- About The Coursera
- About Calculus: Single Variable Part 4 – Applications Course
- Calculus: Single Variable Part 4 – Applications Quiz Answers
- Week 01: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers
- Week 02: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers
- Week 03: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers
- Week 04: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers
- week 05: Calculus: Single Variable Part 4 – Applications Coursera Quiz Answers

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**About Calculus: Single Variable Part 4 – 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 – Applications

**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 x*f*(*x*)=sin3*x* and the x*x*-axis from x=0*x*=0 to \displaystyle x=\frac{\pi}{3}*x*=3*π* ?

- \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^2*y*=*x*2.

**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 x*y*=sin*x* and the x*x*-axis for 0\le x\le \pi0≤*x*≤*π* ?

- \pi
*π* - -1−1
- -\pi−
*π* - 22
- -2−2
- 11

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

- 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)\, dx*G*(*f*)=area between *y*=*x* and *y*=0area between *y*=*x* and *y*=*f*=2∫*x*=01(*x*−*f*(*x*))*d**x*

where f(x)*f*(*x*) is the fraction of total income earned by the lowest x*x* fraction of the populace.

Calculate the Gini index of a country where

f(x) = \frac{2}{5} x^2 + \frac{3}{5} x^3*f*(*x*)=52*x*2+53*x*3

- \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 = 1*y*=1, x = 1*x*=1, and y = \ln x*y*=ln*x*.

**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-1
*e*−1 - 11
- e-2
*e*−2 - e
*e*

Q2 .Find the area of the bounded region enclosed by the x*x*-axis, the lines x=1*x*=1 and x=2*x*=2 and the hyperbola xy = 1*xy*=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 x*x*-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 r*r* (centered at the origin) given by 1 \leq \theta \leq 31≤*θ*≤3 (as usual, \theta*θ* is in radians).

- \displaystyle \frac{\pi r^2}{2}2
*πr*2 - \displaystyle \frac{2}{3}r^332
*r*3 - 2r^22
*r*2 - 2r2
*r* - 2 \pi r^22
*πr*2 - r^2
*r*2

Q5. Compute the area enclosed by the *cardioid* in the figure below. This curve is described by the polar equation r = 1 + \cos\theta*r*=1+cos*θ*

.

- \displaystyle \frac{5\pi}{2}25
*π* - 2\pi2
*π* - 3\pi3
*π* - \displaystyle \frac{3\pi}{2}23
*π* - \pi
*π* - \displaystyle \frac{\pi}{2}2
*π*

#### 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 x*x*-axis is a circular disc of radius e^{-x}*e*−*x*. Choose “+\infty+∞” if the resulting integral diverges.

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

Q2. The base of a solid is given by the region lying between the y*y*-axis, the parabola y=x^2*y*=*x*2, and the line y=16*y*=16 in the first quadrant. Its cross-sections perpendicular to the y*y*-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 y*y*-axis, the parabola y=x^2*y*=*x*2, and the line y=4*y*=4. Its cross-sections perpendicular to the y*y*-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 x*y*=sin*x* and the x*x*-axis from x=0*x*=0 to x=\pi*x*=*π* and whose cross-sections perpendicular to the x*x*-axis are semicircles.

- \pi
*π* - \displaystyle \frac{\pi^2}{16}16
*π*2 - \pi^2
*π*2 - \displaystyle \frac{\pi^2}{4}4
*π*2 - 00
- \displaystyle \frac{\pi^2}{8}8
*π*2

Q5. Consider a cone of height h*h* over a circular base of radius r*r*. 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 \theta*θ* of thickness d\theta*d**θ* ?

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

- dV = \displaystyle \frac{\pi}{3}r^2h
*dV*=3*π**r*2*h* - dV = 2 r^2h\,d\theta
*dV*=2*r*2*hdθ* - dV = \displaystyle \frac{1}{2}r^2\,d\theta
*dV*=21*r*2*dθ* - dV = \displaystyle \frac{1}{3}r^2h\,d\theta
*dV*=31*r*2*hdθ* - dV = \displaystyle r^2h\,d\theta
*dV*=*r*2*hdθ* - dV = \displaystyle \frac{1}{6}r^2h\,d\theta
*dV*=61*r*2*hdθ*

#### Quiz 04:Core Homework: Complex Volumes

Q1. Let D*D* be the region bounded by the curve y = x^3*y*=*x*3, the x*x*-axis, the line x = 0*x*=0 and the line x = 2*x*=2. Find the volume of the region obtained by revolving D*D* about the x*x*-axis.

