Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers 2022 | All Weeks Assessment Answers [💯Correct Answer]

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Calculus: Single Variable Part 4 – Applications Quiz Answers

Week 01: Calculus: Single Variable Part 4 – 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=3π ?

\displaystyle \frac{1}{24}241
\displaystyle \frac{23 – 9\sqrt{3}}{24}2423−93
\displaystyle \frac{5}{24}245
\displaystyle \frac{29}{24}2429
\displaystyle \frac{-25 + 15\sqrt{3}}{24}24−25+153
\displaystyle \frac{19}{24}2419

Q2. Find the area of the bounded region enclosed by the curves y = \sqrt{x}y=x and y = x^2y=x2.

Hint: start by drawing the curves in the plane and identifying the appropriate region.

-3−3
\displaystyle \frac{1}{3}31
11
-1−1
-\displaystyle \frac{1}{3}−31
33

Q3. What is the area between the curve y = \sin xy=sinx and the xx-axis for 0\le x\le \pi0≤x≤π ?

\piπ
-1−1
-\pi−π
22
-2−2
11

Q4. What is the area between the curve y = \sin xy=sinx and the xx-axis for 0\le x \le 2\pi0≤x≤2π ?

00
22
-2−2
11
44
-4−4

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 – f(x)\right)\, dxG(f)=area between y=x and y=0area between y=x and y=f=2∫x=01(x−f(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)=52x2+53x3

\displaystyle G(f) = \frac{13}{60}G(f)=6013
\displaystyle G(f) = \frac{13}{30}G(f)=3013
\displaystyle G(f) = \frac{5}{6}G(f)=65
\displaystyle G(f) = \frac{5}{12}G(f)=125
\displaystyle G(f) = \frac{5}{3}G(f)=35
\displaystyle G(f) = \frac{13}{15}G(f)=1513

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 – \frac{3}{2}e−23
\ln 2ln2
e-1e−1
11
e-2e−2
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}−21
\displaystyle \frac{1}{2}21
\ln 2ln2
22
\ln 3ln3
11

Q3. Compute the area in the bounded — that is, finite— regions between y=x(x-1)(x-2)y=x(x−1)(x−2) and the xx-axis.

11
22
\displaystyle \frac{3}{4}43
\displaystyle \frac{1}{2}21
00
\displaystyle \frac{1}{4}41

Q4. Find the area of the sector of a circular disc of radius rr (centered at the origin) given by 1 \leq \theta \leq 31≤θ≤3 (as usual, \thetaθ is in radians).

\displaystyle \frac{\pi r^2}{2}2πr2
\displaystyle \frac{2}{3}r^332r3
2r^22r2
2r2r
2 \pi r^22πr2
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+cosθ

\displaystyle \frac{5\pi}{2}25π
2\pi2π
3\pi3π
\displaystyle \frac{3\pi}{2}23π
\piπ
\displaystyle \frac{\pi}{2}2π

Quiz 03: Core Homework: Simple Volumes

Q1. Find the volume of the following solid: for 1 \le x \lt +\infty1≤x<+∞, the intersection of this solid with the plane perpendicular to the xx-axis is a circular disc of radius e^{-x}e−x. Choose “+\infty+∞” if the resulting integral diverges.

\displaystyle \frac{\pi-e}{3}3π−e
\piπ
1515
+\infty+∞
\displaystyle \frac{e^2}{2}2e2
\displaystyle \frac{\pi}{2e^2}2e2π

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}643
32\sqrt{3}323
16\sqrt{3}163
11
2
2\sqrt{3}23

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}38
44
1616
\displaystyle\frac{16}{3}316
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=π and whose cross-sections perpendicular to the xx-axis are semicircles.

\piπ
\displaystyle \frac{\pi^2}{16}16π2
\pi^2π2
\displaystyle \frac{\pi^2}{4}4π2
00
\displaystyle \frac{\pi^2}{8}8π2

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 \thetaθ of thickness d\thetadθ ?

Hint: if you like, check to see that integrating over 0\le \theta\le 2\pi0≤θ≤2π gives the correct volume of \pi r^2 h / 3 πr2h/3.

dV = \displaystyle \frac{\pi}{3}r^2hdV=3πr2h
dV = 2 r^2h\,d\thetadV=2r2hdθ
dV = \displaystyle \frac{1}{2}r^2\,d\thetadV=21r2dθ
dV = \displaystyle \frac{1}{3}r^2h\,d\thetadV=31r2hdθ
dV = \displaystyle r^2h\,d\thetadV=r2hdθ
dV = \displaystyle \frac{1}{6}r^2h\,d\thetadV=61r2hdθ

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} \pi7128π
4 \pi4π
\displaystyle \frac{64}{7} \pi764π
2\pi2π
None of these
\displaystyle \frac{64}{4} \pi464π

Q2. Let RR be the region between the curve y = -(x-2)^2+1y=−(x−2)2+1 and the xx-axis. Find the volume of the region obtained by revolving RR about the yy-axis.

