Coursera Calculus I Fake Midterm 2 (March 2013)

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Coursera Calculus I Fake Midterm 2 (March 2013)Jabari Zakiya
Question 1Evaluate Because or state that the limit does not exist. use L' Hopital's Rule: =
Question 2Evaluate Because do: or state that the limit does not exist. = =
= =
= = = =
Question 3Which of the following is the derivative of arcsin(tan(x))? a) b) c) d) none of the above
Question 4Three changing quantities a, b, and c have the property that a4 + b4 + c4 = 4, even as they change. At this particular moment, a is increasing and b is decreasing. What can be said about c? a) The quantity c is decreasing c) The quantity c is increasing b) The quantity c is staying the same d) There is not enough information to determine the answer. where not enough information to characterize c
4a3 da/dt + 4b3 db/dt + 4c3 dc/dt = 0 =
Question 5A giant hailstone gathers even more ice as it hurdles towards me, in such a way that it remains in the shape of a perfect sphere, of radius length r, and the rate of change in its volume is proportional to its surface area. At the beginning of our story, r is half a centimeter, but after one minute has elapsed r is one cm. How much longer (in minutes) must I wait until r is 2 centimeters? a) 4 b) 2 c) 6 d) The answer cannot be determined from information given We need to find an expression of r in terms of t, i.e. r(t). volume of sphere is: at t = 0, r = at t = 1, r = 1 therefore: = equals some value k so: k= and thus: r(t) = t+C and so = and thus
now use initial conditions to solve for k and C: for t = 0, r = for t = 1, r = 1 therefore: now find t when r is 2: = 1= (0) + C (1) + gives gives C= k=
r(t) = t + 2= t+ t= = 3 minutes
Since r was 1 after 1 minute, and 2 after 3 minutes, thus we had to wait 2 more minutes to get there.
Question 6Model a spacefaring civilization's territory as a perfect circle of increasing radius. The area that this spacefaring civilization controls grows at a rate proportional to its frontier. It took this civilization one thousand years to control 100 square lightyears, and then another nine thousand years to control ten thousand square lightyears. How long will it take this civilization to control the whole galaxy, which is a disk of area 1010 square lightyears? a) 100 million years b) 500 million years c) > 100 billion years d) 1 million years e) 10 million years We need to find an expression of r in terms of t, i.e. r(t), which controls the circle's rate of growth. area of circle is: and thus 100 = 10000 = 1010 = gives r = gives r = gives r =
t = 1000, A = 100 t = 10000, A = 10000 t= ? A = 1010
t = 1000, r = t = 10000, r = therefore: = equals some value k so: k= thus, using time shifting: r(t) = and so (t 1000) + C
now use initial conditions to solve for k and C: t = 1000, r = t = 10000, r = therefore: now find t when r is : = = (1000 1000) + C (10000 1000) + r(t) = = t= t = 107 (or 10 million) years (t 1000) + (t 1000) + gives C = gives k =
Question 7I have spilled 2 cm3 of truffle oil on my kitchen floor. This delicious oil slick can be modeled as a cylinder with a very small height; the height is decreasing at a rate of cm/sec and the radius is currently 5 cm. How quickly is the radius of the oil slick increasing, in cm/sec? Your answer should be a fraction of whole numbers. First find cylinder height h of volume of 2 cm3 cylinder with 5 cm radius: V= , h= cm is 0, and because height is decreasing is negative, therefore:
Because V is constant,
and thus: 0= cm/sec So the radius is increasing, while the cylinder is getting flatter.
