- About The Coursera
- About Calculus: Single Variable Part 4 β Applications Course
- Calculus: Single Variable Part 4 β Applications Quiz Answers
- Week 01: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
- Week 02: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
- Week 03: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
- Week 04: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
- week 05: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
- More About This Course
Hello Peers, Today we are going to share all week’s assessment and quiz answers of the Calculus: Single Variable Part 4 β Applications course launched by Coursera totally free of costβ β β . This is a certification course for every interested student.
In case you didn’t find this course for free, then you can apply for financial ads to get this course for totally free.
Check out this article – “How to Apply for Financial Ads?”
About The Coursera
Coursera, India’s biggest learning platform launched millions of free courses for students daily. These courses are from various recognized universities, where industry experts and professors teach in a very well manner and in a more understandable way.
Here, you will find Calculus: Single Variable Part 4 β Applications Exam Answers in Bold Color which are given below.
These answers are updated recently and are 100% correctβ answers of all week, assessment, and final exam answers of Calculus: Single Variable Part 4 β Applications from Coursera Free Certification Course.
Use βCtrl+Fβ To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A βFindβ Option There. Use These Option to Get Any Random Questions Answer.
About Calculus: Single Variable Part 4 β Applications Course
This quick course covers the main ideas of Calculus with one variable, with a focus on understanding the ideas and how to use them. This course is perfect for students who are just starting out in engineering, the physical sciences, or the social sciences.
Course Apply Link – Calculus: Single Variable Part 4 β Applications
Calculus: Single Variable Part 4 β Applications Quiz Answers
Week 01: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
Quiz 01 : Core Homework: Simple Areas
Q1. What is the area between the curve f(x) = \sin^3 xf(x)=sin3x and the xx-axis from x=0x=0 to \displaystyle x=\frac{\pi}{3}x=3Οβ ?
- \displaystyle \frac{1}{24}241β
- \displaystyle \frac{23 β 9\sqrt{3}}{24}2423β93ββ
- \displaystyle \frac{5}{24}245β
- \displaystyle \frac{29}{24}2429β
- \displaystyle \frac{-25 + 15\sqrt{3}}{24}24β25+153ββ
- \displaystyle \frac{19}{24}2419β
Q2. Find the area of the bounded region enclosed by the curves y = \sqrt{x}y=xβ and y = x^2y=x2.
Hint: start by drawing the curves in the plane and identifying the appropriate region.
- -3β3
- \displaystyle \frac{1}{3}31β
- 11
- -1β1
- -\displaystyle \frac{1}{3}β31β
- 33
Q3. What is the area between the curve y = \sin xy=sinx and the xx-axis for 0\le x\le \pi0β€xβ€Ο ?
- \piΟ
- -1β1
- -\piβΟ
- 22
- -2β2
- 11
Q4. What is the area between the curve y = \sin xy=sinx and the xx-axis for 0\le x \le 2\pi0β€xβ€2Ο ?
- 00
- 22
- -2β2
- 11
- 44
- -4β4
Q5. Recall from the Lecture that the Gini index is defined as
G(f) = \frac{\text{area between } y=x \text{ and } y=f}{\text{area between } y=x \text{ and } y=0} = 2 \int_{x=0}^1 \left(x β f(x)\right)\, dxG(f)=area between y=x and y=0area between y=x and y=fβ=2β«x=01β(xβf(x))dx
where f(x)f(x) is the fraction of total income earned by the lowest xx fraction of the populace.
Calculate the Gini index of a country where
f(x) = \frac{2}{5} x^2 + \frac{3}{5} x^3f(x)=52βx2+53βx3
- \displaystyle G(f) = \frac{13}{60}G(f)=6013β
- \displaystyle G(f) = \frac{13}{30}G(f)=3013β
- \displaystyle G(f) = \frac{5}{6}G(f)=65β
- \displaystyle G(f) = \frac{5}{12}G(f)=125β
- \displaystyle G(f) = \frac{5}{3}G(f)=35β
- \displaystyle G(f) = \frac{13}{15}G(f)=1513β
Quiz02 : Core Homework: Complex Areas
Q1. Find the area enclosed by the curves y = 1y=1, x = 1x=1, and y = \ln xy=lnx.
Hint: draw the three curves first and identify the region that they enclose. It should look like a right triangle but with a curved hypotenuse.
- \displaystyle e β \frac{3}{2}eβ23β
- \ln 2ln2
- e-1eβ1
- 11
- e-2eβ2
- ee
Q2 .Find the area of the bounded region enclosed by the xx-axis, the lines x=1x=1 and x=2x=2 and the hyperbola xy = 1xy=1.
- \displaystyle -\frac{1}{2}β21β
- \displaystyle \frac{1}{2}21β
- \ln 2ln2
- 22
- \ln 3ln3
- 11
Q3. Compute the area in the bounded β that is, finiteβ regions between y=x(x-1)(x-2)y=x(xβ1)(xβ2) and the xx-axis.
- 11
- 22
- \displaystyle \frac{3}{4}43β
- \displaystyle \frac{1}{2}21β
- 00
- \displaystyle \frac{1}{4}41β
Q4. Find the area of the sector of a circular disc of radius rr (centered at the origin) given by 1 \leq \theta \leq 31β€ΞΈβ€3 (as usual, \thetaΞΈ is in radians).
- \displaystyle \frac{\pi r^2}{2}2Οr2β
- \displaystyle \frac{2}{3}r^332βr3
- 2r^22r2
- 2r2r
- 2 \pi r^22Οr2
- r^2r2
Q5. Compute the area enclosed by the cardioid in the figure below. This curve is described by the polar equation r = 1 + \cos\thetar=1+cosΞΈ
.
- \displaystyle \frac{5\pi}{2}25Οβ
- 2\pi2Ο
- 3\pi3Ο
- \displaystyle \frac{3\pi}{2}23Οβ
- \piΟ
- \displaystyle \frac{\pi}{2}2Οβ
Quiz 03: Core Homework: Simple Volumes
Q1. Find the volume of the following solid: for 1 \le x \lt +\infty1β€x<+β, the intersection of this solid with the plane perpendicular to the xx-axis is a circular disc of radius e^{-x}eβx. Choose β+\infty+ββ if the resulting integral diverges.
- \displaystyle \frac{\pi-e}{3}3Οβeβ
- \piΟ
- 1515
- +\infty+β
- \displaystyle \frac{e^2}{2}2e2β
- \displaystyle \frac{\pi}{2e^2}2e2Οβ
Q2. The base of a solid is given by the region lying between the yy-axis, the parabola y=x^2y=x2, and the line y=16y=16 in the first quadrant. Its cross-sections perpendicular to the yy-axis are equilateral triangles. Find the volume of this solid.
- 64\sqrt{3}643β
- 32\sqrt{3}323β
- 16\sqrt{3}163β
- 11
- 2
- 2\sqrt{3}23β
Q3. The base of a solid is given by the region lying between the yy-axis, the parabola y=x^2y=x2, and the line y=4y=4. Its cross-sections perpendicular to the yy-axis are squares. Find the volume of this solid.
- 22
- \displaystyle\frac{8}{3}38β
- 44
- 1616
- \displaystyle\frac{16}{3}316β
- 88
Q4. Find the volume of the solid whose base is the region enclosed by the curve y=\sin xy=sinx and the xx-axis from x=0x=0 to x=\pix=Ο and whose cross-sections perpendicular to the xx-axis are semicircles.
- \piΟ
- \displaystyle \frac{\pi^2}{16}16Ο2β
- \pi^2Ο2
- \displaystyle \frac{\pi^2}{4}4Ο2β
- 00
- \displaystyle \frac{\pi^2}{8}8Ο2β
Q5. Consider a cone of height hh over a circular base of radius rr. We computed the volume by slicing parallel to the base. What happens if instead we slice orthogonal to the base? What is the volume element obtained by taking a wedge at angle \thetaΞΈ of thickness d\thetadΞΈ ?
Hint: if you like, check to see that integrating over 0\le \theta\le 2\pi0β€ΞΈβ€2Ο gives the correct volume of \pi r^2 h / 3 Οr2h/3.