\displaystyle \frac{32}{5} \pi532π
8 \pi^28π2
\displaystyle \frac{52}{3} \pi352π
\displaystyle \frac{16}{3} \pi316π
\displaystyle \frac{16}{3} \pi^2316π2
\displaystyle \frac{4}{5} \pi54π

Q3. Find the volume obtained by revolving the region between the curves y = x^3y=x3 and y = \sqrt[3]{x}y=3x in the first quadrant about the xx-axis.

\displaystyle \frac{9}{35} \pi359π
\displaystyle \frac{16}{35} \pi3516π
\displaystyle \frac{1}{11} \pi111π
\displaystyle \frac{26}{35} \pi3526π
\displaystyle \frac{8}{35} \pi358π
\displaystyle \frac{32}{35} \pi3532π

Q4. Let DD be the region under the curve y = \ln \sqrt{x}y=lnx 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)π(e−1)
\displaystyle \frac{\pi(e-1)}{2}2π(e−1)
\displaystyle \frac{\pi(e-2)}{4}4π(e−2)
\pi(e-2)π(e−2)
\displaystyle \frac{\pi(e-1)}{4}4π(e−1)
\displaystyle \frac{\pi(e-2)}{2}2π(e−2)

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 \pi24π
\displaystyle \frac{264}{5} \pi5264π
\displaystyle \frac{184}{3} \pi3184π
48 \pi48π
\displaystyle \frac{56}{5} \pi556π
\displaystyle \frac{216}{5} \pi5216π

Q6. Let DD be the region bounded by the graph of y = 1-x^4y=1−x4, 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) \, dx∫x=012π(5−x)(5−x4)dx
\displaystyle \int_{x=0}^1 \pi (1-x^4)^2 \, dx∫x=01π(1−x4)2dx
\displaystyle \int_{y=1}^15 \pi y\sqrt[3]{y-1} \, dy∫y=115πy3y−1dy
\displaystyle \int_{x=0}^1 2\pi x(x^4-5) \, dx∫x=012πx(x4−5)dx
\displaystyle \int_{x=0}^1 2 \pi (5-x)(1-x^4) \, dx∫x=012π(5−x)(1−x4)dx
\displaystyle \int_{x=0}^1 \pi x^2 (1-x^4) \, dx∫x=01πx2(1−x4)dx

Week 02: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers

Quiz 01 :Core Homework: Volume and Dimension

Q1. Consider a four-dimensional box (or “rectangular prism”) 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}241
\displaystyle \frac{1}{12}121
\displaystyle 11
\displaystyle \frac{1}{2}21
\displaystyle \frac{1}{6}61

Q2. In the 44-d box of Question 1, what is the “diameter” —i.e., the farthest distance between two points in the box?

Hint: think in terms of diagonals.

\displaystyle \frac{5}{2\sqrt{3}}235
\displaystyle \frac{1}{2\sqrt{6}}261
\displaystyle \frac{\sqrt{205}}{12}12205
22
\displaystyle \frac{25}{12}1225

Q3. High-dimensional objects are everywhere and all about. Let’s 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. Let’s 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 “space of all images” that my camera can capture, what does the space look like?

Note: there’s no calculus in this problem…just counting!

A unit ball of dimension 3\times 10^63×106
A unit cube of dimension 3\times 10^63×106
A unit simplex of dimension 3\times 10^63×106
A unit cube of dimension 3\times10^73×107
A unit cube of dimension 3\times 10^{10}3×1010

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+95)3/2 from x = 0x=0 to x = 3x=3.

11
88
77
\displaystyle \frac{63}{4}463
\displaystyle \frac{3}{2}23
\displaystyle \frac{21}{2}221

Q2. Find the arc length of the curve y = -\ln (\cos x)y=−ln(cosx) from x = 0x=0 to x = \displaystyle \frac{\pi}{4}x=4π.

Hint: you may need to use that

\int \sec x\, dx = \ln \left|\sec x + \tan x \right| + C∫secxdx=ln∣secx+tanx∣+C

\ln \sqrt2ln2
\ln (\sqrt{2} – 1)ln(2−1)
\ln (\sqrt2 + 1)ln(2+1)
\displaystyle \ln \frac{\sqrt2}{2}ln22
\displaystyle\ln \left( \frac{\sqrt2}{2} + 1 \right)ln(22+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 -1−1 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=t−kcost,y=t−ksint

where k > 0k>0. Denote by L_kLk its arc length as tt moves from 2\pi2π to +\infty+∞. 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 visualizing…

L_kLk is finite for k \gt 1k>1, and infinite for k \leq 1k≤1
\displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{k^2 + t^2}}{t^{k+1}} \, dtLk=∫t=2π+∞tk+1k2+t2dt
\displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{1 + k^2 t^2}}{t^{k+1}} \, dtLk=∫t=2π+∞tk+11+k2t2dt
L_kLk is finite for k \lt 1k<1, and infinite for k \geq 1k≥1
\displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{1 + t^2}}{k t^{k+1}} \, dtLk=∫t=2π+∞ktk+11+t2dt
\displaystyle L_k = \int_{t=2\pi}^{+\infty} \frac{\sqrt{1 + t^2}}{t^{k+1}} \, dtLk=∫t=2π+∞tk+11+t2dt

Q5. At the close of this lecture we saw an example of a fractal — the so-called Koch snowflake. A similar example is given by the following procedure. Starting with a line segment of length 11 (labelled “1” in the figure below), remove the middle third and replace it by a square hat to obtain the curve “2”. Perform the same operation on each line segment in “2” to obtain “3”.

