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Calculus: Single Variable Part 2 – Differentiation Quiz Answers
Week 01: Calculus: Single Variable Part 2 β Differentiation Coursera Quiz Answers
Main Quiz 01
Q1. Let f(x) = -x^2+6x-3f(x)=βx2+6xβ3. Find f'(-2)fβ²(β2).
- -3β3
- -1β1
- 1010
- 66
- 44
- 00
Q2. Given the position function \displaystyle p(t) = 6 + \frac{1}{2}t +4t^2p(t)=6+21βt+4t2 of a particle as a function of time, what is the particleβs velocity at t=1t=1 ?
- \displaystyle \frac{1}{2}21β
- 66
- 11
- \displaystyle \frac{9}{2}29β
- 88
- \displaystyle \frac{17}{2}217β
Q3. A very rough model of population size PP for an ant species is P(t) = 2\ln(t+2)P(t)=2ln(t+2), where tt is time. What is the rate of change of the population at time t = 2t=2?
- 22
- \displaystyle \frac{1}{4}41β
- 11
- \displaystyle \frac{1}{3}31β
- 44
- \displaystyle \frac{1}{2}21β
Q4. Find \displaystyle \frac{dV}{dt}dtdVβ for \displaystyle V=\frac{1}{4}t^3V=41βt3
- \displaystyle \frac{dV}{dt} = \frac{3}{4}t^3dtdVβ=43βt3
- \displaystyle \frac{dV}{dt} = 3t^2dtdVβ=3t2
- \displaystyle \frac{dV}{dt} = \frac{1}{3}t^2dtdVβ=31βt2
- \displaystyle \frac{dV}{dt} = \frac{3}{4}t^2dtdVβ=43βt2
- \displaystyle \frac{dV}{dt} = 0dtdVβ=0
- \displaystyle \frac{dV}{dt} = \frac{1}{4}t^2dtdVβ=41βt2
Q5. If a carβs position is represented by s(t) = 4t^3s(t)=4t3, what is the carβs change in velocity from t=2t=2 to t=3t=3 ?
- 4848
- 1212
- 6060
- 2020
- 7676
- 108108
Q6. A particleβs position, pp, as a function of time, tt, is represented by \displaystyle p(t) = \frac{1}{3}t^3 β 3t^2 + 9tp(t)=31βt3β3t2+9t. When is the particle at rest?
- Never.
- At t=1t=1.
- At t=3t=3.
- At t = 6t=6.
- At t=0t=0.
- At \displaystyle t = \frac{1}{3}t=31β.
Q7. A rock is dropped from the top of a 320-foot building. The height of the rock at time tt is given s(t)=-8t^2+320s(t)=β8t2+320, where tt is measured in seconds. Find the speed (that is, the absolute value of the velocity) of the rock when it hits the ground in feet per second. Round your answer to one decimal place.
Hookeβs law states that the force FF exerted by an ideal spring displaced a distance xx from its equilibrium point is given by F(x) = -kxF(x)=βkx, where the constant kk is called the spring constant and varies from one spring to another. In real life, many springs are nearly ideal for small displacements; however, for large displacements, they might deviate from what Hookeβs law predicts.
Much of the confusion between nearly-ideal and non-ideal springs is clarified by thinking in terms of series: for xx near zero, F(x) = -kx + O(x^2)F(x)=βkx+O(x2).
Suppose you have a spring whose force follows the equation F(x) = β 2 \tan 3xF(x)=β2tan3x. What is its spring constant?
- 00
- 33
- 1212
- 11
- 22
- 66
Practice Quiz 01
Q1. The profit, PP, of a company that manufactures and sells NN units of a certain product is modeled by the function
P(N) = R(N) β C(N)P(N)=R(N)βC(N)
The revenue function, R(N)=S\cdot NR(N)=Sβ N, is the selling price SS per unit times the number NN of units sold. The companyβs cost, C(N)=C_0+C_\mathrm{op}(N)C(N)=C0β+Copβ(N), is a sum of two terms. The first is a constant C_0C0β describing the initial investment needed to set up production. The other term, C_\mathrm{op}(N)Copβ(N), varies depending on how many units the company produces, and represents the operating costs.
Companies care not only about profit, but also marginal profit, the rate of change of profit with respect to NN.
Assume that S = \$50S=$50, C_0 = \$75,000C0β=$75,000, C_\mathrm{op}(N) = \$50 \sqrt{N}Copβ(N)=$50Nβ, and that the company currently sells N=100N=100 units. Compute the marginal profit at this rate of production. Round your answer to one decimal place.
Q2. In Economics, physical capital represents the buildings or machines used by a business to produce a product. The marginal product of physical capital represents the rate of change of output product with respect to physical capital (informally, if you increase the size of your factory a little, how much more product can you create?).
A particular model tells us that the output product YY is given, as a function of capital KK, by
Y = A K^{\alpha} L^{1-\alpha}Y=AKΞ±L1βΞ±
where AA is a constant, LL is units of labor (assumed to be constant), and \alphaΞ± is a constant between 0 and 1. Determine the marginal product of physical capital predicted by this model.
- \displaystyle \frac{dY}{dK} = \alpha A \frac{L^{1-\alpha}}{K^{\alpha β 1}}dKdYβ=Ξ±AKΞ±β1L1βΞ±β
- \displaystyle \frac{dY}{dK} = (\alpha β 1)A \big( \frac{L}{K} \big)^{\alpha β 1}dKdYβ=(Ξ±β1)A(KLβ)Ξ±β1
- \displaystyle \frac{dY}{dK} = \alpha A K^{\alpha}L^{1-\alpha}dKdYβ=Ξ±AKΞ±L1βΞ±
- \displaystyle \frac{dY}{dK} = \frac{A}{\alpha} \big( \frac{L}{K} \big)^{1-\alpha}dKdYβ=Ξ±Aβ(KLβ)1βΞ±
- None of these.
- \displaystyle \frac{dY}{dK} = (1 β \alpha) A (KL)^{1-\alpha}dKdYβ=(1βΞ±)A(KL)1βΞ±
Main Quiz 02
Q1. Find the derivative of f(x)= \sqrt{x}(2x^2-4x)f(x)=xβ(2x2β4x).