- dV = \displaystyle \frac{\pi}{3}r^2hdV=3Οβr2h
- dV = 2 r^2h\,d\thetadV=2r2hdΞΈ
- dV = \displaystyle \frac{1}{2}r^2\,d\thetadV=21βr2dΞΈ
- dV = \displaystyle \frac{1}{3}r^2h\,d\thetadV=31βr2hdΞΈ
- dV = \displaystyle r^2h\,d\thetadV=r2hdΞΈ
- dV = \displaystyle \frac{1}{6}r^2h\,d\thetadV=61βr2hdΞΈ
Quiz 04:Core Homework: Complex Volumes
Q1. Let DD be the region bounded by the curve y = x^3y=x3, the xx-axis, the line x = 0x=0 and the line x = 2x=2. Find the volume of the region obtained by revolving DD about the xx-axis.
- \displaystyle \frac{128}{7} \pi7128βΟ
- 4 \pi4Ο
- \displaystyle \frac{64}{7} \pi764βΟ
- 2\pi2Ο
- None of these
- \displaystyle \frac{64}{4} \pi464βΟ
Q2. Let RR be the region between the curve y = -(x-2)^2+1y=β(xβ2)2+1 and the xx-axis. Find the volume of the region obtained by revolving RR about the yy-axis.
- \displaystyle \frac{32}{5} \pi532βΟ
- 8 \pi^28Ο2
- \displaystyle \frac{52}{3} \pi352βΟ
- \displaystyle \frac{16}{3} \pi316βΟ
- \displaystyle \frac{16}{3} \pi^2316βΟ2
- \displaystyle \frac{4}{5} \pi54βΟ
Q3. Find the volume obtained by revolving the region between the curves y = x^3y=x3 and y = \sqrt[3]{x}y=3xβ in the first quadrant about the xx-axis.
- \displaystyle \frac{9}{35} \pi359βΟ
- \displaystyle \frac{16}{35} \pi3516βΟ
- \displaystyle \frac{1}{11} \pi111βΟ
- \displaystyle \frac{26}{35} \pi3526βΟ
- \displaystyle \frac{8}{35} \pi358βΟ
- \displaystyle \frac{32}{35} \pi3532βΟ
Q4. Let DD be the region under the curve y = \ln \sqrt{x}y=lnxβ and above the xx-axis from x = 1x=1 to x = ex=e. Find the volume of the region obtained by revolving DD about the xx-axis.
- \pi(e-1)Ο(eβ1)
- \displaystyle \frac{\pi(e-1)}{2}2Ο(eβ1)β
- \displaystyle \frac{\pi(e-2)}{4}4Ο(eβ2)β
- \pi(e-2)Ο(eβ2)
- \displaystyle \frac{\pi(e-1)}{4}4Ο(eβ1)β
- \displaystyle \frac{\pi(e-2)}{2}2Ο(eβ2)β
Q5. Let DD be the region from Question 1. What is the volume of the region formed by rotating DD about the line x = 3x=3?
- 24 \pi24Ο
- \displaystyle \frac{264}{5} \pi5264βΟ
- \displaystyle \frac{184}{3} \pi3184βΟ
- 48 \pi48Ο
- \displaystyle \frac{56}{5} \pi556βΟ
- \displaystyle \frac{216}{5} \pi5216βΟ
Q6. Let DD be the region bounded by the graph of y = 1-x^4y=1βx4, the xx-axis and the yy-axis in the first quadrant. Which of the following integrals can be used to compute the volume of the region obtained by revolving DD around the line x=5x=5?
- \displaystyle\int_{x=0}^1 2\pi (5-x)(5-x^4) \, dxβ«x=01β2Ο(5βx)(5βx4)dx
- \displaystyle \int_{x=0}^1 \pi (1-x^4)^2 \, dxβ«x=01βΟ(1βx4)2dx
- \displaystyle \int_{y=1}^15 \pi y\sqrt[3]{y-1} \, dyβ«y=11β5Οy3yβ1βdy
- \displaystyle \int_{x=0}^1 2\pi x(x^4-5) \, dxβ«x=01β2Οx(x4β5)dx
- \displaystyle \int_{x=0}^1 2 \pi (5-x)(1-x^4) \, dxβ«x=01β2Ο(5βx)(1βx4)dx
- \displaystyle \int_{x=0}^1 \pi x^2 (1-x^4) \, dxβ«x=01βΟx2(1βx4)dx
Week 02: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
Quiz 01 :Core Homework: Volume and Dimension
Q1. Consider a four-dimensional box (or βrectangular prismβ) with side-lengths 11, 1/21/2, 1/31/3, and 1/41/4. What is the 44-dimensional volume of this box?
- \displaystyle \frac{1}{24}241β
- \displaystyle \frac{1}{12}121β
- \displaystyle 11
- \displaystyle \frac{1}{2}21β
- \displaystyle \frac{1}{6}61β
Q2. In the 44-d box of Question 1, what is the βdiameterβ βi.e., the farthest distance between two points in the box?
Hint: think in terms of diagonals.
- \displaystyle \frac{5}{2\sqrt{3}}23β5β
- \displaystyle \frac{1}{2\sqrt{6}}26β1β
- \displaystyle \frac{\sqrt{205}}{12}12205ββ
- 22
- \displaystyle \frac{25}{12}1225β
Q3. High-dimensional objects are everywhere and all about. Letβs consider a very simple model of the space of digital images. Assume a planar digital image (such as that captured by a digital camera), where each pixel is given values that encode color and intensity of light. Letβs assume that this is done via an RGB (red/blue/green) model. Though there are many RGB model specifications, let us use one well-suited for mathematics: to each pixel on associates three numbers (R,G,B)(R,G,B), each taking a value in [0,1][0,1].
Since the red/blue/green values are independent, each pixel has associated to it a 3-d cube of possible color values. Consider a (fairly standard) 10-megapixel camera. If I were to consider the βspace of all imagesβ that my camera can capture, what does the space look like?
Note: thereβs no calculus in this problemβ¦just counting!
- A unit ball of dimension 3\times 10^63Γ106
- A unit cube of dimension 3\times 10^63Γ106
- A unit simplex of dimension 3\times 10^63Γ106
- A unit cube of dimension 3\times10^73Γ107
- A unit cube of dimension 3\times 10^{10}3Γ1010
Quiz 02 :Core Homework: Arclength
Q1. Find the arc length of the curve \displaystyle y = \left( x + \frac{5}{9} \right)^{3/2}y=(x+95β)3/2 from x = 0x=0 to x = 3x=3.
- 11
- 88
- 77
- \displaystyle \frac{63}{4}463β
- \displaystyle \frac{3}{2}23β
- \displaystyle \frac{21}{2}221β
Q2. Find the arc length of the curve y = -\ln (\cos x)y=βln(cosx) from x = 0x=0 to x = \displaystyle \frac{\pi}{4}x=4Οβ.
Hint: you may need to use that
\int \sec x\, dx = \ln \left|\sec x + \tan x \right| + Cβ«secxdx=lnβ£secx+tanxβ£+C
- \ln \sqrt2ln2β
- \ln (\sqrt{2} β 1)ln(2ββ1)
- \ln (\sqrt2 + 1)ln(2β+1)
- \displaystyle \ln \frac{\sqrt2}{2}ln22ββ
- \displaystyle\ln \left( \frac{\sqrt2}{2} + 1 \right)ln(22ββ+1)
- 11
Q3. The so-called cuspidal cubic is given parametrically by the equations
x = t^3, \quad y = t^2x=t3,y=t2
Compute the arc length of this curve as tt goes from -1β1 to 11. Provide a numeric answer rounded to two decimal places.
Q4. Consider the spiral given by the parametric equations
x = t^{-k} \cos t, \quad y = t^{-k} \sin tx=tβkcost,y=tβksint
where k > 0k>0. Denote by L_kLkβ its arc length as tt moves from 2\pi2Ο to +\infty+β. Which of the following statements are true? Select all that apply.
Hint: in Lecture we studied the case k=1k=1: see the figure from the Lecture if you need help visualizingβ¦
- L_kLkβ is finite for k \gt 1k>1, and infinite for k \leq 1kβ€1
- \displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{k^2 + t^2}}{t^{k+1}} \, dtLkβ=β«t=2Ο+ββtk+1k2+t2ββdt
- \displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{1 + k^2 t^2}}{t^{k+1}} \, dtLkβ=β«t=2Ο+ββtk+11+k2t2ββdt
- L_kLkβ is finite for k \lt 1k<1, and infinite for k \geq 1kβ₯1
- \displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{1 + t^2}}{k t^{k+1}} \, dtLkβ=β«t=2Ο+ββktk+11+t2ββdt
- \displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{1 + t^2}}{t^{k+1}} \, dtLkβ=β«t=2Ο+ββtk+11+t2ββdt
Q5. At the close of this lecture we saw an example of a fractal β the so-called Koch snowflake. A similar example is given by the following procedure. Starting with a line segment of length 11 (labelled β1β in the figure below), remove the middle third and replace it by a square hat to obtain the curve β2β. Perform the same operation on each line segment in β2β to obtain β3β.