**\displaystyle \frac{128}{7} \pi7128***π*- 4 \pi4
*π* - \displaystyle \frac{64}{7} \pi764
*π* - 2\pi2
*π* - None of these
- \displaystyle \frac{64}{4} \pi464
*π*

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

- \displaystyle \frac{32}{5} \pi532
*π* - 8 \pi^28
*π*2 - \displaystyle \frac{52}{3} \pi352
*π* - \displaystyle \frac{16}{3} \pi316
*π* - \displaystyle \frac{16}{3} \pi^2316
*π*2 - \displaystyle \frac{4}{5} \pi54
*π*

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

- \displaystyle \frac{9}{35} \pi359
*π* - \displaystyle \frac{16}{35} \pi3516
*π* - \displaystyle \frac{1}{11} \pi111
*π* - \displaystyle \frac{26}{35} \pi3526
*π* - \displaystyle \frac{8}{35} \pi358
*π* - \displaystyle \frac{32}{35} \pi3532
*π*

Q4. Let D*D* be the region under the curve y = \ln \sqrt{x}*y*=ln*x* and above the x*x*-axis from x = 1*x*=1 to x = e*x*=*e*. Find the volume of the region obtained by revolving D*D* about the x*x*-axis.

- \pi(e-1)
*π*(*e*−1) - \displaystyle \frac{\pi(e-1)}{2}2
*π*(*e*−1) - \displaystyle \frac{\pi(e-2)}{4}4
*π*(*e*−2) - \pi(e-2)
*π*(*e*−2) - \displaystyle \frac{\pi(e-1)}{4}4
*π*(*e*−1) - \displaystyle \frac{\pi(e-2)}{2}2
*π*(*e*−2)

Q5. Let D*D* be the region from Question 1. What is the volume of the region formed by rotating D*D* about the line x = 3*x*=3?

- 24 \pi24
*π* - \displaystyle \frac{264}{5} \pi5264
*π* - \displaystyle \frac{184}{3} \pi3184
*π* - 48 \pi48
*π* - \displaystyle \frac{56}{5} \pi556
*π* - \displaystyle \frac{216}{5} \pi5216
*π*

Q6. Let D*D* be the region bounded by the graph of y = 1-x^4*y*=1−*x*4, the x*x*-axis and the y*y*-axis in the first quadrant. Which of the following integrals can be used to compute the volume of the region obtained by revolving D*D* around the line x=5*x*=5?

- \displaystyle\int_{x=0}^1 2\pi (5-x)(5-x^4) \, dx∫
*x*=012*π*(5−*x*)(5−*x*4)*dx* - \displaystyle \int_{x=0}^1 \pi (1-x^4)^2 \, dx∫
*x*=01*π*(1−*x*4)2*dx* - \displaystyle \int_{y=1}^15 \pi y\sqrt[3]{y-1} \, dy∫
*y*=115*πy*3*y*−1*dy* - \displaystyle \int_{x=0}^1 2\pi x(x^4-5) \, dx∫
*x*=012*πx*(*x*4−5)*dx* - \displaystyle \int_{x=0}^1 2 \pi (5-x)(1-x^4) \, dx∫
*x*=012*π*(5−*x*)(1−*x*4)*dx* - \displaystyle \int_{x=0}^1 \pi x^2 (1-x^4) \, dx∫
*x*=01*πx*2(1−*x*4)*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 = 0*x*=0 to x = 3*x*=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(cos*x*) from x = 0*x*=0 to x = \displaystyle \frac{\pi}{4}*x*=4*π*.

**Hint:** you may need to use that

\int \sec x\, dx = \ln \left|\sec x + \tan x \right| + C∫sec*xdx*=ln∣sec*x*+tan*x*∣+*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^2*x*=*t*3,*y*=*t*2

Compute the arc length of this curve as t*t* 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 t*x*=*t*−*k*cos*t*,*y*=*t*−*k*sin*t*

where k > 0*k*>0. Denote by L_k*L**k* its arc length as t*t* moves from 2\pi2*π* to +\infty+∞. Which of the following statements are true? Select all that apply.

**Hint:** in Lecture we studied the case k=1*k*=1: see the figure from the Lecture if you need help visualizing…

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

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 n*n* 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−*x*2 about the x*x*-axis. For any -1 \leq a \lt b \leq 1−1≤*a*<*b*≤1, calculate the surface area of the slice between x=a*x*=*a* and x=b*x*=*b*.