- f'(x) = \sqrt{x}(5x^2-6x)fβ²(x)=xβ(5x2β6x)
- \displaystyle f'(x) = \frac{2x-2}{\sqrt{x}}fβ²(x)=xβ2xβ2β
- f'(x) = 2x^{5/2}-4x^{3/2}fβ²(x)=2x5/2β4x3/2
- f'(x) = \sqrt{x}(5x-6)fβ²(x)=xβ(5xβ6)
- f'(x) = 2x^{3/2}-4x^{1/2}fβ²(x)=2x3/2β4x1/2
- f'(x) = 4\sqrt{x}(x-1)fβ²(x)=4xβ(xβ1)
Q2. Find the derivative of \displaystyle f(x) = 6x^4 -\frac{3}{x^2}-2\pif(x)=6x4βx23ββ2Ο.
- \displaystyle f'(x) = 24x^3 β \frac{3}{x^2}fβ²(x)=24x3βx23β
- \displaystyle f'(x) = 24x^3-\frac{6}{x^2}fβ²(x)=24x3βx26β
- \displaystyle f'(x) = 24x^3-\frac{6}{x^3}fβ²(x)=24x3βx36β
- \displaystyle f'(x) = 24x^3 + \frac{3}{x^2}fβ²(x)=24x3+x23β
- \displaystyle f'(x) = 24x^3 β \frac{3}{2x}fβ²(x)=24x3β2x3β
- \displaystyle f'(x) = 24x^3+\frac{6}{x^3}fβ²(x)=24x3+x36β
Q3. Find the derivative of f(x) = 7(x^3+4x)^5 \cos xf(x)=7(x3+4x)5cosx.
- f'(x) = -45(x^3 + 4x)^4(3x^2 + 4)\sin xfβ²(x)=β45(x3+4x)4(3x2+4)sinx
- f'(x) = 7(x^3+4x)^4 \big[ 5(3x^2+4) \cos x β (x^3+4x)\sin x \big]fβ²(x)=7(x3+4x)4[5(3x2+4)cosxβ(x3+4x)sinx]
- f'(x) = 7(x^3+4x)^4 \big[ 5 \cos x β (x^3+4x)\sin x \big]fβ²(x)=7(x3+4x)4[5cosxβ(x3+4x)sinx]
- f'(x) = 7(x^3+4x)^4 \big[ 5(3x^2+4) \cos x + (x^3+4x)\sin x \big]fβ²(x)=7(x3+4x)4[5(3x2+4)cosx+(x3+4x)sinx]
- f'(x) = 7(x^3+4x)^4 \big[ 5 \cos x + (x^3+4x)\sin x \big]fβ²(x)=7(x3+4x)4[5cosx+(x3+4x)sinx]
- f'(x) = 45(x^3 + 4x)^4(3x^2 + 4)\cos xfβ²(x)=45(x3+4x)4(3x2+4)cosx
Q4. Find the derivative of f(x) = (e^x + \ln x)\sin xf(x)=(ex+lnx)sinx.
- f'(x) = (\sin x )(\ln x) + e^x\cos xfβ²(x)=(sinx)(lnx)+excosx
- \displaystyle f'(x) = \frac{\sin x}{x} + e^x\sin xfβ²(x)=xsinxβ+exsinx
- \displaystyle f'(x) = \frac{\sin x}{x} + (\ln x)(\cos x) + e^x\sin x + e^x\cos xfβ²(x)=xsinxβ+(lnx)(cosx)+exsinx+excosx
- \displaystyle f'(x) = \left( \frac{1}{x}+e^x \right) \cos xfβ²(x)=(x1β+ex)cosx
- \displaystyle f'(x) = \frac{e^x\sin x}{x}fβ²(x)=xexsinxβ
- f'(x) = e^x(\sin x + \cos x)fβ²(x)=ex(sinx+cosx)
Q5. Find the derivative of \displaystyle f(x) = \frac{\sqrt{x+3}}{x^2}f(x)=x2x+3ββ.
- \displaystyle f'(x) = -\frac{3x+12}{2x \sqrt{x+3}}fβ²(x)=β2xx+3β3x+12β
- \displaystyle f'(x) = -\frac{3x+12}{2x^3 \sqrt{x+3}}fβ²(x)=β2x3x+3β3x+12β
- \displaystyle f'(x) = \frac{5x+12}{2x^3 \sqrt{x+3}}fβ²(x)=2x3x+3β5x+12β
- \displaystyle f'(x) = \frac{(x-4)\sqrt{x+3}}{2x^3}fβ²(x)=2x3(xβ4)x+3ββ
- \displaystyle f'(x) = \frac{5x+12}{2x \sqrt{x+3}}fβ²(x)=2xx+3β5x+12β
- \displaystyle f'(x) = \frac{1}{4x\sqrt{x+3}}fβ²(x)=4xx+3β1β
Q6. Find the derivative of \displaystyle f(x) = \frac{\ln x}{\cos x}f(x)=cosxlnxβ.
- \displaystyle f'(x) = \frac{\cos x β \ln x\sin x}{x\sin^2 x}fβ²(x)=xsin2xcosxβlnxsinxβ
- \displaystyle f'(x) = \frac{\ln x\sin x}{x\cos^2 x}fβ²(x)=xcos2xlnxsinxβ
- \displaystyle f'(x) = \frac{\cos x + \ln x\sin x}{x\cos x}fβ²(x)=xcosxcosx+lnxsinxβ
- \displaystyle f'(x) = \frac{\cos x + x\ln x\sin x}{x\cos^2 x}fβ²(x)=xcos2xcosx+xlnxsinxβ
- \displaystyle f'(x) = \frac{\cos x β \ln x\sin x}{x\cos^2 x}fβ²(x)=xcos2xcosxβlnxsinxβ
- \displaystyle f'(x) = \frac{(1 + \ln x)\sin x}{x\cos^2 x}fβ²(x)=xcos2x(1+lnx)sinxβ
Q7. Find the derivative of \displaystyle f(x) = \frac{ \sqrt[3]{x} β 4}{x^3}f(x)=x33xββ4β.
- \displaystyle f'(x) = \frac{10\sqrt[3]{x} β 36}{3x^4}fβ²(x)=3x4103xββ36β
- \displaystyle f'(x) = \frac{10\sqrt[3]{x} β 36}{x^4}fβ²(x)=x4103xββ36β
- \displaystyle f'(x) = \frac{12 β 2\sqrt[3]{x}}{3x^4}fβ²(x)=3x412β23xββ
- \displaystyle f'(x) = \frac{36 β 8\sqrt[3]{x}}{3x^4}fβ²(x)=3x436β83xββ
- \displaystyle f'(x) = \frac{36 β 8\sqrt[3]{x}}{x^4}fβ²(x)=x436β83xββ
- \displaystyle f'(x) = \frac{12 β 2\sqrt[3]{x}}{x^4}fβ²(x)=x412β23xββ
Q8. Find the derivative of f(x)=\sin^3 (x^3)f(x)=sin3(x3).