Doing this ad infinitum yields another fractal β that is, a bounded compact curve of infinite length!
But what is the exact length of the curve obtained after a finite number nn of iterations?
[Images courtesy of Wikimedia Commons]
- \displaystyle \left( \frac{5}{3} \right)^n(35β)n
- \displaystyle \left( \frac{4}{3} \right)^n(34β)n
- \displaystyle \left( \frac{3}{4} \right)^n(43β)n
- \displaystyle \left( \frac{3}{5} \right)^n(53β)n
- \displaystyle \left( \frac{5}{4} \right)^n(45β)n
- \displaystyle \left( \frac{4}{5} \right)^n(54β)n
Quiz 03:Core Homework: Surface Area
Q1. Think of the sphere of radius 11 as obtained by revolving the curve y = \sqrt{1-x^2}y=1βx2β about the xx-axis. For any -1 \leq a \lt b \leq 1β1β€a<bβ€1, calculate the surface area of the slice between x=ax=a and x=bx=b.
- \displaystyle 4\pi \sqrt{\frac{b+a}{2}}4Ο2b+aββ
- 2\pi (b^2 + a^2)2Ο(b2+a2)
- \displaystyle 4\pi \sqrt{\frac{b-a}{2}}4Ο2bβaββ
- 2\pi(b-a)2Ο(bβa)
- 2\pi (b^2 β a^2)2Ο(b2βa2)
- 2\pi(b+a)2Ο(b+a)
Q2. A typical dish antenna is built as a surface of revolution obtained by revolving a parabola about an axis of symmetry. One of the main benefits of this design is that the resulting antenna exhibits very high gains in the direction towards which it points, making it well-suited for applications in which a strong directionality is needed βsuch as TV reception and radar.
We can model such a parabolic antenna as the surface of revolution obtained by revolving the function
y = \sqrt{\frac{K}{4}} x^2, \qquad 0 \leq x \leq Ry=4Kββx2,0β€xβ€R
about the yy-axis. Here RR is the radius of the antenna, and KK βthe curvature at the tipβ controls how flat it is. Compute the surface area of this antenna in terms of the parameters RR and KK.
1 point
- \displaystyle \frac{4\sqrt{2}}{3} R^{3/2}K^{-1/4}342ββR3/2Kβ1/4
- \displaystyle \frac{\pi}{K} \left[ \left( 1 + 2RK \right)^{1/2} β 1 \right]KΟβ[(1+2RK)1/2β1]
- \displaystyle \frac{2\sqrt{2}}{3} R^{1/2} K^{-3/4}322ββR1/2Kβ3/4
- \displaystyle \frac{2\pi}{3K} \left[ \left( 1 + KR^2 \right)^{3/2} β 1 \right]3K2Οβ[(1+KR2)3/2β1]
- \displaystyle \frac{2\pi}{3K} \left[ \left( 1 + 2RK \right)^{3/2} β 1 \right]3K2Οβ[(1+2RK)3/2β1]
- \displaystyle \frac{\pi}{K} \left[ \left( 1 + KR^2 \right)^{1/2} β 1 \right]KΟβ[(1+KR2)1/2β1]
Q3. Consider the truncated circular cone in the figure (just the sides, not including the bottom and top).
It can be modeled as the surface of revolution obtained by revolving the line
x = R_1 + (R_2-R_1)\frac{y}{h}, \qquad 0 \leq y \leq hx=R1β+(R2ββR1β)hyβ,0β€yβ€h
about the yy-axis. Which of the following expressions describes its surface area in terms of the parameters hh, R_1R1β and R_2R2β ?
- \displaystyle \frac{\pi}{2} (R_1 + R_2) \sqrt{h^2 + (R_2-R_1)^2}2Οβ(R1β+R2β)h2+(R2ββR1β)2β
- \displaystyle \pi(R_1 + R_2) \left(h^2 + (R_2-R_1)^2\right)^{3/2}Ο(R1β+R2β)(h2+(R2ββR1β)2)3/2
- \displaystyle \pi(R_1 + R_2) \sqrt{h^2 + (R_2-R_1)^2}Ο(R1β+R2β)h2+(R2ββR1β)2β
- \displaystyle \frac{\pi(R_1 + R_2)}{2\sqrt{h^2 + (R_2-R_1)^2}}2h2+(R2ββR1β)2βΟ(R1β+R2β)β
- \displaystyle \frac{\pi(R_1 + R_2)}{\sqrt{h^2 + (R_2-R_1)^2}}h2+(R2ββR1β)2βΟ(R1β+R2β)β
- \displaystyle \frac{\pi}{2} (R_1 + R_2) \left(h^2 + (R_2-R_1)^2\right)^{3/2}2Οβ(R1β+R2β)(h2+(R2ββR1β)2)3/2
Q4. Consider a circular tent whose roof is made of fabric hanging from the rim of the walls of the tent and supported at a central pole.
If you look at the curve that the fabric roof forms along any radial cross-section, you will discover a catenary βthat is, a hyperbolic cosine. Modeling the roof as the surface of revolution obtained by revolving the curve
y = R \cosh\left( 1 β \frac{x}{R} \right) , \qquad 0 \leq x \leq Ry=Rcosh(1βRxβ),0β€xβ€R
around the yy-axis, which of the following integrals computes its surface area?
- \displaystyle 2\pi \int_{u=0}^1 u \cosh(1-u) \, du2Οβ«u=01βucosh(1βu)du
- \displaystyle 2\pi R \int_{u=0}^1 u \cosh(1-u) \, du2ΟRβ«u=01βucosh(1βu)du
- \displaystyle 2\pi R^2 \int_{u=0}^1 u \cosh(1-u) \, du2ΟR2β«u=01βucosh(1βu)du
- \displaystyle 2\pi R \int_{u=0}^1 u \sinh(1-u) \, du2ΟRβ«u=01βusinh(1βu)du
- \displaystyle 2\pi R^2 \int_{u=0}^1 u \sinh(1-u) \, du2ΟR2β«u=01βusinh(1βu)du
- \displaystyle 2\pi \int_{u=0}^1 u \sinh(1-u) \, du2Οβ«u=01βusinh(1βu)du
Week 03: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
Quiz 01:Core Homework: Work
Q1. How much work is needed to lift a 40 \text{ kg}40 kg television up a height of 22 meters? Take the acceleration of gravity to be g = 10\,\mathrm{m}/\mathrm{s}^2g=10m/s2.
A reminder on units: recall that, in the International System of Units, length is measured in meters (\mathrm{m}m), time in seconds (\mathrm{s}s), and mass in kilograms (\mathrm{kg}kg). The unit of force is called a newton (\mathrm{N}N), and that of work a joule (\mathrm{J}J). Newtonβs Second Law, F = maF=ma, tells us that
1\,\mathrm{N} = 1\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}^21N=1kgm/s2
The basic definition of work as a product of force and distance then yields
1\,\mathrm{J} = 1\,\mathrm{N}\,\mathrm{m} = 1\,\mathrm{kg}\,\mathrm{m}^2/\mathrm{s}^21J=1Nm=1kgm2/s2
- 1,\!000\,\mathrm{J}1,000J
- 80\,\mathrm{J}80J
- 800\,\mathrm{J}800J
- 400\,\mathrm{J}400J
- 1,\!600\,\mathrm{J}1,600J
- 40\,\mathrm{J}40J
Q2. Your swimming pool is 3\,\mathrm{m}3m deep, 10\,\mathrm{m}10m long and 6\,\mathrm{m}6m wide. If the pool is initially full, how much work is required to drain two thirds of the water in the pool (that is, until the water is only 1\,\mathrm{m}1m deep)? Assume that the density of water is 1,\!000\,\mathrm{kg}/\mathrm{m}^31,000kg/m3, and that the acceleration of gravity is g = 10\,\mathrm{m}/\mathrm{s}^2g=10m/s2.
- 1.35\cdot 10^6\,\mathrm{J}1.35β 106J
- 1.25\cdot 10^5\,\mathrm{J}1.25β 105J
- 2.4 \cdot 10^4\,\mathrm{J}2.4β 104J
- 1.2 \cdot 10^6\,\mathrm{J}1.2β 106J
- 1.2\cdot 10^5\,\mathrm{J}1.2β 105J
- 2.7\cdot 10^5\,\mathrm{J}2.7β 105J
Q3. A 100100 meter long cable of linear mass density 0.1\,\mathrm{kg}/\mathrm{m}0.1kg/m hangs over a very high vertical cliff. Assuming that there is no friction, how much work is needed to to lift this cable up to the top of the cliff? Assume that the acceleration due to gravity is g = 10\,\mathrm{m}/\mathrm{s}^2g=10m/s2.
