MOCK FINAL EXAM - SOLUTIONS

Transcription

1 MOCK FINAL EXAM - SOLUTIONS PEYAM RYAN TABRIZIAN Name: Instructions: You have 3 hours to take this eam. It is meant as an opportunity for you to take a real-time practice final and to see which topics you should focus on before the actual final! Even though it counts for 0% of your grade, I will grade it and comment on it overnight, and you can pick up the graded eam tomorrow at noon in my office (830 Evans) Note: Questions are a bit more challenging (although not impossible) than the rest! They are meant to be an etra challenge for people who finish early (and hence they are only worth 5 points each) Note: Please check one of the following boes: I will pick up my eam tomorrow between noon and 5 pm, and I want comments on my eam (Peyam/SkyDrive Tabrizian approves of this choice :) ) I will pick up my eam tomorrow between noon and 5 pm, but I don t want comments on my eam (I only want to know my score) I will not pick up my eam tomorrow, just grade it and enter my score on bspace! Total 00 Date: Monday, May 9th,

2 PEYAM RYAN TABRIZIAN 1. (15 points, 3 points each) Evaluate the following integrals: (a) 1 0 d 1 = [ sin 1 () ] 1 0 = sin 1 (1) sin 1 (0) = π 0 = π (b) 1 sin( 3 )( ) 1 cos()+ d = 0 (because the function inside of the integral is odd) (c) 0 4 d = 1 4 π = π (the integral represents the area of the quarter circle of radius in the upper-left quadrant! (d) cos() sin () d Let u = sin(), then du = cos()d, so: cos() du sin () d = u = 1 u + C = 1 sin() + C = csc() + C (e) 1 ln() d Let u = ln(), then du = 1, and u() = ln(), and u(1) = ln(1) = 0, so: 1 ln() d = ln() 0 [ u udu = ] ln() 0 = (ln())

3 MOCK FINAL EXAM - SOLUTIONS 3. (10 points) (a) (8 points) Show that the function f() = cos() has at least one zero. f(0) = 1 > 0, f( π ) = π < 0, and f is continuous on [ 0, π ], so by the Intermediate Value Theorem, f has at least one zero (in [ 0, π ] ). (b) ( points) Using part (a), show that the function g() = sin() least one critical point. has at Notice that g () = cos() = f(). We ve shown in (a) that f has at least one zero, so g has at least one zero, so g has at least one critical point!

4 4 PEYAM RYAN TABRIZIAN 3. (0 points) Sketch a graph of the function f() = ln(). Your work should include: - Domain - Intercepts - Symmetry - Asymptotes (no Slant asymptotes, though) - Intervals of increase/decrease/local ma/min - Concavity and inflection points (1) Domain: > 0 () No y intercepts, intercept = e (f() = 0 ln() = 0 ln() = ln() = 1 = e) (3) No symmetry (4) NO H.A. or V.A., because: lim ln() (ln() 1) (H means l Hopital s rule), and: ln() H = 0 lim ln() (ln() 1) = = (5) f () = ln() = ln(), f is decreasing on (0, 1) and increasing on (1, ), (1, 1) is a local minimum by the first derivative test. (6) f () = 1 > 0 (if > 0), f is concave up on (0, ), no inflection points (7) Graph: 1A/Practice Eams/Mockgraph.png

5 MOCK FINAL EXAM - SOLUTIONS 5 4. (10 points) Using the definition of the integral, evaluate ( ) d. You may use the fact that n i=1 i = n(n+1) and n i=1 i3 = n (n+1) 4 = n, i = a + i = i n d n i=1 n i=1 n i=1 16 n n 4 n f( i ) ( n (i ) 3 + n n 16i 3 n + n 4 i=1 n i n i=1 16 n (n + 1) n n 4 4 n =4 + =6 4(n + 1) n + ( ) ) i n n 4i n n i=1 i + 4 n(n + 1) n (n + 1) n

6 6 PEYAM RYAN TABRIZIAN 5. (15 points, 5 points each) Evaluate the following limits: (a) lim 0 + sin ( 1 ) 1 sin ( 1 ) 1, so sin ( 1 ). Now lim 0 + = 0 and lim 0 + = 0, so by the squeeze theorem lim 0 + f() = 0 lim (b) lim Where we used the fact that = = (since < 0 here!) ( ) (c) lim 1 + 1) y = ( 1 + ) ) ln(y) = ln(1 + ) 3) lim ln(y) ln(1 + ) ln(1 + ) 1 H = 4) Hence lim y = e = 1