- f'(x) = 9x^2 \sin^3(x^3) \cos^2 (x^3)fβ²(x)=9x2sin3(x3)cos2(x3)
- f'(x) = 3 \sin^2(x^3) \cos(x^3)fβ²(x)=3sin2(x3)cos(x3)
- f'(x) = 9x^2 \sin^2 (x^3) \cos (x^3)fβ²(x)=9x2sin2(x3)cos(x3)
- f'(x) = 3\sin^2 (x^3)fβ²(x)=3sin2(x3)
- f'(x) = 9x^2 \sin^2 (x^2) \cos (3x^2)fβ²(x)=9x2sin2(x2)cos(3x2)
- f'(x) = 3\sin^2(3x^2)fβ²(x)=3sin2(3x2)
Q9. Find the derivative of f(x) = e^{-1/x^2}f(x)=eβ1/x2.
- \displaystyle f'(x) = \frac{2}{x^3} e^{-1/x^2}fβ²(x)=x32βeβ1/x2
- \displaystyle f'(x) = e^{2/x^3}fβ²(x)=e2/x3
- \displaystyle f'(x) = e^{-2/x^3}fβ²(x)=eβ2/x3
- \displaystyle f'(x) = -\frac{1}{x^2} e^{-1/x^2}fβ²(x)=βx21βeβ1/x2
- \displaystyle f'(x) = \frac{1}{x^2} e^{-1/x^2}fβ²(x)=x21βeβ1/x2
- \displaystyle f'(x) = -\frac{2}{x^3} e^{-1/x^2}fβ²(x)=βx32βeβ1/x2
Week 02: Calculus: Single Variable Part 2 β Differentiation Coursera Quiz Answers
Main Quiz 01
Q1. Use a linear approximation to estimate \sqrt[3]{67}367β. Round your answer to four decimal places.
Hint: remember that \sqrt[3]{64} = 4364β=4. You can check the accuracy of this approximation by noting that \sqrt[3]{67} \approx 4.0615367ββ4.0615.
Q2. Use a linear approximation to estimate the cosine of an angle of 66^\mathrm{o}66o. Round your answer to four decimal places.
Hint: remember that \displaystyle 60^\mathrm{o} = \frac{\pi}{3}60o=3Οβ, and hence \displaystyle 6^\mathrm{o} = \frac{\pi}{30}6o=30Οβ. You can check the accuracy of this approximation by noting that \cos 66^\mathrm{o} \approx 0.4067cos66oβ0.4067.
Q3. The golden ratio \displaystyle \varphi = \frac{1+\sqrt{5}}{2}Ο=21+5ββ is a root of the polynomial x^2-x-1x2βxβ1. If you use Newtonβs method to estimate its value, what is the appropriate update rule for the sequence x_nxnβ ?
- \displaystyle x_{n+1} = x_n + \frac{2x_n β 1}{x_n^2 β x_n β 1}xn+1β=xnβ+xn2ββxnββ12xnββ1β
- \displaystyle x_{n+1} = x_n β \frac{x_n^2 β x_n β 1}{2x_n β 1}xn+1β=xnββ2xnββ1xn2ββxnββ1β
- \displaystyle x_{n+1} = x_n β \frac{2x_n β 1}{x_n^2 β x_n β 1}xn+1β=xnββxn2ββxnββ12xnββ1β
- \displaystyle x_{n+1} = \frac{x_n^2 β x_n β 1}{2x_n β 1}xn+1β=2xnββ1xn2ββxnββ1β
- \displaystyle x_{n+1} = x_n + \frac{x_n^2 β x_n β 1}{2x_n β 1}xn+1β=xnβ+2xnββ1xn2ββxnββ1β
- \displaystyle x_{n+1} = \frac{2x_n β 1}{x_n^2 β x_n β 1}xn+1β=xn2ββxnββ12xnββ1β
Q4. To approximate \sqrt{10}10β using Newtonβs method, what is the appropriate update rule for the sequence x_nxnβ ?
- \displaystyle x_{n+1} = \frac{x_n}{2} + \frac{5}{x_n}xn+1β=2xnββ+xnβ5β
- \displaystyle x_{n+1} = \frac{x_n}{2}xn+1β=2xnββ
- \displaystyle x_{n+1} = x_n + \frac{2x_n}{x_n^2 β 10}xn+1β=xnβ+xn2ββ102xnββ
- \displaystyle x_{n+1} = \frac{x_n}{2} β \frac{10}{x_n}xn+1β=2xnβββxnβ10β
- \displaystyle x_{n+1} = \frac{x_n}{2} β \frac{5}{x_n}xn+1β=2xnβββxnβ5β
- \displaystyle x_{n+1} = x_n β \frac{2x_n}{x_n^2 β 10}xn+1β=xnββxn2ββ102xnββ
Q5. You want to build a square pen for your new chickens, with an area of 1200\,\mathrm{ft}^21200ft2. Not having a calculator handy, you decide to use Newtonβs method to approximate the length of one side of the fence. If your first guess is 30\,\mathrm{ft}30ft, what is the next approximation you will get?
- 3535
- 15.0515.05
- 4040
- -5β5
- 3030
- 30.0530.05
Q6. You are in charge of designing packaging materials for your companyβs new product. The marketing department tells you that you must put them in a cube-shaped box. The engineering department says that you will need a box with a volume of 500\,\mathrm{cm}^3500cm3. What are the dimensions of the cubical box? Starting with a guess of 8\,\mathrm{cm}8cm for the length of the side of the cube, what approximation does one iteration of Newtonβs method give you? Round your answer to two decimal places.
Practice Quiz 01
Q1. Without using a calculator, approximate 9.98^{98}9.9898. Here are some hints. First, 9.989.98 is close to 1010, and 10^{98}=1\,{\rm E}\,981098=1E98 in scientific notation. What does linear approximation give as an estimate when we decrease from 10^{98}1098 to 9.98^{98}9.9898?