- 500\,\mathrm{J}500J
- 100\,\mathrm{J}100J
- 2,\!500\,\mathrm{J}2,500J
- 10,\!000\,\mathrm{J}10,000J
- 5,\!000\,\mathrm{J}5,000J
- 50\,\mathrm{J}50J
Q4. Assume that a sports carβs acceleration aa increases linearly with its position xx as a(x) = xa(x)=x. Since the car is burning fuel, its mass mm decreases; assume the decrease is exponential in xx as m(x) = 1 + e^{-x}m(x)=1+eβx. How much work is done in driving the car from x=0x=0 to x = 3x=3 ?
Hint: remember Newtonβs Second Law, F=maF=ma. In our case, both mass and acceleration are functions of xx.
- 3e^2 β 13e2β1
- \displaystyle 3 + \frac{3}{e^3}3+e33β
- \displaystyle 1 β \frac{2}{e}1βe2β
- \displaystyle \frac{9}{2}+\frac{2}{e^3}29β+e32β
- \displaystyle \frac{11}{2}-\frac{4}{e^3}211ββe34β
- \displaystyle \frac{2}{e
Quiz 02:Core Homework: Elements
Q1. Consider a dam of height HH and width WW that has a perfectly vertical face facing the water, which reaches all the way up to the damβs height. If the water has weight density \rhoΟ, what is the total force the water exerts against the face of the dam?
- \displaystyle \frac{1}{4}H^2 W \rho41βH2WΟ
- \displaystyle \frac{1}{2}H W \rho21βHWΟ
- \displaystyle \frac{1}{4}H W^2 \rho41βHW2Ο
- H^2 W \rhoH2WΟ
- \displaystyle \frac{1}{2}H W^2 \rho21βHW2Ο
- \displaystyle \frac{1}{2}H^2 W \rho21βH2WΟ
Q2. Consider two potential income streams, each valued based on an assumption of a constant return on investment at rate r>0r>0. The first, I_1I1β, starts off slow, then peaks, and then decreases. The second, I_2I2β, starts off high, then decreases. Both oscillate eventually with the same period. The specific formulae are:
I_1(t) = I_0 + A\sin\frac{\pi t}{P} \quad ; \quad I_2(t) = I_0 + A\cos\frac{\pi t}{P}I1β(t)=I0β+AsinPΟtβ;I2β(t)=I0β+AcosPΟtβ
Here, I_0>0I0β>0 is a constant (the baseline income), A>0A>0 is a constant (the amplitude of fluctuation) and P>0P>0 is a constant (the half-period). Assume \pi \gt PrΟ>Pr. Which income stream has the greater present value over the time interval [0,P][0,P] ? Which has the greater present value over the time interval [0, +\infty)[0,+β) ?
Hints: (1) Which constants are important? I_0I0β? AA? PP? rr? (2) You may want a reduction formula like that from Lecture 22. (3) If you get stuck in the algebra, try using WolframAlpha.
- On [0,P][0,P], PV_1=PV_2PV1β=PV2β; but on [0,+\infty)[0,+β), PV_1\lt PV_2PV1β<PV2β.
- PV_1 \lt PV_2PV1β<PV2β both on [0, P][0,P] and [0, +\infty)[0,+β).
- On [0,P][0,P], PV_1=PV_2PV1β=PV2β; but on [0,+\infty)[0,+β), PV_1>PV_2PV1β>PV2β.
- On [0,P][0,P], PV_1\lt PV_2PV1β<PV2β, but on [0,+\infty)[0,+β), PV_1\gt PV_2PV1β>PV2β.
- PV_1 = PV_2PV1β=PV2β both on [0, P][0,P] and [0, +\infty)[0,+β).
- PV_1 \gt PV_2PV1β>PV2β both on [0, P][0,P] and [0, +\infty)[0,+β).
Q3 .We have learned about present value of an income stream I(t)I(t); one may also reverse the derivation to determine the future value of the income at a time t=Tt=T. The future value element of I(t)I(t) is
dFV = e^{r(T-t)}I(t)dt,dFV=er(Tβt)I(t)dt,
assuming a continuous compounding at fixed interest rate rr.
If you save for a childβs college at a rate of \$5,\!000 / \mathrm{year}$5,000/year starting at the childβs birth, how much money will be available when she is 2020? Assume a fixed 5\%5% return on investments.
- FV = \$100,\!000eFV=$100,000e
- FV = \$50,\!000 eFV=$50,000e
- FV = \$50,\!000\sqrt{e}FV=$50,000eβ
- FV = \$100,\!000(e-1)FV=$100,000(eβ1)
- FV = \$500,\!000FV=$500,000
- FV = \$100,\!000FV=$100,000
Q4. Consider a cantilever beam of length LL. Suppose that NN people, each of mass m_0m0β, stand on it equally spaced, so that their combined weight is supported uniformly along the beam. If L = 20\,\mathrm{m}L=20m, m_0 = 75\,\mathrm{kg}m0β=75kg and, at the point of attachment, the beam can withstand a maximum torque of \tau_\mathrm{max} = 1.5\cdot 10^6\,\mathrm{N}\cdot\mathrm{m}Οmaxβ=1.5β
106Nβ
m, what is the maximum number of people that can stand on it? Assume the acceleration of gravity to be g = 10\,\mathrm{m}/\mathrm{s}^2g=10m/s2.
- 25 people.
- 50 people.
- 400 people.
- 300 people.
- 200 people.
- 100 people.Suppose that a radiator is turned off at t=0t=0; after that, the amount of heat generated by the radiator is described by the heat flow element
dQ = Q_0 e^{-\lambda t} dtdQ=Q0βeβΞ»tdt
Q5. where both Q_0Q0β and \lambdaΞ» are positive constants. What is the total amount of heat radiated from the moment it is turned off?
- \sqrt{\lambda} Q_0Ξ»βQ0β
- \lambda Q_0Ξ»Q0β
- Q_0 e^\lambdaQ0βeΞ»
- \displaystyle \frac{Q_0}{\lambda}Ξ»Q0ββ
- \lambda^2 Q_0Ξ»2Q0β
- Q_0 e^{-\lambda}Q0βeβΞ»
Week 04: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
Quiz 01:Core Homework: Averagez
Q1. Find the average value of \displaystyle f(x) = \frac{1}{\sqrt{4x β 3}}f(x)=4xβ3β1β from x = 3x=3 to x = 21x=21.
- \displaystyle \frac{1}{12}121β
- \displaystyle -\frac{2}{9}β92β
- 33
- \displaystyle \frac{1}{18}181β
- \displaystyle \frac{3}{2}23β
- \displaystyle \frac{1}{6}61β
Q2. Calculate the average of the function f(x) = x^3 \sqrt{1+x^2}f(x)=x31+x2β over the interval 0 \leq x \leq \sqrt{3}0β€xβ€3β.
- \displaystyle \frac{58}{15}1558β
- \displaystyle \frac{128}{15}15128β
- \displaystyle \frac{128}{15\sqrt{3}}153β128β
- \displaystyle \frac{2}{15\sqrt{3}} \left( 1 + \sqrt{2} \right)153β2β(1+2β)
- \displaystyle \frac{2}{15} \left( 1 + \sqrt{2} \right)152β(1+2β)
- \displaystyle \frac{58}{15\sqrt{3}}153β58β
Q3. It is intuitively clear that the average value of xx over a circle of radius 11 (given by the equation x^2 + y^2 = 1x2+y2=1) is zero. But what is the average value of x^2x2 over this circle?
Hint: notice that this is an average over a curve, so you will need to integrate with respect to the arc length element dLdL. In order to make your calculations easier, use the parametrization
x = \cos t, \quad y = \sin t, \qquad 0 \leq t \leq 2\pix=cost,y=sint,0β€tβ€2Ο
- 00
- \displaystyle \frac{1}{2\pi}2Ο1β
- \displaystyle \frac{1}{4}41β
- \displaystyle \frac{1}{2}21β
- \displaystyle \frac{2}{\pi}Ο2β
- \displaystyle \frac{\pi}{4}4Οβ
Q4. Let us model a mountain as a circular cone of height hh whose base has radius RR. You can see it as the surface obtained by revolving the line
y = h \left( 1 β \frac{x}{R} \right), \qquad 0 \leq x \leq Ry=h(1βRxβ),0β€xβ€R
about the yy-axis. What is the average height of the points on the surface of the mountain?