7 MOCK FINAL EXAM - SOLUTIONS 7 6. (15 points) Find the area between the curves 4 + y = 1 and = y The picture is given as follows (you get it by reflecting the graph 4y + = 1 about the line y = : 1A/Practice Eams/Mockarea.png Here, the rightmost function is = y, and the leftmost function is = y. Now, for the points of intersection, we need to solve: Hence, the area is: 4 + =1 ( + ) =16 + = ± 4 = 6, 6 3 y 4 ydy = [3y y3 1 y ] 6 = = 64 3

8 8 PEYAM RYAN TABRIZIAN 7. (10 points) Suppose f is an odd function and is differentiable everywhere. Prove that, for every positive number b, there eists a number c in ( b, b) such that f (c) = f(b) b (weird question, huh? =) ) Since this is a weird question, it s a Mean Value Theorem question! :) By the Mean Value Theorem applied to [ b, b], we get that, for some c in ( b, b): f(b) f( b) = f (c) b ( b) But f( b) = f(b) since f is odd, and b ( b) = b, so: That is: f(b) b = f (c) f(b) = f (c) b Which is what we wanted to show!

9 MOCK FINAL EXAM - SOLUTIONS 9 1A/Practice Eams/Whale.jpg

10 10 PEYAM RYAN TABRIZIAN 8. (10 points) Find the volume of the solid obtained by rotating the region bounded by y =, y = about y = 1 This is a typical application of the washer method! Here k = 1, Outer = 1, Inner = 1 (see the following picture): 1A/Practice Eams/Mockwasher.png Hence 1 0 π ( ( 1) ( 1) ) d =π =π d 3 + d [ 3 =π =π( ) ] 1 0 = π 6

11 MOCK FINAL EXAM - SOLUTIONS (15 points) A lighthouse is located on a small island 3 km away from the nearest point P on a straight shoreline, and the angular velocity of the light is 8π radians per minute. How fast is the beam of light moving along the shoreline when it is 1 km away from P? This is problem 38 in section 3.9! (in the 6th edition of the tetbook) 1) Again, draw a picture of the situation: 1A/Archive/Solution Bank - Old Edition/Lighthouse.png ) Want to find d dt when = 1 3) tan(θ) = 3 So = 3 tan(θ) 4) Hence d dt = 3 sec (θ) dθ dt 5) We re given that dθ dt = 8π (you put a minus-sign since is decreasing!). Moreover, by drawing the eact same picture as above, ecept with = 1, we can calculate sec (θ), namely:

12 1 PEYAM RYAN TABRIZIAN sec(θ) = hypothenuse 10 = adjacent 3 (and the 10 we get from the Pythagorean theorem!) ( 10 ) It follows that sec (θ) = 3 = Whence d dt = 80π 3 ft/min d dt =3 sec (θ) dθ dt d dt =3(10 9 )( 8π) d dt = 40π 9 d dt = 80π 3

13 MOCK FINAL EXAM - SOLUTIONS (10 points, 5 points each) Find the derivatives of the following functions: (a) f() = sin 1 () 1 f 1 ( ) () = 1 + sin 1 () 1 1 = 1 sin 1 () 1 (b) f() = ln() 1) y = ln() ) ln(y) = ln() ln() = (ln()) 3) y y = ln() ( 4) y = y ln() ) = ln() ( ln() )

14 14 PEYAM RYAN TABRIZIAN 11. (0 points) Find the volume of the donut obtained by rotating the disk of center (3, 0) and radius about the y ais. Picture: 1A/Practice Eams/Mocktorus.png V = For this, let s use the shell method! = 0, so k = 0, and Radius = 0 = =. Also, the equation of the circle is ( 3) +y = 4, so y = ± 4 ( 3), and Outer = 4 ( 3) and Inner = 4 ( 3), so Height = Outer - Inner = 4 ( 3). Hence: V = 5 1 π ( ) 4 ( 3) d = 5 1 4π 4 ( 3) d Now let u = 3, then du = d, = u + 3, and u(1) =, u(5) =, so: 4π(u + 3) 4 u du = 4π u 4 u du + 1π 4 u du However, the first integral is 0 because u 4 u is an odd function, and the second integral is 1 π = π, because it represents the area of a semicircle of radius! Hence, we get: V = 0 + 1π(π) = 4π