- 1.000\,{\rm E}\,981.000E98
- 0.902\,{\rm E}\,980.902E98
- 1.960\,{\rm E}\,981.960E98
- 0.804\,{\rm E}\,980.804E98
- 0.9804\,{\rm E}\,980.9804E98
- 0.822\,{\rm E}\,980.822E98
Q2. A diving-board of length LL bends under the weight of a diver standing on its edge. The free end of the board moves down a distance
D = \frac{P}{3EI} L^3D=3EIPβL3
where PP is the weight of the diver, EE is a constant of elasticity βthat depends on the material from which the board is manufacturedβ and II is a moment of inertia. (These last two quantities will again make an appearance in Lectures 13 and 41, but do not worry about what exactly they mean nowβ¦)
Suppose our board has a length L = 2\,\mathrm{m}L=2m, and that it takes a deflection of D = 20\,\mathrm{cm}D=20cm under the weight of the diver. Use a linear approximation to estimate the deflection that it would take if its length was increased by 20\,\mathrm{cm}20cm
- 20.3\,\mathrm{cm}20.3cm
- 25.7\,\mathrm{cm}25.7cm
- 22\,\mathrm{cm}22cm
- 26\,\mathrm{cm}26cm
- 24.8\,\mathrm{cm}24.8cm
- 26.6\,\mathrm{cm}26.6cm
Main Quiz 02
Q1. You are given the position, velocity and acceleration of a particle at time t = 0t=0. The position is p(0) = 2p(0)=2, the velocity v(0) = 4v(0)=4, and the acceleration a(0) = 3a(0)=3. Using this information, which Taylor series should they use to approximate p(t)p(t), and what is the estimated value of p(4)p(4) using this approximation?
- p(t) = 2 + 2t + 3 t^2 + O(t^3)p(t)=2+2t+3t2+O(t3), p(4) \simeq 58p(4)β58.
- p(t) = 2 + 4t + 6 t^2 + O(t^3)p(t)=2+4t+6t2+O(t3), p(4) \simeq 114p(4)β114.
- p(t) = 2 + 4t + 3 t^2 + O(t^3)p(t)=2+4t+3t2+O(t3), p(4) \simeq 66p(4)β66.
- \displaystyle p(t) = 2 + 2t + \frac{3}{2} t^2 + O(t^3)p(t)=2+2t+23βt2+O(t3), p(4) \simeq 34p(4)β34.
- \displaystyle p(t) = 2 + 4t + \frac{3}{2} t^2 + O(t^3)p(t)=2+4t+23βt2+O(t3), p(4) \simeq 42p(4)β42.
- p(t) = 2 + 2t + 6 t^2 + O(t^3)p(t)=2+2t+6t2+O(t3), p(4) \simeq 106p(4)β106.
Q2. If a particle moves according to the position function s(t) = t^3-6ts(t)=t3β6t, what are its position, velocity and acceleration at t=3t=3 ?
- s(3) = 9s(3)=9, v(3) = 21v(3)=21, a(3) = 18a(3)=18
- s(3) = 9s(3)=9, v(3) = 21v(3)=21, a(3) = 36a(3)=36
- s(3) = 21s(3)=21, v(3) = 18v(3)=18, a(3) = 6a(3)=6
- s(3) = 9s(3)=9, v(3) = 18v(3)=18, a(3) = 18a(3)=18
- s(3) = 9s(3)=9, v(3) = 21v(3)=21, a(3) = 9a(3)=9
- s(3) = 21s(3)=21, v(3) = 18v(3)=18, a(3) = 18a(3)=18
Q3. If the position of a car at time tt is given by the formula p(t) = t^4 β 24t^2p(t)=t4β24t2, for which times tt is its velocity decreasing?
- Never: the velocity always increases.
- -\sqrt[3]{12} < t < \sqrt[3]{12}β312β<t<312β
- t < -2t<β2
- -2 < t < 2β2<t<2
- -\sqrt{24} < t < \sqrt{24}β24β<t<24β
- t > 2t>2
Q4. What is a formula for the second derivative of f(t) = t^2\sin 2tf(t)=t2sin2t? Use this formula to compute fβ(\pi/2)fβ²β²(Ο/2).
- fβ(t) = 4t\cos 2t + (2-4t^2)\sin 2tfβ²β²(t)=4tcos2t+(2β4t2)sin2t, and fβ(\pi/2) = -2\pifβ²β²(Ο/2)=β2Ο
- fβ(t) = -4t^2\sin 2tfβ²β²(t)=β4t2sin2t, and fβ(\pi/2) =0fβ²β²(Ο/2)=0
- fβ(t) = -8\sin 2tfβ²β²(t)=β8sin2t, and fβ(\pi/2) = 0fβ²β²(Ο/2)=0
- fβ(t) = 8t\cos 2t -4t^2\sin 2tfβ²β²(t)=8tcos2tβ4t2sin2t, and fβ(\pi/2) = -4\pifβ²β²(Ο/2)=β4Ο
- fβ(t) = 4t\cos 2tfβ²β²(t)=4tcos2t, and fβ(\pi/2) = -2\pifβ²β²(Ο/2)=β2Ο
- fβ(t) = 8t\cos 2t + (2-4t^2)\sin 2tfβ²β²(t)=8tcos2t+(2β4t2)sin2t, and fβ(\pi/2) = -4\pifβ²β²(Ο/2)=β4Ο
Q5. Use a Taylor series expansion to compute f^{(3)}(0)f(3)(0) for f(x) = \sin^3 \left(\ln(1+x) \right)f(x)=sin3(ln(1+x)).
- -3β3
- 66
- 1212
- 33
- 00
- -6β6
Q6. What is the curvature of the graph of the function f(x) = -2\sin(x^2)f(x)=β2sin(x2) at the point (0,0)(0,0)?
- 00
- 22
- \displaystyle \frac{1}{2}21β
- 11
- 44
- -4β4
Main Quiz 03
Q1. Find all the local maxima and minima of the function y=x e^{-x^2}y=xeβx2.
- The function has local minima at \displaystyle x = \frac{\sqrt{2}}{2}x=22ββ and \displaystyle x = -\frac{\sqrt{2}}{2}x=β22ββ, and a local maximum at x = 0x=0.
- The function has a local maximum at \displaystyle x = -\frac{\sqrt{2}}{2}x=β22ββ, and a local minimum at \displaystyle x = \frac{\sqrt{2}}{2}x=22ββ.
- The function has local minima at \displaystyle x = \frac{\sqrt{2}}{2}x=22ββ and \displaystyle x = -\frac{\sqrt{2}}{2}x=β22ββ, but no local maxima.
- The function has a local maximum at \displaystyle x = \frac{\sqrt{2}}{2}x=22ββ, and a local minimum at \displaystyle x = β \frac{\sqrt{2}}{2}x=β22ββ.
- The function has local maxima at \displaystyle x = \frac{\sqrt{2}}{2}x=22ββ and \displaystyle x = -\frac{\sqrt{2}}{2}x=β22ββ, and a local minimum at x = 0x=0.
- The function has local maxima at \displaystyle x = \frac{\sqrt{2}}{2}x=22ββ and \displaystyle x = -\frac{\sqrt{2}}{2}x=β22ββ, but no local minima.