Hint: This average is an integral with respect to area. You may wish to take as area element an infinitesimal annulus centered at the origin.
- \displaystyle \frac{h}{3}3hβ
- \displaystyle \frac{h}{6}6hβ
- \displaystyle \frac{1}{2} \pi R^2 h21βΟR2h
- \displaystyle \frac{h}{2}2hβ
- \displaystyle \frac{1}{6} \pi R^2 h61βΟR2h
- \displaystyle \frac{1}{3} \pi R^2 h31βΟR2h
Q5. What is the average of (x-1)^2(xβ1)2 over the domain 1\leq \vert x\vert \leq 31β€β£xβ£β€3. Be careful!
- 11
- \displaystyle\frac{16}{3}316β
- \displaystyle\frac{4}{3}34β
- \displaystyle\frac{32}{3}332β
- \displaystyle\frac{8}{3}38β
- 88
Quiz 02:Core Homework: Centroids
Q1. Find the coordinates (\overline{x},\overline{y})(x,yβ) of the centroid of the region bounded by y=\sin xy=sinx and y=\cos xy=cosx for \displaystyle 0 \leq x \leq \frac{\pi}{4}.0β€xβ€4Οβ.
1 point
- \displaystyle \overline{x}=\frac{\pi\sqrt{2}}{\sqrt{2}-1}x=2ββ1Ο2ββ,
- \displaystyle \overline{y}=\frac{1}{\sqrt{2}-1}yβ=2ββ11β
- \displaystyle \overline{x}=\frac{\sqrt 2}{2}x=22ββ,
- \displaystyle \overline{y}=\frac{\sqrt 2}{2}yβ=22ββ
- \displaystyle \overline{x}=\frac{1}{\sqrt 2 -1}x=2ββ11β,
- \displaystyle \overline{y}=\frac{1}{\sqrt 2-1}yβ=2ββ11β
- \displaystyle \overline{x}=\frac{\pi}{8}x=8Οβ,
- \displaystyle \overline{y}=\sqrt \frac{2- \sqrt 2}{2}yβ=22β2βββ
- \displaystyle \overline{x}=\frac{\pi\sqrt{2}-4}{4(\sqrt{2}-1)}x=4(2ββ1)Ο2ββ4β,
- \displaystyle \overline{y}=\frac{1}{4(\sqrt{2}-1)}yβ=4(2ββ1)1β
- \displaystyle \overline{x}=\pi\sqrt 2 x=Ο2β,
- \displaystyle \overline{y}=1yβ=1
Q2. Find the the coordinates (\overline{x},\overline{y})(x,yβ) of the centroid of the region defined by \displaystyle |x+y| \leq 1β£x+yβ£β€1, -1\leq x\leq 1β1β€xβ€1, and -1\leq y\leq 1β1β€yβ€1.
Hint 1: draw a picture!
Hint 2: notice anything interesting about this region?
- \displaystyle (\overline{x},\overline{y})=\left(\frac{1}{\sqrt2},\frac{1}{\sqrt 2}\right)(x,yβ)=(2β1β,2β1β)
- \displaystyle (\overline{x},\overline{y})= \left(\frac{1}{2},\frac{1}{2}\right)(x,yβ)=(21β,21β)
- (\overline{x},\overline{y})=\displaystyle \left(-\frac{1}{\sqrt2},\frac{1}{\sqrt 2}\right)(x,yβ)=(β2β1β,2β1β)
- (\overline{x},\overline{y})=\displaystyle (0,0)(x,yβ)=(0,0)
- (\overline{x},\overline{y})=\displaystyle (1,1)(x,yβ)=(1,1)
- \displaystyle (\overline{x},\overline{y})= \left(-\frac{1}{\sqrt2}, -\frac{1}{\sqrt 2}\right)(x,yβ)=(β2β1β,β2β1β)
Q3. Compute the center of mass of a thin rod with density \rho(x)=e^{-ax}Ο(x)=eβax for a\gt 0a>0 a constant and 0\leq x\lt\infty0β€x<β. (Yes, i know, itβs not-so-physical to talk about infinite rodsβ¦trust me, you will care about this result soon!)
- \displaystyle\overline{x} = e^ax=ea
- \displaystyle\overline{x} = \frac{1}{a}x=a1β
- \displaystyle\overline{x} = ax=a
- \displaystyle\overline{x} = \frac{1}{a^2}x=a21β
- \displaystyle\overline{x}x does not exist (the integral diverges).
- \displaystyle\overline{x} = 1x=1
Q4. Find the coordinates (\overline{x},\overline{y})(x,yβ) of the centroid of the union of the following two discs:
D_1: x^2 + y^2 \leq 4 \qquad\text{and}\qquad D_2: (x β 4)^2 + (y β 2)^2 \leq 1D1β:x2+y2β€4andD2β:(xβ4)2+(yβ2)2β€1
Hint: replace each disc with a vertex at its centroid. What βmassβ should you assign to each vertex?
- \displaystyle (\overline{x},\overline{y})=\left( \frac{4\pi}{5}, \frac{2\pi}{5} \right)(x,yβ)=(54Οβ,52Οβ)
- (\overline{x},\overline{y})=(0,0)(x,yβ)=(0,0)
- (\overline{x},\overline{y})=\displaystyle \left( \frac{4}{5}, \frac{2}{5} \right)(x,yβ)=(54β,52β)
- \displaystyle (\overline{x},\overline{y})=\left( \frac{4}{\pi}, \frac{2}{\pi} \right)(x,yβ)=(Ο4β,Ο2β)
- \displaystyle (\overline{x},\overline{y})= (4\pi,2\pi)(x,yβ)=(4Ο,2Ο)
- \displaystyle (\overline{x},\overline{y})= \left(2,1\right)(x,yβ)=(2,1)
Q5. Find the the coordinates (\overline{x},\overline{y})(x,yβ) of the center of mass of the region between the xx-axis, the yy-axis, and the lines x=2x=2 and \displaystyle y=x+2y=x+2, with density (mass-per-unit-area) \rho=3xΟ=3x.
Hint: remember, this is a center-of-mass, not a centroid, so youβll need to integrate with respect to dM=\rho\cdot dAdM=Οβ dA.
- (\overline{x},\overline{y})=(1,2)(x,yβ)=(1,2)
- (\overline{x},\overline{y})=\displaystyle \left(\frac{7}{5},\frac{17}{5}\right)(x,yβ)=(57β,517β)
- (\overline{x},\overline{y})=\displaystyle \left(\frac{7}{5},\frac{17}{10}\right)(x,yβ)=(57β,1017β)
- (\overline{x},\overline{y})=\displaystyle \left(10, \frac{17}{3}\right)(x,yβ)=(10,317β)
- (\overline{x},\overline{y})=\displaystyle \left(\frac{1}{3},\frac{17}{3}\right)(x,yβ)=(31β,317β)
- (\overline{x},\overline{y})=\displaystyle \left(\frac{14}{3},\frac{17}{3}\right)(x,yβ)=(314β,317β)
Quiz 03:Core Homework: Moments and Gyrations
q1. Three particles, each of mass mm, are located at distances r_1r1β, r_2r2β and r_3r3β respectively from a fixed axis of rotation AA. We now place a fourth particle, also of mass mm, at some distance rr from the axis AA. If the moment of inertia of all four particles is twice as big as the moment of inertia of the first three, what is rr ?
Note: this question doesnβt really use any calculus, but it will give you practice at remembering what moment of inertia means.
- r = ( r_1 + r_2 + r_3 ) \ln 2r=(r1β+r2β+r3β)ln2
- \displaystyle r = \frac{2}{3} ( r_1 + r_2 + r_3 )r=32β(r1β+r2β+r3β)
- \displaystyle r = \sqrt{r_1^2+r_2^2+r_3^2}r=r12β+r22β+r32ββ
- r = \sqrt[3]{r_1 r_2 r_3}r=3r1βr2βr3ββ
- \displaystyle r = \frac{r_1^2}{r_2}+\frac{r_2^2}{r_3}+\frac{r_3^2}{r_1}r=r2βr12ββ+r3βr22ββ+r1βr32ββ
- r = \sqrt[3]{2 \left( r_1^3 + r_2^3 + r_3^3 \right) }r=32(r13β+r23β+r33β)β
Q2. In mathematics, an annulus is defined as the region between two circles with a common center. Assume you are given an annulus with outer radius RR, inner radius rr, and mass MM distributed uniformly. What is its moment of inertia about the central axis shown in the picture below?