Question 8I want to take a one meter long length of wire, and build a "house" shaped figure, like the one shown to the right. The bottom of my house is to be a square, and the roof is to be an isoceles triangle. I must use all of the wire. Which value is closest to the area of the largest such figure I can build, using all of the wire? This problem is somewhat harder than the corresponding problem in the real exam. a) 520 cm2 b) 625 cm2 c) 300 cm2 d) 430 cm2 Let x be length of square sides and y the length of the two roof (triangle) sides, then 1 (meter) = 4x + 2y; y = (1 4x)/2 and from triangle Total Area (TA) = Area of Square + Area of Isoceles Triangle = TA = TA = + + = + = + + ;
now differentiate TA with respect to x, = 2x + (1/4) (1 8x + 15x2)1/2 + (x/8)(1 8x + 15x2)1/2 (30x 8) = 0 2x + (1/4) (1 8x + 15x2)1/2 + x(30x 8)/8(1 8x + 15x2)1/2 = 0 16x(1 8x + 15x2)1/2 + 2(1 8x + 15x2) + x(30x 8) = 0 16x(1 8x + 15x2)1/2 + 2 16x + 30x2 + 30x2 8x = 0 16x(1 8x + 15x2)1/2 + 60x2 24x + 2 = 0 To solve for x (find a root of the expressions) use Newton's Method. Let then y(x) = 16x(1 8x + 15x2)1/2 + 60x2 24x + 2 = 0 y'(x) = 120x 24 + 16(1 8x + 15x2)1/2 + 8x(1 8x + 15x2)1/2 (30x 8) xn+1 = xn y(xn)/y'(xn)
and thus:
Ruby code to perform Newton's Method.defy(x);16*x*(18*x+15*x*x)**(0.5)+60*x*x24*x+2end defyy(x);120*x24+16*(18*x+15*x*x)**(0.5)+8*x*(18*x+15*x*x)**(0.5)*(30*x8)end x=0.17;5.times{x=y(x)/yy(x)};x=>0.19215377322680077
with this value for x, we get TA =
+
= 0.0431 meters2 = 431 cm2
[Can also check graphically by plotting y(x) (say with http://www.geogebra.org, et al) and see that y(x) = 0 close to x = 0.19]
Question 9To build a beautiful trough with a semicircular cross section in celebration of day on March 14, I took a by piece of metal, bent it in semicircle, and welded two semicircles of radius 1 meter to each of the open sides. To keep up the celebration of day, I am pouring water into my trough at a rate of cubic meters per second, and at this moment there is cubic meters of water in my trough. How quickly is the height of the water changing, in meters/sec? This problem is quite a bit more challenging than the corresponding problem on the real exam; it might help to think about how to use trigonometry on this problem. a) 2 m/sec b) 3 m/sec c) 5 m/sec d) 1 m/sec e) m/sec f) 4 m/sec
Fig. 1
The volume of the full trough is half the volume of a cylinder: A line parallel to the top of the semicircle constitutes the current volume of water in the trough. Let h be the height of the amount of water from the bottom of the trough to the waterline and let H be the remaining height from the waterline to the top of the trough. Thus r = h + H and H = r h. A line from the midpoint at the top of the trough to the point on the semicircle where the waterline meets creates a right triangle of height H, and an angle from top of the trough to the waterline of . The area above the waterline is the area of the two right triangle of height H and the area of the two arcsections subtended by angle . Area of Triangle = Area of Arcsection = The area of the crosssection of water in the trough at any time is, thus, the total area of the semicircle minus twice the area of the right triangle of height H and twice the area of the arcsection of angle , A= and because r = 1 meter; A = bH = , so A = 0. When full, and H are 0, so A = = ]
=
where b is the base of the right triangle from the top of the waterline to the edge of the circle in Fig. 1.[When empty, H = r, but b = 0, so bH = 0 and
Now write b, H and
is term of h, using fact the r = 1 meter.