- \displaystyle 4\pi \sqrt{\frac{b+a}{2}}4
*π*2*b*+*a* - 2\pi (b^2 + a^2)2
*π*(*b*2+*a*2) - \displaystyle 4\pi \sqrt{\frac{b-a}{2}}4
*π*2*b*−*a* - 2\pi(b-a)2
*π*(*b*−*a*) - 2\pi (b^2 – a^2)2
*π*(*b*2−*a*2) - 2\pi(b+a)2
*π*(*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 R*y*=4*K**x*2,0≤*x*≤*R*

about the y*y*-axis. Here R*R* is the radius of the antenna, and K*K* —the curvature at the tip— controls how flat it is. Compute the surface area of this antenna in terms of the parameters R*R* and K*K*.

**1 point**

- \displaystyle \frac{4\sqrt{2}}{3} R^{3/2}K^{-1/4}342
*R*3/2*K*−1/4 - \displaystyle \frac{\pi}{K} \left[ \left( 1 + 2RK \right)^{1/2} – 1 \right]
*Kπ*[(1+2*RK*)1/2−1] - \displaystyle \frac{2\sqrt{2}}{3} R^{1/2} K^{-3/4}322
*R*1/2*K*−3/4 - \displaystyle \frac{2\pi}{3K} \left[ \left( 1 + KR^2 \right)^{3/2} – 1 \right]3
*K*2*π*[(1+*KR*2)3/2−1] - \displaystyle \frac{2\pi}{3K} \left[ \left( 1 + 2RK \right)^{3/2} – 1 \right]3
*K*2*π*[(1+2*RK*)3/2−1] - \displaystyle \frac{\pi}{K} \left[ \left( 1 + KR^2 \right)^{1/2} – 1 \right]
*Kπ*[(1+*KR*2)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 h*x*=*R*1+(*R*2−*R*1)*h**y*,0≤*y*≤*h*

about the y*y*-axis. Which of the following expressions describes its surface area in terms of the parameters h*h*, R_1*R*1 and R_2*R*2 ?

- \displaystyle \frac{\pi}{2} (R_1 + R_2) \sqrt{h^2 + (R_2-R_1)^2}2
*π*(*R*1+*R*2)*h*2+(*R*2−*R*1)2 - \displaystyle \pi(R_1 + R_2) \left(h^2 + (R_2-R_1)^2\right)^{3/2}
*π*(*R*1+*R*2)(*h*2+(*R*2−*R*1)2)3/2 **\displaystyle \pi(R_1 + R_2) \sqrt{h^2 + (R_2-R_1)^2}***π*(*R*1+*R*2)*h*2+(*R*2−*R*1)2- \displaystyle \frac{\pi(R_1 + R_2)}{2\sqrt{h^2 + (R_2-R_1)^2}}2
*h*2+(*R*2−*R*1)2*π*(*R*1+*R*2) - \displaystyle \frac{\pi(R_1 + R_2)}{\sqrt{h^2 + (R_2-R_1)^2}}
*h*2+(*R*2−*R*1)2*π*(*R*1+*R*2) - \displaystyle \frac{\pi}{2} (R_1 + R_2) \left(h^2 + (R_2-R_1)^2\right)^{3/2}2
*π*(*R*1+*R*2)(*h*2+(*R*2−*R*1)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 R*y*=*R*cosh(1−*R**x*),0≤*x*≤*R*

around the y*y*-axis, which of the following integrals computes its surface area?

- \displaystyle 2\pi \int_{u=0}^1 u \cosh(1-u) \, du2
*π*∫*u*=01*u*cosh(1−*u*)*du* - \displaystyle 2\pi R \int_{u=0}^1 u \cosh(1-u) \, du2
*πR*∫*u*=01*u*cosh(1−*u*)*du* - \displaystyle 2\pi R^2 \int_{u=0}^1 u \cosh(1-u) \, du2
*πR*2∫*u*=01*u*cosh(1−*u*)*du* - \displaystyle 2\pi R \int_{u=0}^1 u \sinh(1-u) \, du2
*πR*∫*u*=01*u*sinh(1−*u*)*du* - \displaystyle 2\pi R^2 \int_{u=0}^1 u \sinh(1-u) \, du2
*πR*2∫*u*=01*u*sinh(1−*u*)*du* - \displaystyle 2\pi \int_{u=0}^1 u \sinh(1-u) \, du2
*π*∫*u*=01*u*sinh(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}^2*g*=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 = ma*F*=*m**a*, 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}^2*g*=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}^2*g*=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 a*a* increases linearly with its position x*x* as a(x) = x*a*(*x*)=*x*. Since the car is burning fuel, its mass m*m* decreases; assume the decrease is exponential in x*x* as m(x) = 1 + e^{-x}*m*(*x*)=1+*e*−*x*. How much work is done in driving the car from x=0*x*=0 to x = 3*x*=3 ?