Hint: this problem becomes easier if you watch the bonus lecture first!
- \displaystyle I_\text{annulus} = \frac{1}{4}M(R^2-r^2)Iannulusβ=41βM(R2βr2)
- \displaystyle I_\text{annulus} = M(R-r)\sqrt{R^2-r^2}Iannulusβ=M(Rβr)R2βr2β
- \displaystyle I_\text{annulus} = \frac{1}{2}M(R^2+r^2)Iannulusβ=21βM(R2+r2)
- \displaystyle I_\text{annulus} = \frac{1}{4}M(R^2+r^2)Iannulusβ=41βM(R2+r2)
- \displaystyle I_\text{annulus} = \frac{1}{2}M(R^2-r^2)Iannulusβ=21βM(R2βr2)
- \displaystyle I_\text{annulus} = \frac{1}{2}MR^2-\frac{1}{4}Mr^2Iannulusβ=21βMR2β41βMr2
Q3. A hollow cylindrical shell of length LL and radius RR is rotated about the an axis as shown in the picture.
You may assume that this cylindrical shell does not have βcapsβ at either the left or the right edge, and that its mass MM is distributed uniformly along the surface.You may also assume that RR is small enough that the piece of this cylinder at any distance rr from the axis of rotation is a circle. What is its moment of inertia?
HInt: start by computing the area AA and then the density \rho=M/AΟ=M/A. Then, setting rr to be a radial coordinate (distance-to-axis), the moment-of-inertia element is dI=\rho r^2 dAdI=Οr2dA. For dAdA, use the approximation implied by the βRR is smallβ assumption.
- \displaystyle \frac{2}{5}M(L^2+\pi R^2)52βM(L2+ΟR2)
- \displaystyle \frac{2\pi}{3}{MLR}32ΟβMLR
- \displaystyle \frac{1}{4}ML^241βML2
- \displaystyle \frac{2}{3}ML^232βML2
- \displaystyle \frac{1}{4} MR^241βMR2
- \displaystyle \frac{1}{3}ML^231βML2
Q4. You need to install a heavy front door in your home. For simplicity, assume that the door has uniform density, has total mass MM, and fills a rectangular entry of height hh and width \ellβ. You have two choices:
- a single-door, with a single set of hinges on one side; or
- double-doors, meaning spilt down the middle into two rectangular βhalf-doorsβ of height hh and width \ell/2β/2, each with hinges on the side.
You would guess that the single-door option is harder to open. How much more is the moment of inertia II of the single door than the (net) II of the two half-doors?
- Twice as much
- Four times as much
- Six times as much
- Itβs the same
- Three times as much
- Four-thirds as much
week 05: Calculus: Single Variable Part 4 β Applications Coursera Quiz Answers
Quiz 01: Core Homework: Fair Probability
Q1. The result of flipping a single coin is either heads, H, or tails ,T, each one of them with probability 1/21/2 βsuch a coin is said to be fair. If you flip the same coin a second time, there are four possible combinations of the results of both tosses βHH, HT, TH and TTβ, each one of them equally probable. Think of what happens when you do it yet once more: what is the probability of obtaining two heads and one tail, in whatever order?
- \displaystyle \frac{1}{8}81β
- \displaystyle \frac{3}{8}83β
- \displaystyle \frac{1}{4}41β
- \displaystyle \frac{1}{2}21β
- \displaystyle \frac{5}{8}85β
- \displaystyle \frac{7}{8}87β
Q2. Letβs play a game! You toss a (fair) coin. If it comes out heads, you win. Otherwise, the turn passes on to PLAYER 2, who tosses the same coin and wins if it comes out heads. If not, it is PLAYER 3βs turn. If she doesnβt get heads either, it is your turn again. The game goes on until somebody gets heads. What is the probability that you win?
- \displaystyle \frac{1}{2} + \frac{1}{2^4} + \frac{1}{2^7} + \cdots = \frac{4}{7}21β+241β+271β+β―=74β
- \displaystyle \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots = 121β+221β+231β+β―=1
- \displaystyle \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{1}{2}31β+321β+331β+β―=21β
- \displaystyle \frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \cdots = \frac{2}{3}21β+231β+251β+β―=32β
- \displaystyle \frac{1}{3}31β
- 00
Q3. A bus line runs every 30 minutes. If you arrive at a stop randomly, what is the probability that you will have to wait more than 10 minutes for the next bus?
Hint: this probability is a βvolumeβ fraction over some domain. What is the domain, and what is its dimension?
- \displaystyle \frac{1}{4}41β
- \displaystyle \frac{1}{3}31β
- 11
- \displaystyle \frac{2}{3}32β
- \displaystyle \frac{3}{4}43β
- \displaystyle \frac{1}{2}21β
Q4. What is the probability that a randomly chosen point of a square of side length LL is more than a distance rr away from every corner? Suppose r \lt L/2r<L/2.
- \displaystyle L^2 β \frac{\pi r^2}{4}L2β4Οr2β
- \displaystyle \frac{\pi r^2}{L^2}L2Οr2β
- \displaystyle \pi\left(\frac{r}{L}\right)^2 β 1Ο(Lrβ)2β1
- L^2 β \pi r^2L2βΟr2
- \displaystyle 1 β \pi\left(\frac{r}{L}\right)^21βΟ(Lrβ)2
- 1 β \displaystyle\frac{\pi r^2}{L}1βLΟr2β
Q5. In the lecture we found out that the probability that a randomly chosen point in a square lies within its inscribed circle (see the figure on the left) is
P = \frac{\text{area of the disc}}{\text{area of the square}} = \frac{\pi r^2}{(2r)^2} = \frac{\pi}{4},P=area of the squarearea of the discβ=(2r)2Οr2β=4Οβ,
where rr is the radius of the circle. Notice that this probability is independent of rr !
Reasoning in the same way, compute the probability that a randomly chosen point in a disc lies within its inscribed square (see the figure on the right).
- \displaystyle \frac{2}{\pi}Ο2β
- \displaystyle \frac{\pi}{4}4Οβ
- \displaystyle \frac{4}{\pi}Ο4β
- \displaystyle \frac{\pi}{2}2Οβ
- \displaystyle \frac{\sqrt{2}}{\pi r}Οr2ββ
- \displaystyle \frac{\pi r}{\sqrt{2}}2βΟrβ
Quiz 02: Core Homework: Probability Densities
Q1. Which of the following cannot be a probability density function on the domain given? Select all that apply.
- Ο(n)={52ββ51ββif n evenif n oddβ on n = 0, 1, \ldots, 9n=0,1,β¦,9.
- \displaystyle \rho(n) = \frac{1}{10}Ο(n)=101β on n = 0, 1, \ldots, 10n=0,1,β¦,10.
- \displaystyle \rho(n) =
β§β©β¨150if n evenif n odd
- Ο(n)={51β0βif n evenif n oddβ on n = 0, 1, \ldots, 9n=0,1,β¦,9.
\displaystyle \rho(n) = \frac{1}{n}Ο(n)=n1β on n = 1, 2, \ldotsn=1,2,β¦
- \displaystyle \rho(n) = \frac{1}{10}Ο(n)=101β on n = 0, 1, \ldots, 9n=0,1,β¦,9.
\displaystyle \rho(n) =
{10if n=1otherwise
- Ο(n)={10βif n=1otherwiseβ on n = 1, 2, \ldotsn=1,2,β¦
Q2. Which of the following cannot be a probability density function on the domain given? Select all that apply.
- \displaystyle \rho(x) = \frac{1}{10}Ο(x)=101β on [0, 10][0,10]
- \displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{1+x^2}Ο(x)=Ο2β1+x21β on [0, +\infty)[0,+β).
- \displaystyle \rho(x) = \frac{1}{2\pi} + \sin xΟ(x)=2Ο1β+sinx on [0, 2\pi][0,2Ο]
- \displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{1+x^2}Ο(x)=Ο2β1+x21β on \mathbb{R} = (-\infty, +\infty)R=(ββ,+β).
- \displaystyle \rho(x) = \frac{1}{x^2}Ο(x)=x21β on [1, +\infty)[1,+β).