Doing this ad infinitum yields another fractal — 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(35)n
\displaystyle \left( \frac{4}{3} \right)^n(34)n
\displaystyle \left( \frac{3}{4} \right)^n(43)n
\displaystyle \left( \frac{3}{5} \right)^n(53)n
\displaystyle \left( \frac{5}{4} \right)^n(45)n
\displaystyle \left( \frac{4}{5} \right)^n(54)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=1−x2 about the xx-axis. For any -1 \leq a \lt b \leq 1−1≤a<b≤1, calculate the surface area of the slice between x=ax=a and x=bx=b.

\displaystyle 4\pi \sqrt{\frac{b+a}{2}}4π2b+a
2\pi (b^2 + a^2)2π(b2+a2)
\displaystyle 4\pi \sqrt{\frac{b-a}{2}}4π2b−a
2\pi(b-a)2π(b−a)
2\pi (b^2 – a^2)2π(b2−a2)
2\pi(b+a)2π(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 —such 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=4Kx2,0≤x≤R

about the yy-axis. Here RR is the radius of the antenna, and KK —the curvature at the tip— 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}342R3/2K−1/4
\displaystyle \frac{\pi}{K} \left[ \left( 1 + 2RK \right)^{1/2} – 1 \right]Kπ[(1+2RK)1/2−1]
\displaystyle \frac{2\sqrt{2}}{3} R^{1/2} K^{-3/4}322R1/2K−3/4
\displaystyle \frac{2\pi}{3K} \left[ \left( 1 + KR^2 \right)^{3/2} – 1 \right]3K2π[(1+KR2)3/2−1]
\displaystyle \frac{2\pi}{3K} \left[ \left( 1 + 2RK \right)^{3/2} – 1 \right]3K2π[(1+2RK)3/2−1]
\displaystyle \frac{\pi}{K} \left[ \left( 1 + KR^2 \right)^{1/2} – 1 \right]Kπ[(1+KR2)1/2−1]

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=R1+(R2−R1)hy,0≤y≤h

about the yy-axis. Which of the following expressions describes its surface area in terms of the parameters hh, R_1R1 and R_2R2 ?

\displaystyle \frac{\pi}{2} (R_1 + R_2) \sqrt{h^2 + (R_2-R_1)^2}2π(R1+R2)h2+(R2−R1)2
\displaystyle \pi(R_1 + R_2) \left(h^2 + (R_2-R_1)^2\right)^{3/2}π(R1+R2)(h2+(R2−R1)2)3/2
\displaystyle \pi(R_1 + R_2) \sqrt{h^2 + (R_2-R_1)^2}π(R1+R2)h2+(R2−R1)2
\displaystyle \frac{\pi(R_1 + R_2)}{2\sqrt{h^2 + (R_2-R_1)^2}}2h2+(R2−R1)2π(R1+R2)
\displaystyle \frac{\pi(R_1 + R_2)}{\sqrt{h^2 + (R_2-R_1)^2}}h2+(R2−R1)2π(R1+R2)
\displaystyle \frac{\pi}{2} (R_1 + R_2) \left(h^2 + (R_2-R_1)^2\right)^{3/2}2π(R1+R2)(h2+(R2−R1)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 —that is, a hyperbolic cosine. Modeling the roof as the surface of revolution obtained by revolving the curve

y = R \cosh\left( 1 – \frac{x}{R} \right) , \qquad 0 \leq x \leq Ry=Rcosh(1−Rx),0≤x≤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) \, du2π∫u=01ucosh(1−u)du
\displaystyle 2\pi R \int_{u=0}^1 u \cosh(1-u) \, du2πR∫u=01ucosh(1−u)du
\displaystyle 2\pi R^2 \int_{u=0}^1 u \cosh(1-u) \, du2πR2∫u=01ucosh(1−u)du
\displaystyle 2\pi R \int_{u=0}^1 u \sinh(1-u) \, du2πR∫u=01usinh(1−u)du
\displaystyle 2\pi R^2 \int_{u=0}^1 u \sinh(1-u) \, du2πR2∫u=01usinh(1−u)du
\displaystyle 2\pi \int_{u=0}^1 u \sinh(1-u) \, du2π∫u=01usinh(1−u)du

Week 03: Calculus: Single Variable Part 4 – 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). Newton’s 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}1.35⋅106J
1.25\cdot 10^5\,\mathrm{J}1.25⋅105J
2.4 \cdot 10^4\,\mathrm{J}2.4⋅104J
1.2 \cdot 10^6\,\mathrm{J}1.2⋅106J
1.2\cdot 10^5\,\mathrm{J}1.2⋅105J
2.7\cdot 10^5\,\mathrm{J}2.7⋅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 car’s 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+e−x. How much work is done in driving the car from x=0x=0 to x = 3x=3 ?