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

- 3e^2 – 13
*e*2−1 - \displaystyle 3 + \frac{3}{e^3}3+
*e*33 - \displaystyle 1 – \frac{2}{e}1−
*e*2 **\displaystyle \frac{9}{2}+\frac{2}{e^3}29+***e*32- \displaystyle \frac{11}{2}-\frac{4}{e^3}211−
*e*34 - \displaystyle \frac{2}{e

#### Quiz 02:Core Homework: Elements

Q1. Consider a dam of height H*H* and width W*W* 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 \rho*ρ*, what is the total force the water exerts against the face of the dam?

- \displaystyle \frac{1}{4}H^2 W \rho41
*H*2*Wρ* - \displaystyle \frac{1}{2}H W \rho21
*HWρ* - \displaystyle \frac{1}{4}H W^2 \rho41
*HW*2*ρ* - H^2 W \rho
*H*2*Wρ* - \displaystyle \frac{1}{2}H W^2 \rho21
*HW*2*ρ* - \displaystyle \frac{1}{2}H^2 W \rho21
*H*2*Wρ*

Q2. Consider two potential income streams, each valued based on an assumption of a constant return on investment at rate r>0*r*>0. The first, I_1*I*1, starts off slow, then peaks, and then decreases. The second, I_2*I*2, 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}*I*1(*t*)=*I*0+*A*sin*P**π**t*;*I*2(*t*)=*I*0+*A*cos*P**π**t*

Here, I_0>0*I*0>0 is a constant (the baseline income), A>0*A*>0 is a constant (the amplitude of fluctuation) and P>0*P*>0 is a constant (the half-period). Assume \pi \gt Pr*π*>*P**r*. 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_0*I*0? A*A*? P*P*? r*r*? (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_2*PV*1=*PV*2; but on [0,+\infty)[0,+∞), PV_1\lt PV_2*PV*1<*PV*2. **PV_1 \lt PV_2***PV*1<*PV*2 both on [0, P][0,*P*] and [0, +\infty)[0,+∞).- On [0,P][0,
*P*], PV_1=PV_2*PV*1=*PV*2; but on [0,+\infty)[0,+∞), PV_1>PV_2*PV*1>*PV*2. - On [0,P][0,
*P*], PV_1\lt PV_2*PV*1<*PV*2, but on [0,+\infty)[0,+∞), PV_1\gt PV_2*PV*1>*PV*2. - PV_1 = PV_2
*PV*1=*PV*2 both on [0, P][0,*P*] and [0, +\infty)[0,+∞). - PV_1 \gt PV_2
*PV*1>*PV*2 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=T*t*=*T*. The future value element of I(t)*I*(*t*) is

dFV = e^{r(T-t)}I(t)dt,*d**F**V*=*e**r*(*T*−*t*)*I*(*t*)*d**t*,

assuming a continuous compounding at fixed interest rate r*r*.

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,\!000e
*FV*=$100,000*e* **FV = \$50,\!000 e***FV*=$50,000*e*- FV = \$50,\!000\sqrt{e}
*FV*=$50,000*e* - FV = \$100,\!000(e-1)
*FV*=$100,000(*e*−1) - FV = \$500,\!000
*FV*=$500,000 - FV = \$100,\!000
*FV*=$100,000

Q4. Consider a cantilever beam of length L*L*. Suppose that N*N* people, each of mass m_0*m*0, 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}*m*0=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}*τ*max=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}^2*g*=10m/s2.

- 25 people.
- 50 people.
- 400 people.
- 300 people.
- 200 people.
- 100 people.Suppose that a radiator is turned off at t=0
*t*=0; after that, the amount of heat generated by the radiator is described by the*heat flow element*

dQ = Q_0 e^{-\lambda t} dt*d**Q*=*Q*0*e*−*λ**t**d**t*

Q5. where both Q_0*Q*0 and \lambda*λ* are positive constants. What is the total amount of heat radiated from the moment it is turned off?

- \sqrt{\lambda} Q_0
*λ**Q*0 - \lambda Q_0
*λQ*0 - Q_0 e^\lambda
*Q*0*eλ* - \displaystyle \frac{Q_0}{\lambda}
*λQ*0 - \lambda^2 Q_0
*λ*2*Q*0 - Q_0 e^{-\lambda}
*Q*0*e*−*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*)=4*x*−31 from x = 3*x*=3 to x = 21*x*=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*)=*x*31+*x*2 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 x*x* over a circle of radius 11 (given by the equation x^2 + y^2 = 1*x*2+*y*2=1) is zero. But what is the average value of x^2*x*2 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 dL*d**L*. In order to make your calculations easier, use the parametrization

x = \cos t, \quad y = \sin t, \qquad 0 \leq t \leq 2\pi*x*=cos*t*,*y*=sin*t*,0≤*t*≤2*π*