- \displaystyle \rho(x) = \frac{1}{10}Ο(x)=101β on [0, 9][0,9]
Q3. For which value of \lambdaΞ» is \rho(x) = \lambda x^2 e^{-x}Ο(x)=Ξ»x2eβx a probability density function on [0, +\infty)[0,+β) ?
- \rho(x)Ο(x) is not a probability density function for any value of \lambdaΞ»
- \displaystyle \lambda = \frac{e}{2}Ξ»=2eβ
- \displaystyle \lambda = \frac{1}{e}Ξ»=e1β
- \lambda = 1Ξ»=1
- \displaystyle \lambda = \frac{1}{2}Ξ»=21β
- \lambda = 2Ξ»=2
Q4. The amount of time between failures of a printer follows an exponential probability distribution βthat is, right after being repaired, the probability that the printer will fail after a time at most TT is given by
\int_{t=0}^T \alpha e^{-\alpha t} \, dtβ«t=0TβΞ±eβΞ±tdt
for \alpha = 0.01\ln 2\,\,\, \mathrm{h}^{-1}Ξ±=0.01ln2hβ1 (notice that \alphaΞ± has units of inverse time, in this case, inverse hours). What is the probability that the printer does not fail for 200\, \mathrm{h}200h after the last repair?
- 1 β e^{-2}1βeβ2
- \displaystyle \frac{1}{4}41β
- \displaystyle \frac{1}{2}21β
- \displaystyle \frac{3}{4}43β
- e^{-1/2}eβ1/2
- e^{-2}eβ2
Quiz 03: Core Homework: Expectation and Variance
Q1. Find the expectation \mathbb{E}E and variance \mathbb{V}V of xx if its probability density function is \rho(x) = (n+1) x^nΟ(x)=(n+1)xn (nn a positive integer) on [0, 1][0,1].
- \displaystyle \mathbb{E} = \frac{n+1}{n+2}E=n+2n+1β, \displaystyle \mathbb{V} = \frac{n+1}{n+3} β \left( \frac{n+1}{n+2} \right)^2V=n+3n+1ββ(n+2n+1β)2.
- \mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+1}{n+2}V=n+2n+1β.
- \mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+1}{n+2} β 1V=n+2n+1ββ1.
- \mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+2}{n+3}V=n+3n+2β.
- \displaystyle \mathbb{E} = \frac{n+1}{n+2}E=n+2n+1β, \displaystyle \mathbb{V} = \frac{n+2}{n+3}V=n+3n+2β.
- \displaystyle \mathbb{E} = \frac{n+1}{n+2}E=n+2n+1β, \displaystyle \mathbb{V} = \frac{n+2}{n+3} β \left( \frac{n+1}{n+2} \right)^2V=n+3n+2ββ(n+2n+1β)2.
Q2 .Find the expectation \mathbb{E}E and variance \mathbb{V}V of xx if its probability density function is \displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{x^2 + 1}Ο(x)=Ο2βx2+11β on [0, +\infty)[0,+β).
Hint: notice that the integrals calculating the expectation and variance are improper because [0, +\infty)[0,+β) is unbounded. The first thing you should always do when confronted with one of these is check whether it converges or not.
- \displaystyle \mathbb{E} = \frac{2}{\pi}E=Ο2β, but \mathbb{V}V diverges.
- \mathbb{E} = 1E=1, \mathbb{V} = 1V=1.
- \displaystyle \mathbb{E} = \frac{2}{\pi}E=Ο2β, \displaystyle \mathbb{V} = \frac{2}{\pi} β \frac{4}{\pi^2}V=Ο2ββΟ24β.
- \displaystyle \mathbb{E} = \frac{2}{\pi}E=Ο2β, \displaystyle \mathbb{V} = \frac{4}{\pi^2}V=Ο24β.
- \mathbb{E} = 1E=1, but \mathbb{V}V diverges.
- Both \mathbb{E}E and \mathbb{V}V diverge.
Q3. Find the expectation \mathbb{E}E and variance \mathbb{V}V of nn if its probability density function is \displaystyle \rho(n) = \frac{1}{4}Ο(n)=41β on n = 1, 2, 3, 4n=1,2,3,4.
Hint: although we have not talked about expectation and variance for discrete probability distributions, you can do this! Think of the analogy with masses: expectation is center of mass and variance is moment of inertia. This problem hints at the fact that you can think of sums as discrete versions of integrals, opening the door to using Calculus in situations in which inputs are discrete but outputs are continuous. Much more about this in Chapter 5: Discretization!
- \mathbb{E} = 2E=2, \displaystyle \mathbb{V} = \frac{7}{2}V=27β.
- \mathbb{E} = 2E=2, \displaystyle \mathbb{V} = \frac{5}{4}V=45β.
- \displaystyle \mathbb{E} = \frac{5}{2}E=25β, \displaystyle \mathbb{V} = \frac{7}{2}V=27β.
- \displaystyle \mathbb{E} = \frac{5}{2}E=25β, \displaystyle \mathbb{V} = \frac{5}{4}V=45β.
- \displaystyle \mathbb{E} = \frac{5}{2}E=25β, \displaystyle \mathbb{V} = \frac{15}{4}V=415β.
- \mathbb{E} = 2E=2, \displaystyle \mathbb{V} = \frac{1}{2}V=21β.
Q4. The median mm of a (one-dimensional) continuous probability distribution on [a,b][a,b] is defined to be the value of xx for which the probability of x \lt mx<m is equal to the probability of x \gt mx>m βthat is, 1/21/2. In the language of integrals, this is:
\int_a^m\rho(x)dx = \frac{1}{2} = \int_m^b\rho(x)dxβ«amβΟ(x)dx=21β=β«mbβΟ(x)dx
Find the value of the median for an exponential distribution with probability density function
\rho(x) = \alpha e^{-\alpha x} \qquad \text{on}\:\:[0, +\infty)Ο(x)=Ξ±eβΞ±xon[0,+β)
- m = \alpham=Ξ±
- \displaystyle m = \alpha \ln 2m=Ξ±ln2
- \displaystyle m = \frac{\alpha}{\ln 2}m=ln2Ξ±β
- \displaystyle m = \frac{1}{\alpha \ln 2}m=Ξ±ln21β
- \displaystyle m = \frac{\ln 2}{\alpha}m=Ξ±ln2β
- \displaystyle m = \frac{1}{\alpha}m=Ξ±1β
Q5. There is a host of other numbers that one can associate to a probability distribution that generalize the median: e.g., the so-called quantiles. Let us consider an example βthe quartiles:
- the first (or lower) quartile is the unique value Q_1Q1β for which the probability of x \lt Q_1x<Q1β is 1/41/4;
- the second quartile (really the median) is the unique value Q_2Q2β for which the probability of x \lt Q_2x<Q2β is 2/42/4;
- the third (or upper) quartile is the unique value Q_3Q3β for which the probability of x \lt Q_3x<Q3β is 3/43/4;
You also have quintiles, deciles, the ubiquitous percentiles, etc.
Find the value of the first and third quartiles of the exponential distribution of the previous problem with \rho(x)=\alpha e^{-\alpha x}Ο(x)=Ξ±eβΞ±x.
- \displaystyle Q_1 = \frac{1}{\alpha \ln 4}Q1β=Ξ±ln41β, \displaystyle Q_3 = \frac{1}{\alpha \ln (4/3)}Q3β=Ξ±ln(4/3)1β.
- \displaystyle Q_1 = \frac{\alpha}{\ln 4}Q1β=ln4Ξ±β, \displaystyle Q_3 = \frac{\alpha}{\ln (4/3)}Q3β=ln(4/3)Ξ±β.
- \displaystyle Q_1 = \frac{\ln (4/3)}{\alpha}Q1β=Ξ±ln(4/3)β, \displaystyle Q_3 = \frac{\ln 4}{\alpha}Q3β=Ξ±ln4β.
- \displaystyle Q_1 = \frac{\alpha}{\ln (4/3)}Q1β=ln(4/3)Ξ±β, \displaystyle Q_3 = \frac{\alpha}{\ln 4}Q3β=ln4Ξ±β.
- \displaystyle Q_1 = \frac{1}{\alpha \ln (4/3)}Q1β=Ξ±ln(4/3)1β, \displaystyle Q_3 = \frac{1}{\alpha \ln 4}Q3β=Ξ±ln41β.
- \displaystyle Q_1 = \frac{\ln 4}{\alpha}Q1β=Ξ±ln4β, \displaystyle Q_3 = \frac{\ln (4/3)}{\alpha}Q3β=Ξ±ln(4/3)β.