Hint: remember Newton’s Second Law, F=maF=ma. In our case, both mass and acceleration are functions of xx.

3e^2 – 13e2−1
\displaystyle 3 + \frac{3}{e^3}3+e33
\displaystyle 1 – \frac{2}{e}1−e2
\displaystyle \frac{9}{2}+\frac{2}{e^3}29+e32
\displaystyle \frac{11}{2}-\frac{4}{e^3}211−e34
\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 dam’s height. If the water has weight density \rhoρ, what is the total force the water exerts against the face of the dam?

\displaystyle \frac{1}{4}H^2 W \rho41H2Wρ
\displaystyle \frac{1}{2}H W \rho21HWρ
\displaystyle \frac{1}{4}H W^2 \rho41HW2ρ
H^2 W \rhoH2Wρ
\displaystyle \frac{1}{2}H W^2 \rho21HW2ρ
\displaystyle \frac{1}{2}H^2 W \rho21H2Wρ

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_1I1, starts off slow, then peaks, and then decreases. The second, I_2I2, 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}I1(t)=I0+AsinPπt;I2(t)=I0+AcosPπt

Here, I_0>0I0>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 Prπ>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,+∞) ?

Hints: (1) Which constants are important? I_0I0? 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_2PV1=PV2; but on [0,+\infty)[0,+∞), PV_1\lt PV_2PV1<PV2.
PV_1 \lt PV_2PV1<PV2 both on [0, P][0,P] and [0, +\infty)[0,+∞).
On [0,P][0,P], PV_1=PV_2PV1=PV2; but on [0,+\infty)[0,+∞), PV_1>PV_2PV1>PV2.
On [0,P][0,P], PV_1\lt PV_2PV1<PV2, but on [0,+\infty)[0,+∞), PV_1\gt PV_2PV1>PV2.
PV_1 = PV_2PV1=PV2 both on [0, P][0,P] and [0, +\infty)[0,+∞).
PV_1 \gt PV_2PV1>PV2 both on [0, P][0,P] and [0, +\infty)[0,+∞).

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(T−t)I(t)dt,

assuming a continuous compounding at fixed interest rate rr.

If you save for a child’s college at a rate of \$5,\!000 / \mathrm{year}$5,000/year starting at the child’s 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=$50,000e
FV = \$100,\!000(e-1)FV=$100,000(e−1)
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_0m0, 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}m0=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}τmax=1.5⋅106N⋅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=Q0e−λtdt

Q5. where both Q_0Q0 and \lambdaλ are positive constants. What is the total amount of heat radiated from the moment it is turned off?

\sqrt{\lambda} Q_0λQ0
\lambda Q_0λQ0
Q_0 e^\lambdaQ0eλ
\displaystyle \frac{Q_0}{\lambda}λQ0
\lambda^2 Q_0λ2Q0
Q_0 e^{-\lambda}Q0e−λ

Week 04: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers

Quiz 01:Core Homework: Averagez

Q1. Find the average value of \displaystyle f(x) = \frac{1}{\sqrt{4x – 3}}f(x)=4x−31 from x = 3x=3 to x = 21x=21.

\displaystyle \frac{1}{12}121
\displaystyle -\frac{2}{9}−92
33
\displaystyle \frac{1}{18}181
\displaystyle \frac{3}{2}23
\displaystyle \frac{1}{6}61

Q2. Calculate the average of the function f(x) = x^3 \sqrt{1+x^2}f(x)=x31+x2 over the interval 0 \leq x \leq \sqrt{3}0≤x≤3.

\displaystyle \frac{58}{15}1558
\displaystyle \frac{128}{15}15128
\displaystyle \frac{128}{15\sqrt{3}}153128
\displaystyle \frac{2}{15\sqrt{3}} \left( 1 + \sqrt{2} \right)1532(1+2)
\displaystyle \frac{2}{15} \left( 1 + \sqrt{2} \right)152(1+2)
\displaystyle \frac{58}{15\sqrt{3}}15358

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,0≤t≤2π

\displaystyle \frac{1}{2\pi}2π1
\displaystyle \frac{1}{4}41
\displaystyle \frac{1}{2}21
\displaystyle \frac{2}{\pi}π2
\displaystyle \frac{\pi}{4}4π

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 – \frac{x}{R} \right), \qquad 0 \leq x \leq Ry=h(1−Rx),0≤x≤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}3h
\displaystyle \frac{h}{6}6h
\displaystyle \frac{1}{2} \pi R^2 h21πR2h
\displaystyle \frac{h}{2}2h
\displaystyle \frac{1}{6} \pi R^2 h61πR2h
\displaystyle \frac{1}{3} \pi R^2 h31πR2h

Q5. What is the average of (x-1)^2(x−1)2 over the domain 1\leq \vert x\vert \leq 31≤∣x∣≤3. Be careful!

11
\displaystyle\frac{16}{3}316
\displaystyle\frac{4}{3}34
\displaystyle\frac{32}{3}332
\displaystyle\frac{8}{3}38
88

Quiz 02:Core Homework: Centroids

Q1. Find the coordinates (\overline{x},\overline{y})(x,y) 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}.0≤x≤4π.

