1}{4}q^{2}.\)
(B)\(10,000=100q-\frac{1}{4}q^{2}\Rightarrow q^{2}-400q+40,000=0\Rightarrow (q-200)(q-200)=0\Rightarrow q=0\)
(C)Revenue begins to decrease when \(MR_{total}=0 or 100-0.5q=0 q=\frac{100}{0.5}=200\).
Question
The density r of a 6-meter-long metal rod of nonuniform density is given by \(\rho (x)=\frac{3}{2\sqrt{x}}\) in units of kg/m, and x is given as the distance along the rod measuring from the left end (x = 0).
(A) Find m(x), the mass, as the antiderivative of r(x).
(B) What is the mass 2 m from the left end to the nearest tenth of a kilogram?
(C) What is the total mass of the rod to the nearest tenth of a kilogram?
Answer/Explanation
(A)\(m(x)=\int \frac{3}{2}\sqrt{x}dx=\frac{3}{2}\int x^{1/2}dx=\frac{3}{2}\frac{x^{3/2}}{\frac{3}{2}}+C=x^{3/2}+C\).Since there is no mass zero distance from the left side of the rod,m(0)=0.Thus,C=0.
(B)\(m(2)=(2)^{3/2}=\sqrt{8}g\approx 2.83\)kg
(C) When x=6 metres,\(m(x)=(6)^{3/2}=\sqrt{216}kg\approx 14.70\)kg.
Question
A butterfly population is modeled by a function P that satisfies the
logistic differential equation:\(\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{1}{7}P\left ( 1-\frac{P}{21} \right )\)
(A) If P(0) = 15, what is \(\lim_{t\rightarrow \infty }P(t)\) ?
(B) For what value of P is the population growing fastest?
(C) A different population is modeled by the function Q that satisfies
the separable differential equation: \(\frac{\mathrm{d} Q}{\mathrm{d} t}=\frac{1}{7}Q(1-\frac{t}{21})\)
Find Q(t) if Q(0) = 10.
(D) For the function found in part (c), what is \(\lim_{t\rightarrow \infty }Q(t)\) ?
Answer/Explanation
(A) Separate the variables of the differential equation and integrate both sides.
$
\begin{aligned}
\frac{d P}{d t} & =\frac{1}{7} P\left(1-\frac{P}{21}\right) \Rightarrow \frac{1}{P\left(1-\frac{P}{21}\right)} d P=\frac{1}{7} d t \Rightarrow \frac{21}{P(21-P)} d P=\frac{1}{7} d t \\
\frac{d P}{P}+\frac{d P}{21-P}= & \frac{1}{7} d t \Rightarrow \int \frac{d P}{P}+\int \frac{d P}{21-P}=\int \frac{1}{7} d t \Rightarrow \ln |P|-\ln |21-P|=\frac{t}{7}+c_1 \\
\ln \left|\frac{P}{21-P}\right| & =\frac{t}{7}+c_1
\end{aligned}
$
Exponentiate both sides, then isolate $P$.
$
\frac{P}{21-P}=e^{\frac{t}{7}} \cdot c_2 \Rightarrow P=\frac{21 c_2 e^{\frac{t}{7}}}{1+c_2 e^{\frac{t}{7}}}
$
Use both possible initial conditions to solve for $c_2$ and evaluate the limits.
$
\begin{aligned}
& P(0)=15 \Rightarrow 15=\frac{21 c_2}{1+c_2} \Rightarrow 21 c_2=15\left(1+c_2\right) \Rightarrow 6 c_2=15 \Rightarrow c_2=2.5 \\
& P(t)=\frac{21(2.5) e^{\frac{t}{7}}}{1+2.5 e^{\frac{t}{7}}} \Rightarrow P(t)=\frac{52.5 e^{\frac{t}{7}}}{1+2.5 e^{\frac{t}{7}}}
\end{aligned}
$
To calculate the limit as $t$ tends to infinity, first factor out $e^{t / 7}$ from the numerator and denominator.
$
\begin{aligned}
P(t) & =\frac{e^{t / 7}(52.5)}{e^{t / 7}\left(e^{-t / 7}+2.5\right)}=\frac{52.5}{\left(e^{-t / 7}+2.5\right)} \\
\lim _{x \rightarrow \infty} \frac{52.5}{e^{-t / 7}+2.5} & =\frac{52.5}{2.5}=21
\end{aligned}
$
(B) The rate of growth of the population is given by
$
\frac{d P}{d t}=\frac{1}{7} P\left(1-\frac{P}{21}\right)
$
To find the maximum value of this function, calculate the derivative, set it equal to zero, and solve.
$
\begin{aligned}
\frac{d^2 P}{d t^2} & =\frac{1}{7} P\left(-\frac{1}{21}\right)+\frac{1}{7}\left(1-\frac{P}{21}\right)=\frac{1}{7}-\frac{2 P}{147} \\
\frac{1}{7}-\frac{2 P}{147} & =0 \Rightarrow P=10.5
\end{aligned}
$
So the maximum of $\frac{d P}{d t}$ occurs when $P=10.5$. Evaluate $\frac{d P}{d t}$ at $P=10.5$ to find the maximum value.
$
\frac{d P}{d t}=\frac{1}{7} P\left(1-\frac{P}{21}\right) \Rightarrow \frac{d P}{d t}=\frac{1}{7}(10.5)\left(1-\frac{10.5}{21}\right) \Rightarrow \frac{d P}{d t}=0.75
$
The population is growing fastest, at a rate of $0.75$ butterflies per unit time, when the population is $10.5$.
(C) Separate the variables and integrate both sides.
$
\begin{aligned}
\frac{d Q}{d t} & =\frac{1}{7} Q\left(1-\frac{t}{21}\right) \Rightarrow \frac{1}{Q} d Q=17\left(\frac{21-t}{21}\right) d t \Rightarrow \int \frac{1}{Q} d Q=\frac{1}{147} \int(21-t) d t \\
\ln |Q| & =\frac{1}{147}\left(21 t-\frac{t^2}{2}\right)+c_1 \Rightarrow \ln |Q|=\frac{t}{7}-\frac{t^2}{294}+c_1
\end{aligned}
$
Now exponentiate both sides to solve for $Q$.
$
Q=c_2 e^{\left(t / 7-t^2 / 294\right)}
$
Use the initial condition, $Q(0)=10$, to solve for $c_2$.
$
10=c_2 e^0 \Rightarrow c_2=10 \Rightarrow Q(t)=10 e^{\left(t / 7-t^2 / 294\right)}
$
(D) For the function found in part (c), as $t$ increases notice that
$
\frac{t}{7}-\frac{t^2}{294} \rightarrow-\infty \text { so } e^{\left(t / 7-t^2 / 294\right)} \rightarrow 0
$
Therefore $\lim _{t \rightarrow \infty} Q(t)=0$.