- 00

- \displaystyle \frac{1}{2\pi}2
*π*1 - \displaystyle \frac{1}{4}41
- \displaystyle \frac{1}{2}21
- \displaystyle \frac{2}{\pi}
*π*2 - \displaystyle \frac{\pi}{4}4
*π*

Q4. Let us model a mountain as a circular cone of height h*h* whose base has radius R*R*. 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 R*y*=*h*(1−*R**x*),0≤*x*≤*R*

about the y*y*-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}3***h*- \displaystyle \frac{h}{6}6
*h* - \displaystyle \frac{1}{2} \pi R^2 h21
*πR*2*h* - \displaystyle \frac{h}{2}2
*h* - \displaystyle \frac{1}{6} \pi R^2 h61
*πR*2*h* - \displaystyle \frac{1}{3} \pi R^2 h31
*πR*2*h*

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 x*y*=sin*x* and y=\cos x*y*=cos*x* for \displaystyle 0 \leq x \leq \frac{\pi}{4}.0≤*x*≤4*π*.

**1 point**

- \displaystyle \overline{x}=\frac{\pi\sqrt{2}}{\sqrt{2}-1}
*x*=2−1*π*2, - \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*=8*π*, - \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)*π*2−4, - \displaystyle \overline{y}=\frac{1}{4(\sqrt{2}-1)}
*y*=4(2−1)1 - \displaystyle \overline{x}=\pi\sqrt 2
*x*=*π*2, - \displaystyle \overline{y}=1
*y*=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 0*a*>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^a
*x*=*ea* - \displaystyle\overline{x} = \frac{1}{a}
*x*=*a*1 **\displaystyle\overline{x} = a***x*=*a*- \displaystyle\overline{x} = \frac{1}{a^2}
*x*=*a*21 - \displaystyle\overline{x}
*x*does not exist (the integral diverges). - \displaystyle\overline{x} = 1
*x*=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 1*D*1:*x*2+*y*2≤4and*D*2:(*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*)=(54*π*,52*π*) - (\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*)=(*π*4,*π*2) - \displaystyle (\overline{x},\overline{y})= (4\pi,2\pi)(
*x*,*y*)=(4*π*,2*π*) - \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 x*x*-axis, the y*y*-axis, and the lines x=2*x*=2 and \displaystyle y=x+2*y*=*x*+2, with density (mass-per-unit-area) \rho=3x*ρ*=3*x*.

**Hint:** remember, this is a center-of-mass, not a centroid, so you’ll need to integrate with respect to dM=\rho\cdot dA*dM*=*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 m*m*, are located at distances r_1*r*1, r_2*r*2 and r_3*r*3 respectively from a fixed axis of rotation A*A*. We now place a fourth particle, also of mass m*m*, at some distance r*r* from the axis A*A*. If the moment of inertia of all four particles is twice as big as the moment of inertia of the first three, what is r*r* ?

**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 2
*r*=(*r*1+*r*2+*r*3)ln2 - \displaystyle r = \frac{2}{3} ( r_1 + r_2 + r_3 )
*r*=32(*r*1+*r*2+*r*3) - \displaystyle r = \sqrt{r_1^2+r_2^2+r_3^2}
*r*=*r*12+*r*22+*r*32 - r = \sqrt[3]{r_1 r_2 r_3}
*r*=3*r*1*r*2*r*3 - \displaystyle r = \frac{r_1^2}{r_2}+\frac{r_2^2}{r_3}+\frac{r_3^2}{r_1}
*r*=*r*2*r*12+*r*3*r*22+*r*1*r*32 - r = \sqrt[3]{2 \left( r_1^3 + r_2^3 + r_3^3 \right) }
*r*=32(*r*13+*r*23+*r*33)

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 R*R*, inner radius r*r*, and mass M*M* 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)
*I*annulus=41*M*(*R*2−*r*2) - \displaystyle I_\text{annulus} = M(R-r)\sqrt{R^2-r^2}
*I*annulus=*M*(*R*−*r*)*R*2−*r*2 - \displaystyle I_\text{annulus} = \frac{1}{2}M(R^2+r^2)
*I*annulus=21*M*(*R*2+*r*2) - \displaystyle I_\text{annulus} = \frac{1}{4}M(R^2+r^2)
*I*annulus=41*M*(*R*2+*r*2) - \displaystyle I_\text{annulus} = \frac{1}{2}M(R^2-r^2)
*I*annulus=21*M*(*R*2−*r*2) - \displaystyle I_\text{annulus} = \frac{1}{2}MR^2-\frac{1}{4}Mr^2
*I*annulus=21*MR*2−41*Mr*2

Q3. A hollow cylindrical shell of length L*L* and radius R*R* 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 M*M* is distributed uniformly along the surface.You may also assume that R*R* is small enough that the piece of this cylinder at any distance r*r* from the axis of rotation is a circle. What is its moment of inertia?