Quiz 04:Chapter 4: Applications β Exam
Q1. Compute the expectation \mathbb{E}E of xx with the probability density function
\rho(x) = \frac{3}{2}\sqrt{x}Ο(x)=23βxβ
on 0 \leq x \leq 10β€xβ€1.
- \displaystyle \frac{5}{3}35β
- \displaystyle \frac{2}{3}32β
- \displaystyle \frac{1}{2}21β
- \displaystyle \frac{2}{5}52β
- \displaystyle \frac{4}{15}154β
- \displaystyle \frac{3}{5}53β
Q2. An aerosol spray releases spherical droplets whose radii are distributed randomly by a uniform distribution between 11 and 33 micrometers.
What is the average volume of such an aerosol droplet (in units of cubic micrometers)?
Hint: The volume of the average-radius droplet is not necessarily the average volumeβ¦
- \displaystyle\frac{80}{3}\pi380βΟ
- 9\pi9Ο
- \displaystyle\frac{13}{3}\pi313βΟ
- \displaystyle\frac{26}{3}\pi326βΟ
- \displaystyle\frac{80}{9}\pi980βΟ
- \displaystyle\frac{40}{3}\pi340βΟ
Q3. Find the yy-coordinate of the center of mass of a thin sheet of metal of constant density of a shape bounded by the xx-axis and the parabola
y= 1 β \frac{x^2}{25}y=1β25x2β
- \displaystyle \frac{2}{5}52β
- \displaystyle \frac{8}{3}38β
- 00
- \displaystyle \frac{4}{5}54β
- \displaystyle \frac{4}{3}34β
- \displaystyle \frac{8}{5}58β
Q4. Consider a swimming pool of some shape (with vertical sides, so that horizontal cross-sections have the same shape). Assume that it is completely full of water, and that it takes WW units of work to pump out all the water from the pool (pumping out to the elevation at the top of the pool). How much work did it take to pump out the first half of the water from the pool?
- \displaystyle\frac{1}{3}W31βW
- \displaystyle\frac{2}{\sqrt{2}}W2β2βW
- \displaystyle\frac{1}{8}W81βW
- There is not enough information to answer this question.
- \displaystyle\frac{1}{\sqrt{2}}W2β1βW
- \displaystyle\frac{1}{4}W41βW
Q5. Compute the moment of inertia II of a solid cylinder of mass MM, radius RR, and height hh about the central axis (passing through the centers of the cross-sectional discs).
- I = \displaystyle \frac{1}{2}MR^2I=21βMR2
- I = \displaystyle \frac{2}{3}MR^2I=32βMR2
- I = \displaystyle \frac{1}{2}MR^2hI=21βMR2h
- I = \displaystyle \frac{2}{3}MRhI=32βMRh
- I = \displaystyle \frac{1}{3}MRhI=31βMRh
- I = \displaystyle \frac{1}{2}MRhI=21βMRh
Q6. Find the volume of the body obtained by rotating about the xx-axis the region between the cuspidal cubic x^2 = y^3x2=y3, the xx-axis and the lines x=0x=0 and x=1x=1. Hint: you do not need a picture to solve this problemβ¦
- \displaystyle \frac{3\pi}{5}53Οβ
- \displaystyle \frac{\pi}{5}5Οβ
- \displaystyle \frac{3\pi}{7}73Οβ
- \displaystyle \frac{\pi}{8}8Οβ
- \displaystyle \frac{9\pi}{7}79Οβ
- \displaystyle \frac{\pi}{7}7Οβ
Q7. What is the area in the plane enclosed by the graph of the function r(\theta) = \cos \theta + \sin \thetar(ΞΈ)=cosΞΈ+sinΞΈ (defined using polar coordinates) for \thetaΞΈ between 00 and 3\pi/43Ο/4?
- \displaystyle \frac{1}{4}41β
- \displaystyle \frac{1 + \sqrt{2}}{2}21+2ββ
- 1 + \sqrt{2}1+2β
- \displaystyle \frac{3\pi}{4}+\frac{1}{2}43Οβ+21β
- \displaystyle \frac{3\pi}{8}+\frac{1}{4}83Οβ+41β
- \piΟ
Q8. Which one of the following integrals computes the surface area of the surface obtained by rotating a quarter-circle
x^2 + y^2 = 4, \qquad x, y \geq 0x2+y2=4,x,yβ₯0
about the line x=-1x=β1?
Hint 1: slice into horizontal strips.
Hint 2: donβt integrate this! (though you could if you had toβ¦)
- \displaystyle \int_{x=0}^2 \sqrt{\frac{4}{4-x^2}} \, dxβ«x=02β4βx24ββdx
- \displaystyle \int_{x=0}^2 2\pi x \sqrt{\frac{4}{4-x^2}} \, dxβ«x=02β2Οx4βx24ββdx
- \displaystyle \int_{x=-1}^2 2\pi x \sqrt{\frac{4}{4-x^2}} \, dxβ«x=β12β2Οx4βx24ββdx
- \displaystyle \int_{x=-1}^1 2\pi x \sqrt{\frac{4}{4-x^2}} \, dxβ«x=β11β2Οx4βx24ββdx
- \displaystyle \int_{x=0}^2 2\pi(x+1)\sqrt{\frac{4}{4-x^2}} \, dxβ«x=02β2Ο(x+1)4βx24ββdx
- \displaystyle \int_{x=0}^2 2\pi(x+1)\sqrt{1+4x^2} \, dxβ«x=02β2Ο(x+1)1+4x2βdx
Q9. Find the arc length of the curve \displaystyle y = \frac{x^2}{4} β \frac{\ln x}{2}y=4x2ββ2lnxβ between x=1x=1 and x=ex=e.
Hint: if you compute the length element correctly, a miraculous simplification should occur, making the integral doable.
- \displaystyle \frac{e^2 + 1}{4}4e2+1β
- \displaystyle \frac{2\pi e}{3}32Οeβ
- \displaystyle \frac{e^2 β 2}{4}4e2β2β
- \displaystyle \frac{e^2 + 2}{4}4e2+2β
- \displaystyle \frac{e^2 β 1}{4}4e2β1β
- \displaystyle \frac{e^2}{4}4e2β
Q10. the present value PVPV of the following income stream I(t)I(t), assuming an continuously-compounding interest rate of 55 per cent (r=0.05r=0.05). The income stream is the following: for the first 1010 years, you get nothing: I(t)=0I(t)=0 for 0\leq t\leq 100β€tβ€10. Then, you get income at a constant rate of ten-thousand (10,\!00010,000) dollars-per-year in perpetuity (that is, you get money at that rate for all future time).
- PV = \displaystyle \frac{200,\!000}{e}PV=e200,000β
- PV = \displaystyle \frac{500}{\sqrt{e}}PV=eβ500β
- PV = \displaystyle 100,\!000 e^2PV=100,000e2
- PV = 200,\!000PV=200,000
- PV = \displaystyle 5,\!000 ePV=5,000e
- PV = \displaystyle \frac{200,\!000}{\sqrt{e}}PV=eβ200,000β
More About This Course
Calculus is one of the greatest things that people have thought of. It helps us understand everything from the orbits of planets to the best size for a city to how often a heart beats. This quick course covers the main ideas of Calculus with one variable, with a focus on understanding the ideas and how to use them. This course is perfect for students who are just starting out in engineering, the physical sciences, or the social sciences.
The course is different because:
1) Taylor series and approximations are introduced and used from the start;
2) a new way of combining discrete and continuous forms of calculus is used;
3) the emphasis is on the ideas rather than the calculations; and
4) the course is taught in a clear, dynamic, and unified way.
In this fourth part, part four of five, we talk about computing areas and volumes, other geometric applications, physical applications, averages and mass, and probability.
Conclusion
Hopefully, this article will be useful for you to find all theΒ Week, final assessment, and Peer Graded Assessment Answers of Calculus: Single Variable Part 4 β Applications Quiz of CourseraΒ and grab some premium knowledge with less effort. If this article really helped you in any way then make sure to share it with your friends on social media and let them also know about this amazing training. You can also check out our other courseΒ Answers.Β So, be with us guys we will share a lot more free courses and their exam/quiz solutions also, and follow ourΒ Techno-RJΒ BlogΒ for more updates.
You have brought up a very superb points, thankyou for the post.
Very interesting information!Perfect just what I was searching for! “The only gift is a portion of thyself.” by Ralph Waldo Emerson.
Thank you for every other excellent post. The place else may just anybody get that kind of info in such an ideal approach of writing? I’ve a presentation subsequent week, and I am on the look for such info.