Q2. Which of the following statements is true about the function f(x) = e^{\sin(x^4)}\cos(x^2)f(x)=esin(x4)cos(x2) ?
- Its Taylor series expansion about x=0x=0 is \displaystyle 1 β \frac{x}{3} + O(x^2)1β3xβ+O(x2). Hence x=0x=0 is not a critical point of f(x)f(x).
- Its Taylor series expansion about x=0x=0 is \displaystyle 1 + \frac{x^3}{2} + O(x^4)1+2x3β+O(x4). Hence x=0x=0 is a critical point of f(x)f(x) that is neither a local maximum nor a local minimum.
- Its Taylor series expansion about x=0x=0 is \displaystyle 1 + \frac{x^4}{2} + O(x^5)1+2x4β+O(x5). Hence it has a local minimum at x=0x=0.
- Its Taylor series expansion about x=0x=0 is \displaystyle 1 β \frac{x^2}{2} + O(x^5)1β2x2β+O(x5). Hence it has a local minimum at x=0x=0.
- Its Taylor series expansion about x=0x=0 is \displaystyle 1 β \frac{x^2}{2} + O(x^5)1β2x2β+O(x5). Hence it has a local maximum at x=0x=0.
- Its Taylor series expansion about x=0x=0 is \displaystyle 1 + \frac{x^4}{2} + O(x^5)1+2x4β+O(x5). Hence it has a local maximum at x=0x=0.
Q3. Use a Taylor series about x=0x=0 to determine whether the function f(x) = \sin^3(x^3)f(x)=sin3(x3) has a local maximum or local minimum at the origin.
- x=0x=0 is a critical point of ff, but it is neither a local maximum nor a local minimum.
- x=0x=0 is not a critical point of ff.
- x=0x=0 is a local minimum of ff.
- x=0x=0 is a local maximum of ff.
Q4. Find the location of the global maximum and minimum of f(x) = x^3-6x^2+1f(x)=x3β6x2+1 on the interval [-1,7][β1,7].
- The global maximum is attained at x = 0x=0 and the global minimum at x = -1x=β1.
- The global maximum is attained at x = 0x=0 and the global minimum at x = 4x=4.
- The global maximum is attained at x = 7x=7, but there is no global minimum.
- The global maximum is attained at x = 7x=7 and the global minimum at x = 4x=4.
- The global maximum is attained at x = 7x=7 and the global minimum at x = -1x=β1.
- The global maximum is attained at x = 0x=0, but there is no global minimum.
Q5. Which of the following statements are true for the function \displaystyle f(x) = x^3 + \frac{48}{x^2}f(x)=x3+x248β ? Select all that apply.
- x=-2x=β2 is the global maximum of ff in [-3, -1][β3,β1]
- x=-1x=β1 is the global maximum of ff in [-3, -1][β3,β1]
- x=2x=2 is the global maximum of ff in [-3, 3][β3,3]
- x=1x=1 is the global minimum of ff in [1, 3][1,3]
- x=2x=2 is the global minimum of ff in [1, 3][1,3]
- x=1x=1 is the global maximum of ff in [1, 3][1,3]
Week 03: Calculus: Single Variable Part 2 β Differentiation Coursera Quiz Answers
Main Quiz 01
Q1. Use implicit differentiation to find \displaystyle \frac{dy}{dx}dxdyβ from the equation y^2 β y = \sin 2xy2βy=sin2x.
- \displaystyle \frac{dy}{dx} = \frac{y^2 β y}{2\cos 2x}dxdyβ=2cos2xy2βyβ
- \displaystyle \frac{dy}{dx} = \frac{\sin 2x}{2y β 1}dxdyβ=2yβ1sin2xβ
- \displaystyle \frac{dy}{dx} = \frac{2\cos 2x}{2y β 1}dxdyβ=2yβ12cos2xβ
- \displaystyle \frac{dy}{dx} = \frac{2\cos 2x}{y^2 β y}dxdyβ=y2βy2cos2xβ
- \displaystyle \frac{dy}{dx} = \frac{2y β 1}{\sin 2x}dxdyβ=sin2x2yβ1β
- \displaystyle \frac{dy}{dx} = \frac{2y β 1}{2\cos 2x}dxdyβ=2cos2x2yβ1β
Q2. Find the derivative \displaystyle \frac{dy}{dx}dxdyβ if xx and yy are related through xy = e^yxy=ey.
- \displaystyle \frac{dy}{dx} = \frac{e^y + x}{y}dxdyβ=yey+xβ
- \displaystyle \frac{dy}{dx} = \frac{x β e^y}{y}dxdyβ=yxβeyβ
- \displaystyle \frac{dy}{dx} = \frac{y}{e^y + x}dxdyβ=ey+xyβ
- \displaystyle \frac{dy}{dx} = \frac{y}{x β e^y}dxdyβ=xβeyyβ
- \displaystyle \frac{dy}{dx} = \frac{y}{e^y β x}dxdyβ=eyβxyβ
- \displaystyle \frac{dy}{dx} = \frac{e^y β x}{y}dxdyβ=yeyβxβ
Q3. Use implicit differentiation to find \displaystyle \frac{dy}{dx}dxdyβ if \sin x = e^{-y\cos x}sinx=eβycosx.
- \displaystyle \frac{dy}{dx} = y\cos x β e^{y\cos x}\sin xdxdyβ=ycosxβeycosxsinx
- \displaystyle \frac{dy}{dx} = \frac{y\sin x β e^{-y\cos x}}{\cos x}dxdyβ=cosxysinxβeβycosxβ
- \displaystyle \frac{dy}{dx} = y\tan x β e^{y\cos x}dxdyβ=ytanxβeycosx
- \displaystyle \frac{dy}{dx} = -y\sin x + e^{-y\cos x}\cos xdxdyβ=βysinx+eβycosxcosx
- \displaystyle \frac{dy}{dx} = e^{-y\cos x}(\cos x β y\sin x)dxdyβ=eβycosx(cosxβysinx)
- \displaystyle \frac{dy}{dx} = \frac{y β e^{y\cos x}\tan x}{\sin x}dxdyβ=sinxyβeycosxtanxβ
Q4. Find the derivative \displaystyle \frac{dy}{dx}dxdyβ from the equation x\tan y β y^2\ln x = 4xtanyβy2lnx=4.