**HInt:** start by computing the area A*A* and then the density \rho=M/A*ρ*=*M*/*A*. Then, setting r*r* to be a radial coordinate (distance-to-axis), the moment-of-inertia element is dI=\rho r^2 dA*dI*=*ρr*2*dA*. For dA*dA*, use the approximation implied by the “R*R* is small” assumption.

- \displaystyle \frac{2}{5}M(L^2+\pi R^2)52
*M*(*L*2+*πR*2) **\displaystyle \frac{2\pi}{3}{MLR}32***π**MLR*- \displaystyle \frac{1}{4}ML^241
*ML*2 - \displaystyle \frac{2}{3}ML^232
*ML*2 - \displaystyle \frac{1}{4} MR^241
*MR*2 - \displaystyle \frac{1}{3}ML^231
*ML*2

Q4. You need to install a heavy front door in your home. For simplicity, assume that the door has uniform density, has total mass M*M*, and fills a rectangular entry of height h*h* 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 h
*h*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 I*I* of the single door than the (net) I*I* 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 L*L* is more than a distance r*r* away from every corner? Suppose r \lt L/2*r*<*L*/2.

- \displaystyle L^2 – \frac{\pi r^2}{4}
*L*2−4*πr*2 - \displaystyle \frac{\pi r^2}{L^2}
*L*2*πr*2 **\displaystyle \pi\left(\frac{r}{L}\right)^2 – 1***π*(*Lr*)2−1- L^2 – \pi r^2
*L*2−*πr*2 - \displaystyle 1 – \pi\left(\frac{r}{L}\right)^21−
*π*(*Lr*)2 - 1 – \displaystyle\frac{\pi r^2}{L}1−
*Lπr*2

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=(2*r*)2*π**r*2=4*π*,

where r*r* is the radius of the circle. Notice that this probability is *independent* of r*r* !

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}***π*2- \displaystyle \frac{\pi}{4}4
*π* - \displaystyle \frac{4}{\pi}
*π*4 - \displaystyle \frac{\pi}{2}2
*π* - \displaystyle \frac{\sqrt{2}}{\pi r}
*πr*2 - \displaystyle \frac{\pi r}{\sqrt{2}}2
*πr*

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

⎧⎩⎨150if *n* evenif *n* odd

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

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

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

\displaystyle \rho(n) =

{10if *n*=1otherwise

*ρ*(*n*)={10if*n*=1otherwise on n = 1, 2, \ldots*n*=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*)=*π*21+*x*21 on [0, +\infty)[0,+∞). - \displaystyle \rho(x) = \frac{1}{2\pi} + \sin x
*ρ*(*x*)=2*π*1+sin*x*on [0, 2\pi][0,2*π*] - \displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{1+x^2}
*ρ*(*x*)=*π*21+*x*21 on \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞). - \displaystyle \rho(x) = \frac{1}{x^2}
*ρ*(*x*)=*x*21 on [1, +\infty)[1,+∞). - \displaystyle \rho(x) = \frac{1}{10}
*ρ*(*x*)=101 on [0, 9][0,9]

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

- \rho(x)
*ρ*(*x*) is not a probability density function for any value of \lambda*λ* - \displaystyle \lambda = \frac{e}{2}
*λ*=2*e* **\displaystyle \lambda = \frac{1}{e}***λ*=*e*1- \lambda = 1
*λ*=1 - \displaystyle \lambda = \frac{1}{2}
*λ*=21 - \lambda = 2
*λ*=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 T*T* is given by

\int_{t=0}^T \alpha e^{-\alpha t} \, dt∫*t*=0*T**α**e*−*α**t**d**t*

for \alpha = 0.01\ln 2\,\,\, \mathrm{h}^{-1}*α*=0.01ln2h−1 (notice that \alpha*α* 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 x*x* if its probability density function is \rho(x) = (n+1) x^n*ρ*(*x*)=(*n*+1)*xn* (n*n* a positive integer) on [0, 1][0,1].