1 point

\displaystyle \overline{x}=\frac{\pi\sqrt{2}}{\sqrt{2}-1}x=2−1π2,
\displaystyle \overline{y}=\frac{1}{\sqrt{2}-1}y=2−11
\displaystyle \overline{x}=\frac{\sqrt 2}{2}x=22,
\displaystyle \overline{y}=\frac{\sqrt 2}{2}y=22
\displaystyle \overline{x}=\frac{1}{\sqrt 2 -1}x=2−11,
\displaystyle \overline{y}=\frac{1}{\sqrt 2-1}y=2−11
\displaystyle \overline{x}=\frac{\pi}{8}x=8π,
\displaystyle \overline{y}=\sqrt \frac{2- \sqrt 2}{2}y=22−2
\displaystyle \overline{x}=\frac{\pi\sqrt{2}-4}{4(\sqrt{2}-1)}x=4(2−1)π2−4,
\displaystyle \overline{y}=\frac{1}{4(\sqrt{2}-1)}y=4(2−1)1
\displaystyle \overline{x}=\pi\sqrt 2 x=π2,
\displaystyle \overline{y}=1y=1

Q2. Find the the coordinates (\overline{x},\overline{y})(x,y) of the centroid of the region defined by \displaystyle |x+y| \leq 1∣x+y∣≤1, -1\leq x\leq 1−1≤x≤1, and -1\leq y\leq 1−1≤y≤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,y)=(21,21)
\displaystyle (\overline{x},\overline{y})= \left(\frac{1}{2},\frac{1}{2}\right)(x,y)=(21,21)
(\overline{x},\overline{y})=\displaystyle \left(-\frac{1}{\sqrt2},\frac{1}{\sqrt 2}\right)(x,y)=(−21,21)
(\overline{x},\overline{y})=\displaystyle (0,0)(x,y)=(0,0)
(\overline{x},\overline{y})=\displaystyle (1,1)(x,y)=(1,1)
\displaystyle (\overline{x},\overline{y})= \left(-\frac{1}{\sqrt2}, -\frac{1}{\sqrt 2}\right)(x,y)=(−21,−21)

Q3. Compute the center of mass of a thin rod with density \rho(x)=e^{-ax}ρ(x)=e−ax for a\gt 0a>0 a constant and 0\leq x\lt\infty0≤x<∞. (Yes, i know, it’s not-so-physical to talk about infinite rods…trust me, you will care about this result soon!)

\displaystyle\overline{x} = e^ax=ea
\displaystyle\overline{x} = \frac{1}{a}x=a1
\displaystyle\overline{x} = ax=a
\displaystyle\overline{x} = \frac{1}{a^2}x=a21
\displaystyle\overline{x}x does not exist (the integral diverges).
\displaystyle\overline{x} = 1x=1

Q4. Find the coordinates (\overline{x},\overline{y})(x,y) 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 – 4)^2 + (y – 2)^2 \leq 1D1:x2+y2≤4andD2:(x−4)2+(y−2)2≤1

Hint: replace each disc with a vertex at its centroid. What “mass” should you assign to each vertex?

\displaystyle (\overline{x},\overline{y})=\left( \frac{4\pi}{5}, \frac{2\pi}{5} \right)(x,y)=(54π,52π)
(\overline{x},\overline{y})=(0,0)(x,y)=(0,0)
(\overline{x},\overline{y})=\displaystyle \left( \frac{4}{5}, \frac{2}{5} \right)(x,y)=(54,52)
\displaystyle (\overline{x},\overline{y})=\left( \frac{4}{\pi}, \frac{2}{\pi} \right)(x,y)=(π4,π2)
\displaystyle (\overline{x},\overline{y})= (4\pi,2\pi)(x,y)=(4π,2π)
\displaystyle (\overline{x},\overline{y})= \left(2,1\right)(x,y)=(2,1)

Q5. Find the the coordinates (\overline{x},\overline{y})(x,y) 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=3xρ=3x.

Hint: remember, this is a center-of-mass, not a centroid, so you’ll need to integrate with respect to dM=\rho\cdot dAdM=ρ⋅dA.

(\overline{x},\overline{y})=(1,2)(x,y)=(1,2)
(\overline{x},\overline{y})=\displaystyle \left(\frac{7}{5},\frac{17}{5}\right)(x,y)=(57,517)
(\overline{x},\overline{y})=\displaystyle \left(\frac{7}{5},\frac{17}{10}\right)(x,y)=(57,1017)
(\overline{x},\overline{y})=\displaystyle \left(10, \frac{17}{3}\right)(x,y)=(10,317)
(\overline{x},\overline{y})=\displaystyle \left(\frac{1}{3},\frac{17}{3}\right)(x,y)=(31,317)
(\overline{x},\overline{y})=\displaystyle \left(\frac{14}{3},\frac{17}{3}\right)(x,y)=(314,317)

