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Coursera Calculus I Fake Midterm 2 problems solutions.

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 space-faring civilization's territory as a perfect circle of increasing radius. The area that this space-faring civilization controls grows at a rate proportional to its frontier. It took this civilization one thousand years to control 100 square light-years, and then another nine thousand years to control ten thousand square light-years. How long will it take this civilization to control the whole galaxy, which is a disk of area 1010 square light-years? 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 cross-section 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

+ arc