- \displaystyle \mathbb{E} = \frac{n+1}{n+2}E=
*n*+2*n*+1, \displaystyle \mathbb{V} = \frac{n+1}{n+3} – \left( \frac{n+1}{n+2} \right)^2V=*n*+3*n*+1−(*n*+2*n*+1)2. - \mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+1}{n+2}V=
*n*+2*n*+1. - \mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+1}{n+2} – 1V=
*n*+2*n*+1−1. - \mathbb{E} = 1E=1, \displaystyle \mathbb{V} = \frac{n+2}{n+3}V=
*n*+3*n*+2. - \displaystyle \mathbb{E} = \frac{n+1}{n+2}E=
*n*+2*n*+1, \displaystyle \mathbb{V} = \frac{n+2}{n+3}V=*n*+3*n*+2. - \displaystyle \mathbb{E} = \frac{n+1}{n+2}E=
*n*+2*n*+1, \displaystyle \mathbb{V} = \frac{n+2}{n+3} – \left( \frac{n+1}{n+2} \right)^2V=*n*+3*n*+2−(*n*+2*n*+1)2.

Q2 .Find the expectation \mathbb{E}E and variance \mathbb{V}V of x*x* if its probability density function is \displaystyle \rho(x) = \frac{2}{\pi} \frac{1}{x^2 + 1}*ρ*(*x*)=*π*2*x*2+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=
*π*2, but \mathbb{V}V diverges. - \mathbb{E} = 1E=1, \mathbb{V} = 1V=1.
- \displaystyle \mathbb{E} = \frac{2}{\pi}E=
*π*2, \displaystyle \mathbb{V} = \frac{2}{\pi} – \frac{4}{\pi^2}V=*π*2−*π*24. - \displaystyle \mathbb{E} = \frac{2}{\pi}E=
*π*2, \displaystyle \mathbb{V} = \frac{4}{\pi^2}V=*π*24. - \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 n*n* if its probability density function is \displaystyle \rho(n) = \frac{1}{4}*ρ*(*n*)=41 on n = 1, 2, 3, 4*n*=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* m*m* of a (one-dimensional) continuous probability distribution on [a,b][*a*,*b*] is defined to be the value of x*x* for which the probability of x \lt m*x*<*m* is equal to the probability of x \gt m*x*>*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∫*a**m**ρ*(*x*)*d**x*=21=∫*m**b**ρ*(*x*)*d**x*

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*−*αx*on[0,+∞)

- m = \alpha
*m*=*α* - \displaystyle m = \alpha \ln 2
*m*=*α*ln2 - \displaystyle m = \frac{\alpha}{\ln 2}
*m*=ln2*α* - \displaystyle m = \frac{1}{\alpha \ln 2}
*m*=*α*ln21 - \displaystyle m = \frac{\ln 2}{\alpha}
*m*=*α*ln2 - \displaystyle m = \frac{1}{\alpha}
*m*=*α*1

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_1
*Q*1 for which the probability of x \lt Q_1*x*<*Q*1 is 1/41/4; - the second quartile (really the median) is the unique value Q_2
*Q*2 for which the probability of x \lt Q_2*x*<*Q*2 is 2/42/4; - the third (or upper) quartile is the unique value Q_3
*Q*3 for which the probability of x \lt Q_3*x*<*Q*3 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}
*Q*1=*α*ln41, \displaystyle Q_3 = \frac{1}{\alpha \ln (4/3)}*Q*3=*α*ln(4/3)1. - \displaystyle Q_1 = \frac{\alpha}{\ln 4}
*Q*1=ln4*α*, \displaystyle Q_3 = \frac{\alpha}{\ln (4/3)}*Q*3=ln(4/3)*α*. - \displaystyle Q_1 = \frac{\ln (4/3)}{\alpha}
*Q*1=*α*ln(4/3), \displaystyle Q_3 = \frac{\ln 4}{\alpha}*Q*3=*α*ln4. - \displaystyle Q_1 = \frac{\alpha}{\ln (4/3)}
*Q*1=ln(4/3)*α*, \displaystyle Q_3 = \frac{\alpha}{\ln 4}*Q*3=ln4*α*. - \displaystyle Q_1 = \frac{1}{\alpha \ln (4/3)}
*Q*1=*α*ln(4/3)1, \displaystyle Q_3 = \frac{1}{\alpha \ln 4}*Q*3=*α*ln41. - \displaystyle Q_1 = \frac{\ln 4}{\alpha}
*Q*1=*α*ln4, \displaystyle Q_3 = \frac{\ln (4/3)}{\alpha}*Q*3=*α*ln(4/3).