15 MOCK FINAL EXAM - SOLUTIONS (15 points) Show that the equation of the tangent line to the ellipse a at ( 0, y 0 ) is: 0 a + y 0y b = 1 + y b = 1 First of all, by implicit differentiation: a + yy b =0 ) y ( y b = a y = b a y y = b a y It follows that the tangent line to the ellipse at ( 0, y 0 ) has slope b a 0 y 0, and since it goes through ( 0, y 0 ), its equation is: y y 0 = ( b 0 a y 0 ) ( 0 ) And the rest of the problem is just a little algebra! First of all, by multiplying both sides by a y 0, we get: Epanding out, we get: Now rearranging, we have: (y y 0 )(a y 0 ) = b 0 ( 0 ) ya y 0 a (y 0 ) = b 0 + b ( 0 ) ya y 0 + b 0 = a (y 0 ) + b ( 0 ) Now dividing both sides by a, we get: yy 0 + b a 0 = (y 0 ) + b a ( 0) And dividing both sides by b, we get: yy 0 b + 0 a = (y 0) b + ( 0) a But now, since ( 0, y 0 ) is on the ellipse, (y0) b 0 a + y 0y b = 1 + (0) a = 1, we get:

16 16 PEYAM RYAN TABRIZIAN 1A/Practice Eams/Snake.jpg

17 MOCK FINAL EXAM - SOLUTIONS (15 points) Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r. 1) First draw a good picture! 1A/Practice Eams/Mockrectangle.png ) Based on your picture, the length of the rectangle is and the width is y, and the area is A = ()(y) = 4y. But since (, y) is on the circle, + y = r, so y = r, so A() = 4 r. 3) The constraint is 0 r 4) A () = 4 r + 4 r = ( 4 r r ) = 0 r = 0 = r. Also, A(0) = A(r) = 0, and A( r ) > 0 (we don t really care what it is, as long as it s positive), so by the closed interval method, = r is an absolute maimizer. So our answer is: Length = = r, W idth = y = So the optimal rectangle is a square!!! r r = r = r =

18 18 PEYAM RYAN TABRIZIAN Note: I would like to remind you that questions are more challenging than the rest, but you can give them a try if you want to, they are not impossible to do! 14. (5 points) Solve the differential equation T = T 5. Hint: Let y = T 5. What differential equation does y solve? y = T = T 5 = y, so y = y, so y = Ce t, so T 5 = Ce kt, so T = 5 + Ce t

19 MOCK FINAL EXAM - SOLUTIONS (5 points) If f is continuous on [0, 1], show that 1 f()d is finite. 0 Since f is continuous on [0, 1], by the etreme value theorem, f attains an absolute maimum M and an absolute minimum m, so m f() M. Integrating, we get: 1 0 md 1 0 f()d 1 0 Md, so m 1 0 f()d M, so 1 f()d is finite because both m and M are finite! 0

20 0 PEYAM RYAN TABRIZIAN 16. (5 points) (a) Use l Hopital s rule to show: f( + h) f() + f( h) lim h 0 h = f () f( + h) f() + f( h) H f ( + h) f ( h) lim h 0 h h 0 h H f ( + h) + f ( h) h 0 = f () + f () =f () (The tricky part is that we re differentiating with respect to h here, not with respect to ) (b) Use (a) to answer the following question: If f() = sin( 1 ) with f(0) = 0, does f (0) eist? The above formula with = 0 gives: f f(h) f(0) + f( h) (0) h 0 h But here f(0) = 0, f(h) = h sin( 1 h ), and f( h) = ( h) sin( 1 h ) = h sin( 1 h ) = f(h). Hence: So f (0) = 0 f f(h) 0 f(h) 0 (0) h 0 h h 0 h = 0

21 MOCK FINAL EXAM - SOLUTIONS (5 points) If f is differentiable (ecept possibly at 0) and lim f() = 0, is it true that lim f () = 0? Prove it or give an eplicit countereample! You might think this is true, but it is actually FALSE! Let f() = sin( ). Then f is differentiable ecept at 0, lim f() = 0 by the squeeze theorem. Moreover: f () = cos( )()() sin( ) = cos( ) sin( ) And lim f () 0. In fact, the limit does not eist! Although lim sin( ) = 0 by the squeeze theorem, lim cos( ) does not eist, which causes lim f () not to eist! The interesting thing about this function is that although it goes to 0 at, it oscillates wildly, and the oscillations are faster than rate of convergence of the function at 0 (that s why I chose the factors and, this wouldn t work with sin(), its oscillations are not that bad!)

22 PEYAM RYAN TABRIZIAN You re done!!! Any comments about this eam? (too long? too hard?) 1A/Practice Eams/Soccer.jpg