Question
Researchers on a boat are investigating plankton cells in a sea. At a depth of h meters, the density of plankton cells, in millions of cells per cubic meter, is modeled by \(p(h)=0.2h^{2}e^{-0.0025h^{2}}\) for 0 ≤ h ≤ 30 and is modeled by f(h) for h ≥ 30. The continuous function f is not explicitly given.
(a) Find p'(25 ). Using correct units, interpret the meaning of p ‘(25) in the context of the problem.
(b) Consider a vertical column of water in this sea with horizontal cross sections of constant area 3 square meters. To the nearest million, how many plankton cells are in this column of water between h = 0 and h = 30 meters?
(c) There is a function u such that 0 ≤ f(h) ≤ u(h) for all h ≥ 30 and \(\int_{30}^{\infty }u(h)dh=105.\) The column of water in part (b) is K meters deep, where K > 30. Write an expression involving one or more integrals that gives the number of plankton cells, in millions, in the entire column. Explain why the number of plankton cells in the column is less than or equal to 2000 million.
(d) The boat is moving on the surface of the sea. At time t ≥ 0, the position of the boat is (x(t), y(t)), where x'(t)= 662 sin (5t) and y ‘ (t) = 880 cos (6t). Time t is measured in hours, and x(t) and y (t) are measured in meters. Find the total distance traveled by the boat over the time interval 0 ≤ t ≤ 1.
Answer/Explanation
Ans:
(a) p′(25) = – 1.179
At a depth of 25 meters, the density of plankton cells is changing at a rate of −1.179 million cells per cubic meter per meter.
(b) \(\int_{0}^{30}3p(h)dh=1675.414936\)
There are 1675 million plankton cells in the column of water between h = 0 and h = 30 meters.
(c) \(\int_{30}^{k}3f(h)dh\) represents the number of plankton cells, in millions, in the column of water from a depth of 30 meters to a depth of K meters. The number of plankton cells, in millions, in the entire column of water is given by \(\int_{0}^{30}3p(h)dh+\int_{30}^{k}3f(h)dh.\)
Because 0 ≤ f(h) ≤ u(h) for all h ≥30,
\(3\int_{30}^{k}f(h)dh\leq 3\int_{30}^{k}u(h)dh\leq 3\int_{30}^{\infty }=3\cdot 105=315.\)
The total number of plankton cells in the column of water is bounded by 1675.415 +315 =1990.415 ≤ 2000 million.
(d) \(\int_{0}^{1}\sqrt{(x'(t))^{2}+(y'(t))}dt=757.455862\)
The total distance traveled by the boat over the time interval 0 ≤ t ≤ 1 is 757.456 (or 757.455) meters.
Question
The rate at which rainwater flows into a drainpipe is modeled by the function R, where \(R(t)=20 sin \left ( \frac{t^{2}}{35} \right )\) cubic feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. out the other end of the pipe at a rate modeled by D(t) = -0.04t3 + 0.4t2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8. There are 30 cubic feet of water in the pipe at time t = 0.
(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8 ?
(b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your answer.
(c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer.
(d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the pipe will begin to overflow.
Answer/Explanation
Ans:
(a) \(\int_{0}^{8}R(t)dt=76.570\)
(b) R(3) – D(3) = -0.313632 < 0
Since R(3) < D(3), the amount of water in the pipe is decreasing at time t = 3 hours.
(c) The amount of water in the pipe at time t, 0 ≤ t ≤ 8, is
\(30+\int_{0}^{t}\left [ R(x)-D(x) \right ]dx.\)
R(t) – D(t) = 0 ⇒ t = 0, 3.271658
| t | Amount of water in the pipe |
0 3.271658 8 | 30 27.964561 48.543686 |
The amount of water in the pipe is a minimum at time t = 3.272 (or 3.271) hours.
(d) \(30+\int_{0}^{w}\left [ R(t)-D(t) \right ]dt.=50\)
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
At time t, a particle moving in the xy-plane is at position ( x(t), y(t)), where x(t) and y(t) are not explicitly given. For t ≥ 0, \(\frac{dx}{dt}=4t + 1 and \frac{dy}{dt}= sin(t^{2}).\) At time t = 0, x(0) = 0 and y(0) = -4.
(a) Find the speed of the particle at time t = 3, and find the acceleration vector of the particle at time t = 3.
(b) Find the slope of the line tangent to the path of the particle at time t = 3.
(c) Find the position of the particle at time t = 3.
(d) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 3.
Answer/Explanation
Ans:
(a)
\(Speed =\sqrt{(\frac{dy}{dt})^{2}+(\frac{dx}{dt})^{2}}\)
\(\frac{dy}{dt}|_{t=3}=sin(3^{2})=sin9\)
\(\frac{dx}{dt}|_{t=3}=4(3)+1=13\)
\(Speed|_{t=3}=\sqrt{(sin9)^{2}+(13)^{2}}\)
Speed = 13.007
Acceleration Vector = \(\left \langle \frac{d^{2}x}{dt^{2}},{\frac{d^{2}y}{dt^{2}}} \right \rangle\)
\(\frac{dx}{dt}=4t +1; \frac{d^{2}x}{dt^{2}}=4\)
\(\frac{dy}{dt}=sin(t^{2}); \frac{d^{2}y}{dt^{2}}=2t cos (t^{2})\)
\(\frac{d^{2}x}{dt^{2}}|_{t=3}=4; \frac{d^{2}y}{dt^{2}}|_{t=3}=-5.467\)
Acceleration vector = <4, -5.467>
(b)
From part A,
\(\frac{dx}{dt}=\frac{\left ( \frac{dy}{dt} \right )}{\left ( \frac{dx}{dt} \right )}\) \(\frac{dy}{dt}|_{t=3}=sin9\)
\(\frac{dx}{dt}|_{t=3}=13\)
\(\frac{dy}{dx}= \frac{sin9}{13}=0.0317\)
(c)
x position at t = 3 is equal to x(0) + \(\int_{0}^{3}\frac{dx}{dt}dt\)
y position at t = 3 is equal to y(0) + \(\int_{0}^{3}\frac{dy}{dt}dt\)
\(x(3)=0+\int_{0}^{3}(4t+1)dt=21\)
\(y(3)=-4+\int_{0}^{3}sin(t^{2})dt=-3.226\)
At t = 3, the particle’s position is (21, -3.226)
(d)
Distance Travelled = \(\int_{0}^{3}\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt\)
\(=\int_{0}^{3}\sqrt{(4t+1)^{2}+sin^{2}(t^{2})}dt\)
Distance Travelled = 21.091