Quiz 03:Core Homework: Moments and Gyrations

q1. Three particles, each of mass mm, are located at distances r_1r1, r_2r2 and r_3r3 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 doesn’t 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=(r1+r2+r3)ln2
\displaystyle r = \frac{2}{3} ( r_1 + r_2 + r_3 )r=32(r1+r2+r3)
\displaystyle r = \sqrt{r_1^2+r_2^2+r_3^2}r=r12+r22+r32
r = \sqrt[3]{r_1 r_2 r_3}r=3r1r2r3
\displaystyle r = \frac{r_1^2}{r_2}+\frac{r_2^2}{r_3}+\frac{r_3^2}{r_1}r=r2r12+r3r22+r1r32
r = \sqrt[3]{2 \left( r_1^3 + r_2^3 + r_3^3 \right) }r=32(r13+r23+r33)

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)Iannulus=41M(R2−r2)
\displaystyle I_\text{annulus} = M(R-r)\sqrt{R^2-r^2}Iannulus=M(R−r)R2−r2
\displaystyle I_\text{annulus} = \frac{1}{2}M(R^2+r^2)Iannulus=21M(R2+r2)
\displaystyle I_\text{annulus} = \frac{1}{4}M(R^2+r^2)Iannulus=41M(R2+r2)
\displaystyle I_\text{annulus} = \frac{1}{2}M(R^2-r^2)Iannulus=21M(R2−r2)
\displaystyle I_\text{annulus} = \frac{1}{2}MR^2-\frac{1}{4}Mr^2Iannulus=21MR2−41Mr2

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 “caps” 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/Aρ=M/A. Then, setting rr to be a radial coordinate (distance-to-axis), the moment-of-inertia element is dI=\rho r^2 dAdI=ρr2dA. For dAdA, use the approximation implied by the “RR is small” assumption.

\displaystyle \frac{2}{5}M(L^2+\pi R^2)52M(L2+πR2)
\displaystyle \frac{2\pi}{3}{MLR}32πMLR
\displaystyle \frac{1}{4}ML^241ML2
\displaystyle \frac{2}{3}ML^232ML2
\displaystyle \frac{1}{4} MR^241MR2
\displaystyle \frac{1}{3}ML^231ML2

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 \ellℓ. 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 “half-doors” of height hh and width \ell/2ℓ/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
It’s the same
Three times as much
Four-thirds as much

week 05: Calculus: Single Variable Part 4 – 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 —such 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 —HH, HT, TH and TT—, 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}81
\displaystyle \frac{3}{8}83
\displaystyle \frac{1}{4}41
\displaystyle \frac{1}{2}21
\displaystyle \frac{5}{8}85
\displaystyle \frac{7}{8}87

Q2. Let’s 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 3’s turn. If she doesn’t 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}21+241+271+⋯=74
\displaystyle \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots = 121+221+231+⋯=1
\displaystyle \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{1}{2}31+321+331+⋯=21
\displaystyle \frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \cdots = \frac{2}{3}21+231+251+⋯=32
\displaystyle \frac{1}{3}31
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 “volume” fraction over some domain. What is the domain, and what is its dimension?

\displaystyle \frac{1}{4}41
\displaystyle \frac{1}{3}31
11
\displaystyle \frac{2}{3}32
\displaystyle \frac{3}{4}43
\displaystyle \frac{1}{2}21

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 – \frac{\pi r^2}{4}L2−4πr2
\displaystyle \frac{\pi r^2}{L^2}L2πr2
\displaystyle \pi\left(\frac{r}{L}\right)^2 – 1π(Lr)2−1
L^2 – \pi r^2L2−πr2
\displaystyle 1 – \pi\left(\frac{r}{L}\right)^21−π(Lr)2
1 – \displaystyle\frac{\pi r^2}{L}1−Lπr2

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 disc=(2r)2πr2=4π,

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}π2
\displaystyle \frac{\pi}{4}4π
\displaystyle \frac{4}{\pi}π4
\displaystyle \frac{\pi}{2}2π
\displaystyle \frac{\sqrt{2}}{\pi r}πr2
\displaystyle \frac{\pi r}{\sqrt{2}}2πr

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.

ρ(n)={52−51if n evenif n odd on n = 0, 1, \ldots, 9n=0,1,…,9.
\displaystyle \rho(n) = \frac{1}{10}ρ(n)=101 on n = 0, 1, \ldots, 10n=0,1,…,10.
\displaystyle \rho(n) =

⎧⎩⎨150if n evenif n odd

ρ(n)={510if n evenif n odd on n = 0, 1, \ldots, 9n=0,1,…,9.

\displaystyle \rho(n) = \frac{1}{n}ρ(n)=n1 on n = 1, 2, \ldotsn=1,2,…

\displaystyle \rho(n) = \frac{1}{10}ρ(n)=101 on n = 0, 1, \ldots, 9n=0,1,…,9.