- \displaystyle \frac{dy}{dx} = \frac{-y^2}{x^2\sec^2 y}dxdyβ=x2sec2yβy2β
- \displaystyle \frac{dy}{dx} = \tan y β \frac{y^2}{\sec^2 y}dxdyβ=tanyβsec2yy2β
- \displaystyle \frac{dy}{dx} = \frac{x\tan y β y^2}{2xy\ln x β x^2\sec^2 y}dxdyβ=2xylnxβx2sec2yxtanyβy2β
- \displaystyle \frac{dy}{dx} = \frac{2xy\ln x β x^2\sec^2 y}{x\tan y β y^2}dxdyβ=xtanyβy22xylnxβx2sec2yβ
- \displaystyle \frac{dy}{dx} = \frac{y^2 β \tan y}{x^2\sec^2 y β 2xy\ln x}dxdyβ=x2sec2yβ2xylnxy2βtanyβ
- \displaystyle \frac{dy}{dx} = \frac{x\tan y}{2xy\ln x}dxdyβ=2xylnxxtanyβ
Q5. Model a hailstone as a round ball of radius RR. As the hailstone falls from the sky, its radius increases at a constant rate CC. At what rate does the volume VV of the hailstone change?
- \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi C R^3dtdVβ=34βΟCR3
- \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi C^3dtdVβ=34βΟC3
- \displaystyle \frac{dV}{dt} = 8\pi C RdtdVβ=8ΟCR
- \displaystyle \frac{dV}{dt} = 4\pi C R^2dtdVβ=4ΟCR2
- \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi R^3dtdVβ=34βΟR3
- \displaystyle \frac{dV}{dt} = 4\pi R^2dtdVβ=4ΟR2
Q6. The volume of a cubic box of side-length LL is V = L^3V=L3. How are the relative rates of change of LL and VV related?
- \displaystyle \frac{dL}{L} = \frac{dV}{V}LdLβ=VdVβ
- \displaystyle \frac{dV}{V} = 3 L^3 \frac{dL}{L}VdVβ=3L3LdLβ
- \displaystyle \frac{dL}{L} = 3 \frac{dV}{V}LdLβ=3VdVβ
- \displaystyle \frac{dV}{V} = -\frac{dL}{L}VdVβ=βLdLβ
- \displaystyle \frac{dV}{V} = 0VdVβ=0
- \displaystyle \frac{dV}{V} = 3 \frac{dL}{L}VdVβ=3LdLβ
Practice Quiz 01
Q1. Consider a box of height hh with a square base of side length LL. Assume that LL is increasing at a rate of 10\%10% per day, but hh is decreasing at a rate of 10\%10% per day. Use a linear approximation to find at what (approximate) rate the volume of the box changing?
Hint: consider the relative rate of change of the volume of the box.
Hint^\mathbf{2}2: in this case you can very easily calculate the exact rate of change β8.9%β, so using linearization might seem like overkill. However, if you set up things right, you donβt even need a calculator to find out the approximate rate of change! Do you see why?
- Increasing at a rate of 5\%5% per day.
- Increasing at a rate of 10\%10% per day.
- Decreasing at a rate of 10\%10% per day.
- Increasing at a rate of 2.5\%2.5% per day.
- It does not change.
- Decreasing at a rate of 5\%5% per day.
Q2. A large tank of oil is slowly leaking oil into a containment tank surrounding it. The oil tank is a vertical cylinder with a diameter of 10 meters. The containment tank has a square base with side length of 15 meters and tall vertical walls. The bottom of the oil tank and the bottom of the containment tank are concentric (the round base inside the square base). Denote by h_ohoβ the height of the oil inside of the oil tank, and by h_chcβ the height of the oil in the containment tank. How are the rates of change of these two quantities related?
Q2. \displaystyle dh_c = -\frac{225-25\pi}{25\pi} dh_odhcβ=β25Ο225β25Οβdhoβ
dh_c = (25\pi β 225) dh_odhcβ=(25Οβ225)dhoβ
\displaystyle dh_c = -\frac{25\pi}{225} dh_odhcβ=β22525Οβdhoβ
\displaystyle dh_c = (225 β 25\pi) dh_odhcβ=(225β25Ο)dhoβ
\displaystyle dh_c = -\frac{25\pi}{225-25\pi} dh_odhcβ=β225β25Ο25Οβdhoβ
\displaystyle dh_c = -\frac{225}{25\pi} dh_odhcβ=β25Ο225βdhoβ
Q3. The stopping distance D_\mathrm{stop}Dstopβ is the distance traveled by a vehicle from the moment the driver becomes aware of an obstacle in the road until the car stops completely. This occurs in two phases.
The first one, the reaction phase, spans from the moment the driver sees the obstacle until he or she has completely depressed the brake pedal. This entails taking the decision to stop the vehicle, lifting the foot from the gas pedal and onto the brake pedal, and pressing the latter down its full distance to obtain maximum braking power. The amount of time necessary to do all this is called the reaction time t_\mathrm{react}treactβ, and is independent of the speed at which the vehicle was traveling. Although this quantity varies from driver to driver, it is typically between 1.5\,\mathrm{s}1.5s and 2.5\,\mathrm{s}2.5s. For the purposes of this problem, we will use an average value of 2\,\mathrm{s}2s. The distance traversed by the vehicle in this time is unsurprisingly called reaction distance D_\mathrm{react}Dreactβ and is given by the formula
D_\mathrm{react} = v t_\mathrm{react}Dreactβ=vtreactβ
where vv is the initial speed of the vehicle.
In the braking phase, the vehicle decelerates and comes to a complete stop. The braking distance D_\mathrm{brake}Dbrakeβ that the vehicle covers in this phase is proportional to the square of the initial speed of the vehicle:
D_\mathrm{brake} = \alpha v^2Dbrakeβ=Ξ±v2
The constant of proportionality \alphaΞ± depends on the vehicle type and condition, as well as on the road conditions. Consider a typical value of 10^{-2}\,\mathrm{s^2/m}10β2s2/m.
If the initial speed of the vehicle is 108\,\mathrm{km/h} = 30\,\mathrm{m/s}108km/h=30m/s, what is the ratio between the relative rate of change of the stopping distance and the relative rate of change of the initial speed?
- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{26}{23}dv/vdDstopβ/Dstopββ=2326β
- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{24}{23}dv/vdDstopβ/Dstopββ=2324β
- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{27}{23}dv/vdDstopβ/Dstopββ=2327β
- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = 1dv/vdDstopβ/Dstopββ=1
- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{28}{26}dv/vdDstopβ/Dstopββ=2628β
- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{25}{23}dv/vdDstopβ/Dstopββ=2325β
Q4. Assume that you possess equal amounts of a product XX and YY, but you value them differently. Specifically, your utility function is of the form
U(X,Y) = C X^\alpha Y^\betaU(X,Y)=CXΞ±YΞ²
for \alphaΞ±, \betaΞ², and CC positive constants. What is your marginal rate of substitution (MRS) of YY for XX?