h + H = r; h + H = 1; H = 1 h and therefore and also so
So now the volume of the water in the trough at any height is: V = A * length = A* 1) 2) V=( V=( bH ) (1 h) =( = 3) Notice in 3) that the + = bH + bH ) bH arcsin(1 h))
gives
Using the initial volume condition V =
equate this to expression 1)
[Here is where mathematical insight and inspiration comes into play to make this problem simple.] term is the value of an angle (in radians), therefore equate: = Therefore must equal bH. Remember, we are trying to find h (the height at this given volume of water). There are two ways to do this, the hard way and the easy way! Let's do the easy way first. Since we know 4) = , then from Fig. 1, b = cos = cos( = bH = H/2 = (1 h)/2 =1 h h=1 Now let's find h the hard(er) way. Using 5) = bH = (1 h) 3/16 = (1 h)2 (2h h2) 3/16 = 2h 5h2 + 4h3 h4 Newton's Method can now be used to find h from y(h) (a 4th degree polynomial with 4 roots). Let and then and to find h iterate: y(h) = 2h 5h2 + 4h3 h4 3/16 = 0 y'(h) = 2 10h + 12h2 4h3 hn+1 = hn y(hn)/y'(hn) = 1 0.87 = 0.13 and H = (1 h) we get 4) to be: ) = 1/2
Below is Ruby code to perform Newton's Method with h0 that give correct root value of h.defy(h);2*h5*h*h+4*h**3h**43.0/16end defyy(h);210*h+12*h*h4*h**3end h=0.0;10.times{h=hy(h)/yy(h)};h=>0.13397459621556132 h=0.25;10.times{h=hy(h)/yy(h)};h=>0.13397459621556135 h=1(3**0.5)/2=>0.1339745962155614
[Other h0 values < 2 give all 4 roots of y(h): h1,2,3,4 = 1 0.5 (0.5,1.5) and 1 (0.13,1.87) but h must be < r=1, so h = 1.5 or 1.87 are invalid and h = 0.5 would reduce the value of from .] The really interesting question may be how would you find h if you didn't see (have the insight) to equate = and then = bH ? (This is frequently done with complex numbers, where with an expression like (3 x) + (5 + y)i = 7 + 4i you equate the real and imaginary parts of both sides to solve for x and y). The answer is YES!, we can find h using Newton's Method.
We can brute force our way to finding h from 3) using Newton's Method. + + Let and then y(h) = y'(h) = y'(h) = and to find h iterate: hn+1 = hn y(hn)/y'(hn) The Ruby code to do this is below:includeMath defy(h);(2*hh*h)**0.5h*(2*hh*h)**0.5+asin(1h)(PI/3+(3**0.5)/4)end defyy(h);2*(2*hh*h)**0.5end h=0.0;5.times{h=hy(h)/yy(h)};h=>(NaN+NaN*i) h=0.1;5.times{h=hy(h)/yy(h)};h=>0.13397459621556132 h=0.25;5.times{h=hy(h)/yy(h)};h=>0.13397459621556143
= bH + = (1 h) + arcsin(1 h) + )=0
h
+ arcsin(1 h) ( = +
=
=
Thus we see, we can find the height h for any initial volume of water by brute force using Newton's Method. In fact, this becomes the generic process for solving for h for any initial volume amount without the need of applying any mathematical insight to the form or value of the volume expression. Now that we have determined h we are ready to find dh/dt for the given dV/dt, using 2) V(h). V= V= dV/dt = d h{ =0 1 = d h[ 1 = [ 1 = [ 1 = 1 = = = = = = 0.99999999 = 1 meter/sec d h[ h [ (1 h) [(1 h) [(1 h) h arcsin(1 h)] + arcsin(1 h)] + arcsin(1 h)] } /dt + arcsin(1 h)] /dt + arcsin(1 h)] /dt + ] ]
Question 10Consider the ellipse plotted by 3x2 + 2y2 = 4. Find a polynomial with a root at the xcoordinate of the point on the ellipse that is closest to the point (2,2). This problem is quite a bit harder than the corresponding problem on the real exam. a) 3x4 + 24x3 + 116x2 31x 64 c) 3x4 + 24x3 + 116x2 30x 64 b) 3x4 + 24x3 + 116x2 33x 64 d) 3x4 + 24x3 + 116x2 32x 64
To find the xcoordinate that minimizes h, the distance from the point (2,2) to the ellipse, write h in terms of x, then differentiate it and set to zero. 3x2 + 2y2 = 4 and h2 = (2 y)2 + (2 x)2 = 4 4y + y2 + 4 4x + x2 = 8 4(x +y) + y2 + x2 h2 = h2 = 10 4x 4 2h = 4 2 = (4 + 6x 4 + 6x 6x 36x2 72x2 (3x) x x)/2h = 0 x=0 =x+4 = (x + 4)2
= (x + 4)2 72x2 = (x + 4)2 (4 3x2) 2 2 72x (x + 8x + 16)(4 3x2) = 0 2 2 72x (4x + 32x + 64 3x4 24x3 48x2 ) = 0 72x2 4x2 32x 64 + 3x4 + 24x3 + 48x2 = 0 3x4 + 24x3 + 116x2 32x 64 = 0 Newton's Method can now be used to find x from y(x) (a 4th degree polynomial with 4 roots). Let and then and to find x iterate: y(x) = 3x4 + 24x3 + 116x2 32x 64 = 0 y'(x) = 12x3 + 72x2 + 232x 32 xn+1 = xn y(xn)/y'(xn) with x0 = 0.5
The best root is x = 0.80888 which makes y = 1.00924 and then h = 1.54931 is shortest distance.