\displaystyle \rho(n) =

{10if n=1otherwise

ρ(n)={10if n=1otherwise on n = 1, 2, \ldotsn=1,2,…

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}ρ(x)=101 on [0, 10][0,10]
\displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{1+x^2}ρ(x)=π21+x21 on [0, +\infty)[0,+∞).
\displaystyle \rho(x) = \frac{1}{2\pi} + \sin xρ(x)=2π1+sinx on [0, 2\pi][0,2π]
\displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{1+x^2}ρ(x)=π21+x21 on \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞).
\displaystyle \rho(x) = \frac{1}{x^2}ρ(x)=x21 on [1, +\infty)[1,+∞).
\displaystyle \rho(x) = \frac{1}{10}ρ(x)=101 on [0, 9][0,9]

Q3. For which value of \lambdaλ is \rho(x) = \lambda x^2 e^{-x}ρ(x)=λx2e−x a probability density function on [0, +\infty)[0,+∞) ?

\rho(x)ρ(x) is not a probability density function for any value of \lambdaλ
\displaystyle \lambda = \frac{e}{2}λ=2e
\displaystyle \lambda = \frac{1}{e}λ=e1
\lambda = 1λ=1
\displaystyle \lambda = \frac{1}{2}λ=21
\lambda = 2λ=2

Q4. The amount of time between failures of a printer follows an exponential probability distribution —that 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} \, dt∫t=0Tαe−αtdt

for \alpha = 0.01\ln 2\,\,\, \mathrm{h}^{-1}α=0.01ln2h−1 (notice that \alphaα 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 – e^{-2}1−e−2
\displaystyle \frac{1}{4}41
\displaystyle \frac{1}{2}21
\displaystyle \frac{3}{4}43
e^{-1/2}e−1/2
e^{-2}e−2

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^nρ(x)=(n+1)xn (nn a positive integer) on [0, 1][0,1].

\displaystyle \mathbb{E} = \frac{n+1}{n+2}E=n+2n+1, \displaystyle \mathbb{V} = \frac{n+1}{n+3} – \left( \frac{n+1}{n+2} \right)^2V=n+3n+1−(n+2n+1)2.
\mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+1}{n+2}V=n+2n+1.
\mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+1}{n+2} – 1V=n+2n+1−1.
\mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+2}{n+3}V=n+3n+2.
\displaystyle \mathbb{E} = \frac{n+1}{n+2}E=n+2n+1, \displaystyle \mathbb{V} = \frac{n+2}{n+3}V=n+3n+2.
\displaystyle \mathbb{E} = \frac{n+1}{n+2}E=n+2n+1, \displaystyle \mathbb{V} = \frac{n+2}{n+3} – \left( \frac{n+1}{n+2} \right)^2V=n+3n+2−(n+2n+1)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}ρ(x)=π2x2+11 on [0, +\infty)[0,+∞).

Hint: notice that the integrals calculating the expectation and variance are improper because [0, +\infty)[0,+∞) 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=π2, but \mathbb{V}V diverges.
\mathbb{E} = 1E=1, \mathbb{V} = 1V=1.
\displaystyle \mathbb{E} = \frac{2}{\pi}E=π2, \displaystyle \mathbb{V} = \frac{2}{\pi} – \frac{4}{\pi^2}V=π2−π24.
\displaystyle \mathbb{E} = \frac{2}{\pi}E=π2, \displaystyle \mathbb{V} = \frac{4}{\pi^2}V=π24.
\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}ρ(n)=41 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=27.
\mathbb{E} = 2E=2, \displaystyle \mathbb{V} = \frac{5}{4}V=45.
\displaystyle \mathbb{E} = \frac{5}{2}E=25, \displaystyle \mathbb{V} = \frac{7}{2}V=27.
\displaystyle \mathbb{E} = \frac{5}{2}E=25, \displaystyle \mathbb{V} = \frac{5}{4}V=45.
\displaystyle \mathbb{E} = \frac{5}{2}E=25, \displaystyle \mathbb{V} = \frac{15}{4}V=415.
\mathbb{E} = 2E=2, \displaystyle \mathbb{V} = \frac{1}{2}V=21.

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 —that 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)dx∫amρ(x)dx=21=∫mbρ(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)ρ(x)=αe−αxon[0,+∞)

m = \alpham=α
\displaystyle m = \alpha \ln 2m=αln2
\displaystyle m = \frac{\alpha}{\ln 2}m=ln2α
\displaystyle m = \frac{1}{\alpha \ln 2}m=αln21
\displaystyle m = \frac{\ln 2}{\alpha}m=αln2
\displaystyle m = \frac{1}{\alpha}m=α1

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 —the quartiles:

the first (or lower) quartile is the unique value Q_1Q1 for which the probability of x \lt Q_1x<Q1 is 1/41/4;
the second quartile (really the median) is the unique value Q_2Q2 for which the probability of x \lt Q_2x<Q2 is 2/42/4;
the third (or upper) quartile is the unique value Q_3Q3 for which the probability of x \lt Q_3x<Q3 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}ρ(x)=αe−αx.