#### Quiz 04:Chapter 4: Applications – Exam

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

\rho(x) = \frac{3}{2}\sqrt{x}*ρ*(*x*)=23*x*

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}\pi380
*π* - 9\pi9
*π* - \displaystyle\frac{13}{3}\pi313
*π* - \displaystyle\frac{26}{3}\pi326
*π* - \displaystyle\frac{80}{9}\pi980
*π* - \displaystyle\frac{40}{3}\pi340
*π*

Q3. Find the y*y*-coordinate of the center of mass of a thin sheet of metal of constant density of a shape bounded by the x*x*-axis and the parabola

y= 1 – \frac{x^2}{25}*y*=1−25*x*2

- \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 W*W* 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}W31
*W* - \displaystyle\frac{2}{\sqrt{2}}W22
*W* - \displaystyle\frac{1}{8}W81
*W* - There is not enough information to answer this question.
- \displaystyle\frac{1}{\sqrt{2}}W21
*W* - \displaystyle\frac{1}{4}W41
*W*

Q5. Compute the moment of inertia I*I* of a solid cylinder of mass M*M*, radius R*R*, and height h*h* about the central axis (passing through the centers of the cross-sectional discs).

**I = \displaystyle \frac{1}{2}MR^2***I*=21*MR*2- I = \displaystyle \frac{2}{3}MR^2
*I*=32*MR*2 - I = \displaystyle \frac{1}{2}MR^2h
*I*=21*MR*2*h* - I = \displaystyle \frac{2}{3}MRh
*I*=32*MRh* - I = \displaystyle \frac{1}{3}MRh
*I*=31*MRh* - I = \displaystyle \frac{1}{2}MRh
*I*=21*MRh*

Q6. Find the volume of the body obtained by rotating about the x*x*-axis the region between the *cuspidal cubic* x^2 = y^3*x*2=*y*3, the x*x*-axis and the lines x=0*x*=0 and x=1*x*=1. **Hint:** you do not need a picture to solve this problem…

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

Q7. What is the area in the plane enclosed by the graph of the function r(\theta) = \cos \theta + \sin \theta*r*(*θ*)=cos*θ*+sin*θ* **(defined using polar coordinates)** for \theta*θ* between 00 and 3\pi/43*π*/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}43
*π*+21 - \displaystyle \frac{3\pi}{8}+\frac{1}{4}83
*π*+41 - \pi
*π*

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 0*x*2+*y*2=4,*x*,*y*≥0

about the line x=-1*x*=−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−*x*24*dx* - \displaystyle \int_{x=0}^2 2\pi x \sqrt{\frac{4}{4-x^2}} \, dx∫
*x*=022*πx*4−*x*24*dx* - \displaystyle \int_{x=-1}^2 2\pi x \sqrt{\frac{4}{4-x^2}} \, dx∫
*x*=−122*πx*4−*x*24*dx* - \displaystyle \int_{x=-1}^1 2\pi x \sqrt{\frac{4}{4-x^2}} \, dx∫
*x*=−112*πx*4−*x*24*dx* - \displaystyle \int_{x=0}^2 2\pi(x+1)\sqrt{\frac{4}{4-x^2}} \, dx∫
*x*=022*π*(*x*+1)4−*x*24*dx* - \displaystyle \int_{x=0}^2 2\pi(x+1)\sqrt{1+4x^2} \, dx∫
*x*=022*π*(*x*+1)1+4*x*2*dx*

Q9. Find the arc length of the curve \displaystyle y = \frac{x^2}{4} – \frac{\ln x}{2}*y*=4*x*2−2ln*x* between x=1*x*=1 and x=e*x*=*e*.

**Hint:** if you compute the length element correctly, a miraculous simplification should occur, making the integral doable.

**\displaystyle \frac{e^2 + 1}{4}4***e*2+1- \displaystyle \frac{2\pi e}{3}32
*πe* - \displaystyle \frac{e^2 – 2}{4}4
*e*2−2 - \displaystyle \frac{e^2 + 2}{4}4
*e*2+2 - \displaystyle \frac{e^2 – 1}{4}4
*e*2−1 - \displaystyle \frac{e^2}{4}4
*e*2

Q10. the present value PV*PV* of the following income stream I(t)*I*(*t*), assuming an continuously-compounding interest rate of 55 per cent (r=0.05*r*=0.05). The income stream is the following: for the first 1010 years, you get nothing: I(t)=0*I*(*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*=*e*200,000 - PV = \displaystyle \frac{500}{\sqrt{e}}
*PV*=*e*500 - PV = \displaystyle 100,\!000 e^2
*PV*=100,000*e*2 - PV = 200,\!000
*PV*=200,000 - PV = \displaystyle 5,\!000 e
*PV*=5,000*e* - PV = \displaystyle \frac{200,\!000}{\sqrt{e}}
*PV*=*e*200,000

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