Question 11You have a 10 cm by 15 cm sheet of cardboard that you wish to fold into a box with an open top. You plan to cut x centimeter squares out of each of the four corners, and fold up the sides. Find the value of x that maximizes the volume. a) cm b) cm c) cm d) cm
V = (10 2x)(15 2x)x = (150 50x + 4x2)x = 4x3 50x2 + 150x = 12x2 100x + 150 = 0 6x2 50x + 75 = 0 and x = x= x= = = = cm, because twice ( + ) is > 10 cm. or gives
Question 12A certain acting troupe puts on plays inspired by calculus. The troupe agrees to put on a show in your home town, provided that at least one person shows up. The troupe charges $10 if only one person is in the audience, $9.75 per person if there are two audience members, $9.50 per person if there are three members, and so on. At most how much money can the troupe expect to bring in? a) $75.50 b) $105.06 c) $20.50 d) $21.00 e) $105.00 f) $21.00
You can create a chart to find the revenue per number of people, as below: number of people $ price per person total revenue 1 2 3 4 10 9.75 9.50 9.25 10 19.50 28.50 37.00 5 ......... 9.00 ......... 45.00 .........
We can translate this chart into a curve. Let p be the number of people attending a show. Total Revenue (TR) = p(10 (p 1)(0.25)) = p(10 0.25p + 0.25) = 10.25p 0.25p2 = 10.25 0.5p = 0, so p = 10.25/0.5 = 20.5 to maximum TR Since there can only be whole numbers of people 20 or 21 people will maximize revenue: Maximum TR is: TR(20) = TR(21) = 10.25(20) 0.25(20)2 = 10.25(21) 0.25(21)2 = $105.00
Question 13Find the x coordinate of the point on the curve that is closest to the origin.
This question is quite a bit harder than the corresponding question on the real exam; to make it a bit easier, remember that minimizing distance to (0,0) is the same as minimizing the value of x2 + y2, which is a bit easier to deal with than . a) 0 b) 2 c) 3 d) 1 Want to find the x value that minimizes distance h from origin to curve y.
=0 Set numerator to zero and solve for x using Newton's Method: xn+1 = xn y(xn)/y'(xn) Let and then y(x) = y'(x) = 1 + =0
Ruby code to perform Newton's Method with x0 = 0:defy(x);x+(3.3*x*x48*x+36043.0/435)*(6.6*x48)end defyy(x);1+(6.6*x48)**2+(3.3*x*x48*x+36043.0/435)*6.6end x=0;5.times{x=xy(x)/yy(x)};x=>1.9999999999997298 x=0;6.times{x=xy(x)/yy(x)};x=>2.0
When x = 2.0 then y = 0.05747 and h = 2.00083 is shortest distance.
Question 14Which of the following could be a slope field for = cos x + sin y?
a
b
c The answer is b.
d
Question 15Which of the following could be a slope field for ?
a
b
c The answer is d.
d
Question 16For which of the following functions f is it the case that f '(x) = 2x + 2f(x)x? In other words, which function satisfies the differential equation a) f(x) = ex b) f(x) = sin(x) c) f(x) = e2x , y' = = d) f(x) = ?
The answer is d): y =
Question 17Suppose f is a function for which f(2) = 0, and which satisfies the equation f ' (x) = f(x)2 4 To get a sense of the function f, you may wish to consider the slope field , since you can recover f by starting at (2,0) and following the lines. Place a checkmark next to the true statements below.
a) f is increasing at x = 4 b) c) f(x)2 is < 4 for all x d) f is decreasing at x = 2 e)
The answer is: c), d), and e) are true.