\displaystyle Q_1 = \frac{1}{\alpha \ln 4}Q1=αln41, \displaystyle Q_3 = \frac{1}{\alpha \ln (4/3)}Q3=αln(4/3)1.
\displaystyle Q_1 = \frac{\alpha}{\ln 4}Q1=ln4α, \displaystyle Q_3 = \frac{\alpha}{\ln (4/3)}Q3=ln(4/3)α.
\displaystyle Q_1 = \frac{\ln (4/3)}{\alpha}Q1=αln(4/3), \displaystyle Q_3 = \frac{\ln 4}{\alpha}Q3=αln4.
\displaystyle Q_1 = \frac{\alpha}{\ln (4/3)}Q1=ln(4/3)α, \displaystyle Q_3 = \frac{\alpha}{\ln 4}Q3=ln4α.
\displaystyle Q_1 = \frac{1}{\alpha \ln (4/3)}Q1=αln(4/3)1, \displaystyle Q_3 = \frac{1}{\alpha \ln 4}Q3=αln41.
\displaystyle Q_1 = \frac{\ln 4}{\alpha}Q1=αln4, \displaystyle Q_3 = \frac{\ln (4/3)}{\alpha}Q3=αln(4/3).

Quiz 04:Chapter 4: Applications – Exam

Q1. Compute the expectation \mathbb{E}E of xx with the probability density function

\rho(x) = \frac{3}{2}\sqrt{x}ρ(x)=23x

on 0 \leq x \leq 10≤x≤1.

\displaystyle \frac{5}{3}35
\displaystyle \frac{2}{3}32
\displaystyle \frac{1}{2}21
\displaystyle \frac{2}{5}52
\displaystyle \frac{4}{15}154
\displaystyle \frac{3}{5}53

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 volume…

\displaystyle\frac{80}{3}\pi380π
9\pi9π
\displaystyle\frac{13}{3}\pi313π
\displaystyle\frac{26}{3}\pi326π
\displaystyle\frac{80}{9}\pi980π
\displaystyle\frac{40}{3}\pi340π

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 – \frac{x^2}{25}y=1−25x2

\displaystyle \frac{2}{5}52
\displaystyle \frac{8}{3}38
00
\displaystyle \frac{4}{5}54
\displaystyle \frac{4}{3}34
\displaystyle \frac{8}{5}58

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}W31W
\displaystyle\frac{2}{\sqrt{2}}W22W
\displaystyle\frac{1}{8}W81W
There is not enough information to answer this question.
\displaystyle\frac{1}{\sqrt{2}}W21W
\displaystyle\frac{1}{4}W41W

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=21MR2
I = \displaystyle \frac{2}{3}MR^2I=32MR2
I = \displaystyle \frac{1}{2}MR^2hI=21MR2h
I = \displaystyle \frac{2}{3}MRhI=32MRh
I = \displaystyle \frac{1}{3}MRhI=31MRh
I = \displaystyle \frac{1}{2}MRhI=21MRh

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 problem…

\displaystyle \frac{3\pi}{5}53π
\displaystyle \frac{\pi}{5}5π
\displaystyle \frac{3\pi}{7}73π
\displaystyle \frac{\pi}{8}8π
\displaystyle \frac{9\pi}{7}79π
\displaystyle \frac{\pi}{7}7π

Q7. What is the area in the plane enclosed by the graph of the function r(\theta) = \cos \theta + \sin \thetar(θ)=cosθ+sinθ (defined using polar coordinates) for \thetaθ between 00 and 3\pi/43π/4?

\displaystyle \frac{1}{4}41
\displaystyle \frac{1 + \sqrt{2}}{2}21+2
1 + \sqrt{2}1+2
\displaystyle \frac{3\pi}{4}+\frac{1}{2}43π+21
\displaystyle \frac{3\pi}{8}+\frac{1}{4}83π+41
\piπ

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,y≥0

about the line x=-1x=−1?

Hint 1: slice into horizontal strips.

Hint 2: don’t integrate this! (though you could if you had to…)

\displaystyle \int_{x=0}^2 \sqrt{\frac{4}{4-x^2}} \, dx∫x=024−x24dx
\displaystyle \int_{x=0}^2 2\pi x \sqrt{\frac{4}{4-x^2}} \, dx∫x=022πx4−x24dx
\displaystyle \int_{x=-1}^2 2\pi x \sqrt{\frac{4}{4-x^2}} \, dx∫x=−122πx4−x24dx
\displaystyle \int_{x=-1}^1 2\pi x \sqrt{\frac{4}{4-x^2}} \, dx∫x=−112πx4−x24dx
\displaystyle \int_{x=0}^2 2\pi(x+1)\sqrt{\frac{4}{4-x^2}} \, dx∫x=022π(x+1)4−x24dx
\displaystyle \int_{x=0}^2 2\pi(x+1)\sqrt{1+4x^2} \, dx∫x=022π(x+1)1+4x2dx

Q9. Find the arc length of the curve \displaystyle y = \frac{x^2}{4} – \frac{\ln x}{2}y=4x2−2lnx 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+1
\displaystyle \frac{2\pi e}{3}32πe
\displaystyle \frac{e^2 – 2}{4}4e2−2
\displaystyle \frac{e^2 + 2}{4}4e2+2
\displaystyle \frac{e^2 – 1}{4}4e2−1
\displaystyle \frac{e^2}{4}4e2

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 100≤t≤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=e200,000
PV = \displaystyle \frac{500}{\sqrt{e}}PV=e500
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=e200,000

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.

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