Hint: recall that the MRS is equal to \displaystyle -\frac{dY}{dX}βdXdYβ along the indifference curve where UU is constant.
- \displaystyle \frac{\beta}{\alpha}Ξ±Ξ²β
- \displaystyle \frac{C}{\alpha\beta}Ξ±Ξ²Cβ
- \displaystyle \frac{\alpha}{\beta}Ξ²Ξ±β
- 11
- \displaystyle C\frac{\beta}{\alpha}CΞ±Ξ²β
- \displaystyle \frac{\alpha Y}{\beta X}Ξ²XΞ±Yβ
Main Quiz 02
Q1. Find the derivative of f(x) = (\cos x)^xf(x)=(cosx)x.
- f'(x) = \ln\cos x β x\tan xfβ²(x)=lncosxβxtanx
- f'(x) = (\ln\cos x + x\cot x)(\cos x)^xfβ²(x)=(lncosx+xcotx)(cosx)x
- f'(x) = (\ln\cos x β x\tan x)(\cos x)^{x-1}fβ²(x)=(lncosxβxtanx)(cosx)xβ1
- f'(x) = β x (\cos x)^{x-1}\sin xfβ²(x)=βx(cosx)xβ1sinx
- f'(x) = (\ln\cos x β x\tan x)(\cos x)^xfβ²(x)=(lncosxβxtanx)(cosx)x
- f'(x) = -(\cos x)^{x-1}\sin xfβ²(x)=β(cosx)xβ1sinx
Q2. Find the derivative of f(x) = (\ln x)^xf(x)=(lnx)x.
- \displaystyle f'(x) = (\ln x)^x \left(\frac{1}{\ln x} + \ln(\ln x) \right)fβ²(x)=(lnx)x(lnx1β+ln(lnx))
- \displaystyle f'(x) = \frac{1}{\ln x} + \ln(\ln x)fβ²(x)=lnx1β+ln(lnx)
- \displaystyle f'(x) = (\ln x)^x \left(\frac{1}{e^x} + e^x\ln x \right)fβ²(x)=(lnx)x(ex1β+exlnx)
- f'(x) = (\ln x)^x \ln(\ln x)fβ²(x)=(lnx)xln(lnx)
- \displaystyle f'(x) = (\ln x)^x \frac{\ln x}{x}fβ²(x)=(lnx)xxlnxβ
- \displaystyle f'(x) = \frac{1}{e^x} + e^x\ln xfβ²(x)=ex1β+exlnx
Q3. Find the derivative of f(x) = x^{\ln x}f(x)=xlnx.
- f'(x) = 2\ln xfβ²(x)=2lnx
- f'(x) = 2x^{\ln x} \ln xfβ²(x)=2xlnxlnx
- f'(x) = x^{\ln x} \ln xfβ²(x)=xlnxlnx
- f'(x) = x^{\ln(x) β 1} \ln xfβ²(x)=xln(x)β1lnx
- f'(x) = 2x^{\ln(x) β 1} \ln xfβ²(x)=2xln(x)β1lnx
- f'(x) = (\ln x + x) x^{\ln x}fβ²(x)=(lnx+x)xlnx
Q4. \displaystyle \lim_{x \to +\infty} \left( \frac{x+2}{x+3} \right)^{2x} =xβ+βlimβ(x+3x+2β)2x=
Hint: write the fraction \displaystyle \frac{x+2}{x+3}x+3x+2β as 1 + \text{something}1+something.
- e^{3/2}e3/2
- e^2e2
- e^{-2}eβ2
- e^{4/3}e4/3
- e^{2/3}e2/3
- 11
Q5. \displaystyle \lim_{x \to 0^+} \left[ \ln(1+x) \right]^{x} =xβ0+limβ[ln(1+x)]x=
- 11
- e^2e2
- The limit does not exist.
- 00
- \sqrt{e}eβ
- ee
Q6. \displaystyle \lim_{x \to 0} \left(1 + \arctan\frac{x}{2} \right)^{2/x} =xβ0limβ(1+arctan2xβ)2/x=
- e^2e2
- \sqrt{e}eβ
- 00
- 11
- ee
- +\infty+β
Main Quiz 03
Q1. If f(x) = x^{2x}f(x)=x2x, compute \displaystyle \frac{df}{dx}dxdfβ.
- 2 \ln \left( x^{2x} β 2x \right)2ln(x2xβ2x)
- 2x^{2x}\left(1 + \ln x\right)2x2x(1+lnx)
- 2 \left[ x^x β \ln(2x-1) + 1 \right]2[xxβln(2xβ1)+1]
- x^{2x} \ln \left( x^{2x}+1 \right)x2xln(x2x+1)
- x^2 + (e^x)^2x2+(ex)2
- 2x^{2x-1}2x2xβ1
- x^{2\ln x} β 2x^2x2lnxβ2x2
- x^{2x} \ln 2xx2xln2x
Q2. Consider the function f(x) = \sqrt{3}\,x^2\,e^{1-x}f(x)=3βx2e1βx. Use the formula for curvature,
\kappa = \frac{|fβ|}{ \left( 1+|fβ|^2 \right)^{3/2}}ΞΊ=(1+β£fβ²β£2)3/2β£fβ²β²β£β
to compute the curvature of the graph of ff at the point (1,\sqrt{3})(1,3β).
- \displaystyle -\frac{\sqrt{3}}{9}β93ββ
- \displaystyle \frac{\sqrt{3}}{\left(\sqrt{1+\sqrt{3}}\right)^3}(1+3ββ)33ββ
- \displaystyle \frac{\sqrt{3}}{64}643ββ
- \displaystyle \frac{2\sqrt{3}}{27}2723ββ
- \displaystyle \frac{x^2-4x+2}{2x-x^2}2xβx2x2β4x+2β
- \displaystyle \frac{2}{x} β 1x2ββ1
- \sqrt{3}3β
- \displaystyle \frac{\sqrt{3}}{8}83ββ
Q3. Assume that xx and yy are related by the equation y \ln x = e^{1-x} + y^3ylnx=e1βx+y3. Compute \displaystyle \frac{dy}{dx}dxdyβ evaluated at x = 1x=1.
- -3β3
- \displaystyle -\frac{1}{3}β31β
- \displaystyle \frac{e^2}{6}6e2β
- \displaystyle \frac{2 + e^2}{3}32+e2β
- \displaystyle \frac{-2 + e^{-2}}{6}6β2+eβ2β
- 00
- \displaystyle \frac{2-e^2}{3}32βe2β
- \displaystyle \frac{1}{3}31β
Q4. Use the linear approximation of the function f(x) = \arctan\left(e^{3x}\right)f(x)=arctan(e3x) at x = 0x=0 to estimate the value of f(0.01)f(0.01).