Question 18Consider the interval I = [8,14]. Break I into three subintervals of equal length, namely the three subintervals [8,10], [10,12], [12,14]. Suppose that f(8) = 2, f '(8) = 3, f '(10) = 4, f '(12) = 5. What is the approximate value of f(14) by applying Euler's method? for h =2: yn+1 = yn + h*f '(yn) f(10) = f(8) + 2*f '(8) = 2 + 2(3) = 2 + 6 = 8 f(12) = f(10) + 2*f '(10) = 8 + 2(4) = 8 + 8 = 16 f(14) = f(12) + 2*f '(12) = 16 + 2(5) = 16 10 = 6
Question 19The function f satisfies the differential equation , meaning f ' = x2 and also suppose f(0) = 1. Will applying Euler's method overapproximate or underapproximate the actual value f(2)? a) overapproximate try h = 0.5: b) underapproximate
yn+1 = yn + h*f '(yn) with x0 = 0 and f(0) = 1. f(0.5) = f(0) + (0.5)*f '(0) = 1 + (0.5)((0)2) = 1 0 = 1 f(1.0) = f(0.5) + (0.5)*f '(0.5) = 1 + (0.5)((0.5)2) = 1 (0.125) = 0.875 f(1.5) = f(1.0) + (0.5)*f '(1.0) = 0.875 + (0.5)((1.0)2) = 0.875 0.5 = 0.375 f(2.0) = f(1.5) + (0.5)*f '(1.5) = 0.375 + (0.5)((1.5)2)= 0.375 1.25 = 0.75
Because f '(x) = x2 then f (x)= x3/3 + C and f(0) = 1 = (0)3/3 + C makes C = 1 Then f(2) = (2)3/3 + 1 = 1 8/3 = 1.6666 is < 0.75 so Euler's Method overapproximates f(2).
Question 20Suppose f satisfies the differential equation f '(x) = f(x) + f(x)2 and f(1) = 3. Use Euler's method with a step size of 0.01 to approximate f(1.02). a) b) c) d) and f(1) = 3. In other words,
for h = 0.01: yn+1 = yn + h*f '(yn), with x0 = 1 and f(1) = 3. f(1.01) = f(1.00) + (0.01)*f '(1.00) = 3 + (0.01)(3+32) = 3.12 f(1.02) = f(1.01) + (0.01)*f '(1.01) = 3.12 + (0.01)(3.12+(3.12)2) = 3.248544 therefore, f(1.02) = 3.248544 =
Question 21Let f(x) = x5 x 2 . Then f has a root somewhere between x = 0 and x = 2. Using x0 = 2 as your first value for Newton's method, compute the third approximation x2. a) b) c) with x0 = 2
Newton's Method gives: xn+1 = xn f(xn)/f '(xn) = xn Ruby code to perform this is given below:
x=2.0;2.times{x=x(x**5x2)/(5*x**41)};x=>1.4094484436882508
Therefore, x2 = 1.4094484436882508 =
Question 22How many iterations may Newton's method require to approximate a root of a polynomial to within 1/100 ? a) 100 iterations b) 17 iterations c) Newton's method may never come within of a root.
d) Newton's method is guaranteed to find a root to within 1/100 as long as you repeat until two successive approximations are within, say, 1/1000.
Question 23Which of the following is an antiderivative of x3cos(x) + 3x2sin(x) with regard to x ? a) sin2(x) + 2 b) sin(x3) + cos(3) c) x2sin(x) + e2 d) x3sin(x) +
Question 24Is x3/3 an antiderivative of (x sin(x))2 + (x cos(x))2 with respect to x? a) No b) Yes (x sin(x))2 + (x cos(x))2 = x2(sin2x + cos2x) = x2(1) = x2 which is the derivative of x3/3.
Question 25Suppose F(x) is an antiderivative of f(x) = log log x. Is F increasing or decreasing near x = 10? a) Decreasing b) Increasing Since F(x) is an antiderivative of f(x) = log log x then f(x) is the derivative of F(x). Therefore, the sign of f(x) indicates whether F(x) is increasing (positive) or decreasing (negative). At x = 10 the derivative f(10) = log log (10) = 0.834032445247956 is positive so F(x) is increasing.