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 T-shirt maker estimates that the weekly cost of making x shirts is \(C(x)=50+2x+\frac{x^{2}}{20}\). The weekly revenue from selling x shirts is given by the function: \(R(x)=20+\frac{x^{2}}{200}\). (Show your work.)
(A) Derive the weekly profit function.
(B) What is the maximum weekly profit?
Answer/Explanation
(A)\(P(x)=R(x)-C(x)\)
\(R(x)=20x+\frac{x^{2}}{200}\)
\(C(x)=50+2x+\frac{x^{2}}{20}\)
\(P(x)=20x+\frac{x^{2}}{200}-\left ( 50+2x+\frac{x^{2}}{20} \right )\)
\(P(x)=20x+\frac{x^{2}}{200}-50-2x-\frac{x^{2}}{20}\)
\(P(x)=18x-50-\frac{9x^{2}}{200}\)
(B)\(P(x)=18x-50-\frac{9x^{2}}{200}\)
\(P'(x)=18-(2)\left ( \frac{9x}{200} \right )\)
\(P'(x)=18-0.09x\)
\(0=18-0.09x\)
\(0.09x=18\)
x=200 shirts
\(P(200)=18(200)-50-\frac{9(200)^{2}}{200}\)
\(P(200)=3,600-50-1,800\)
\(P(200)=$1,750\)
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
| t (minutes) | 0 | 4 | 9 | 15 | 20 |
| W(t) (degrees Fahrenheit) | 55.0 | 57.1 | 61.8 | 67.9 | 71.0 |
The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W, where W(t) is measured in degrees Fahrenheit and t is measured in minutes. At time t =0, the temperature of the water is 55 0F. The water is heated for 30 minutes, beginning at time t = 0. Values of W (t) at selected times t for the first 20 minutes are given in the table above.
(a) Use the data in the table to estimate W'(12) . Show the computations that lead to your answer. Using correct units, interpret the meaning of your answer in the context of this problem.
(b) Use the data in the table to evaluate \(\int_{0}^{20}W'(t)dt.\) Using correct units, interpret the meaning of \(\int_{0}^{20}W'(t)dt\) in the context of this problem.
(c) For 0 ≤ t ≤ 20, the average temperature of the water in the tub is \(\frac{1}{20}\int_{0}^{20}W'(t)dt.\) Use a left Riemann sum with the four subintervals indicated by the data in the table to approximate \(\frac{1}{20}\int_{0}^{20}W'(t)dt.\) Does this approximation overestimate or underestimate the average temperature of the water over these 20 minutes? Explain your reasoning.
(d) For 20 ≤ t ≤ 25, the function W that models the water temperature has first derivative given by \(W'(t)=0.4\sqrt{t}cos (0.06t).\) Based on the model, what is the temperature of the water at time t=25 ?
Answer/Explanation
Ans:
(a) \(W'(12)\approx \frac{W(15)-W(9)}{15-9}=\frac{67.9-61.8}{6}\)
= 1.017 (or 1.016)
The water temperature is increasing at a rate of approximately 1.017 F° per minute at time t = 12 minutes.
(b) \(\int_{0}^{20}W'(t)dt=W(20)-W(0)=71.0-55.0=16\)
The water has warmed by 16 F° over the interval from t = 0 to t = 20 minutes.
(c) \(\frac{1}{20}\int_{0}^{20}W(t)dt\approx \frac{1}{20}(4\cdot W(0)+5\cdot W(4)+6\cdot W(9)+5\cdot W(15))\)
\(= \frac{1}{20}(4\cdot 55.0+5\cdot 57.1+6\cdot 61.8+5\cdot 67.9)\)
\(= \frac{1}{20}\cdot 1215.8=60.79\)
This approximation is an underestimate, because a left Riemann sum is used and the function W is strictly increasing.
(d) \(W(25)=71.0+\int_{20}^{25}W'(t)dt\)
=71.0+2.043155 = 73.043