Hint: remember that \displaystyle \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}dxdβarctan(x)=1+x21β.
- \displaystyle \frac{\pi}{4} + \frac{3}{2}4Οβ+23β
- \displaystyle \frac{\pi}{4} β \frac{3}{2}4Οββ23β
- \displaystyle \frac{\pi}{4} + \frac{3}{200}4Οβ+2003β
- \displaystyle \frac{\pi}{4} β \frac{1}{20}4Οββ201β
- \displaystyle \frac{\pi}{4} + \frac{1}{20}4Οβ+201β
- \displaystyle \frac{\pi}{4} β \frac{1}{200}4Οββ2001β
- \displaystyle \frac{\pi}{4} β \frac{3}{200}4Οββ2003β
- \displaystyle \frac{\pi}{4} + \frac{1}{200}4Οβ+2001β
Q5. A rectangular picture frame with total area 50000 \text{ cm}^250000 cm2 includes a border which is 1\text{ cm}1 cm thick at the top and the bottom and 5 \text{ cm}5 cm thick at the left and right side. What is the largest possible area of a picture that can be displayed in this frame?
- 85\text{ cm} \times 470\text{ cm}85 cmΓ470 cm
- 98\text{ cm} \times 490\text{ cm}98 cmΓ490 cm
- 80\text{ cm} \times 460\text{ cm}80 cmΓ460 cm
- 94\text{ cm} \times 475\text{ cm}94 cmΓ475 cm
- 96\text{ cm} \times 485\text{ cm}96 cmΓ485 cm
- 95\text{ cm} \times 499\text{ cm}95 cmΓ499 cm
- 99\text{ cm} \times 495\text{ cm}99 cmΓ495 cm
- 110\text{ cm} \times 450\text{ cm}110 cmΓ450 cm
Q6. Which of the following statements are true for the function \displaystyle f(x) = \frac{4}{x} + x^4f(x)=x4β+x4? In order to receive full credit for this problem, you must select all the true statements (there may be many) and none of the false statements.
1 point
- The global minimum of ff for \displaystyle \frac{1}{2}\leq x \leq 221ββ€xβ€2 is at x = 1x=1.
- ff is not differentiable at x=0x=0.
- The global maximum of ff for \displaystyle -1\leq x \leq-\frac{1}{2}β1β€xβ€β21β is at x = -1x=β1.
- The critical points of ff are at x = -1x=β1 and x = 1x=1.
- The global maximum of ff for -2\leq x \leq -1β2β€xβ€β1 is at x = -2x=β2.
- The global maximum of ff for \displaystyle -\frac{3}{2}\leq x \leq 2β23ββ€xβ€2 is at x = -1x=β1.
- The global minimum of ff for -1\leq x \leq 2β1β€xβ€2 is at x = 1x=1.
- The global minimum of ff for -2\leq x \leq 2β2β€xβ€2 is at x = 1x=1.
Q7. To approximate \sqrt[3]{15}315β (the cube root of 1515) using Newtonβs method, what is the appropriate update rule for the sequence x_nxnβ?
- \displaystyle x_{n+1} = x_n + 3x_n^2xn+1β=xnβ+3xn2β
- \displaystyle x_{n+1} = x_n + \frac{5}{x_n^2}xn+1β=xnβ+xn2β5β
- \displaystyle x_{n+1} = \frac{2x_n}{3} β \frac{5}{x_n^2}xn+1β=32xnβββxn2β5β
- \displaystyle x_{n+1} = \frac{2x_n}{3} + \frac{5}{x_n^2}xn+1β=32xnββ+xn2β5β
- \displaystyle x_{n+1} = x_n β \frac{3x_n^2}{x_n^3-15}xn+1β=xnββxn3ββ153xn2ββ
- \displaystyle x_{n+1} = \frac{4x_n}{3} β \frac{5}{x_n^2}xn+1β=34xnβββxn2β5β
- \displaystyle x_{n+1} = x_n + \frac{3x_n^2}{x_n^3-15}xn+1β=xnβ+xn3ββ153xn2ββ
- \displaystyle x_{n+1} = \frac{2}{3}x_nxn+1β=32βxnβ
Q8. Fill in the blank:
\ln^2(x+h) = \ln^2 x + \underline{\qquad}\cdot h + O(h^2)ln2(x+h)=ln2x+ββ h+O(h2)
Here, \ln^2 xln2x means \left(\ln x\right)^2(lnx)2.
- \displaystyle \frac{2}{x+h}\ln(x+h)x+h2βln(x+h)
- 2\ln x2lnx
- \displaystyle \frac{2}{x}x2β
- \displaystyle \ln \frac{2}{x}lnx2β
- 22
- \displaystyle \ln \frac{1}{x}lnx1β
- \displaystyle 2\frac{\ln x}{x}2xlnxβ
- 2\ln(x+h)2ln(x+h)
Q9. Recall that the kinetic energy of a body is
K = \frac{1}{2}mv^2K=21βmv2
where mm is mass and vv is velocity. Compute the relative rate of change of kinetic energy, \displaystyle\frac{dK}{K}KdKβ, given that the relative rate of change of mass is -7β7 and the relative rate of change of velocity is +5+5.
- \displaystyle\frac{dK}{K}=-2KdKβ=β2
- \displaystyle\frac{dK}{K}=-\frac{7}{2}KdKβ=β27β
- Not enough information is given to solve the problem.
- \displaystyle\frac{dK}{K}=\frac{3}{2}KdKβ=23β
- \displaystyle\frac{dK}{K}=5KdKβ=5
- \displaystyle\frac{dK}{K}=-7KdKβ=β7
- \displaystyle\frac{dK}{K}=3KdKβ=3
- \displaystyle\frac{dK}{K}=-9KdKβ=β9
Q10. Compute the ninth derivative of (x-3)^{10}(xβ3)10 with respect to xx.
- 10(x-3)^910(xβ3)9
- \displaystyle\frac{1}{9!}(x-3)^99!1β(xβ3)9
- 9!9!
- 11
- 9!(x-3)9!(xβ3)
- 10!10!
- 10!(x-3)10!(xβ3)
- 00
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:
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In this second part, the second of five, we talk about derivatives, differentiation rules, linearization, higher derivatives, optimization, differentials, and differentiation operators.
SKILLS YOU WILL GAIN
- Differential (Mathematics)
- Newton’S Method
- Linear Approximation
- Differential Calculus
- Derivative
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