SOLUTION - BANK PEYAM RYAN TABRIZIAN

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1 SOLUTION - BANK PEYAM RYAN TABRIZIAN This document contains solutions to various Math A homework problems I ve compiled, from Fall 00 and Spring 0. At first, the solutions might seem disappointingly short, but don t worry! As the course progresses, the solutions become longer and more detailed! It contains solutions to the following problems: Section Problems. 6, 8, 3, 45, 57, 6., 4, 8, 6.3, 7, 4, 30, , 5, 6, 6, 7, 8.6 3, 5, 6, 8, 6, 35, 36, 39, 48, 58, 60, 64, 65, 66., 6, 8, 9, , 0, 3, 7, 8, 6, 9, 38, 47, 58, 59.4, 4, 9, 3, 37, 4, , 8, 9, 7, 34, 40, 47, 56, ,, 6, 3, 34, ,, 7, 8, 3, 34, 40, ,, 38, 43, 5 3., 3, 0, 3, 35, 49, , 7, 33, 4, 47, , 3, 37, 39, 40, , 5, 39, 45, 49, 63, 66, 70, 83, 84, , 9, 7, 8, 30, 40, 46, 49, 67, ,, , 0, 7, 4, , 9,, , 3, 5, 7, 38, ,, 5,, 5, 35, ,, 5,, 3(a), 6, 9(a)(b), 3, , 38, 39, 5, , 3, 8, 3, 9, 3, 34, , 3, 7, 43, 45, , 4,, 4, 5, 3, 7, 9, 40, 47, 49, 5, 56, 59, 7 4.5, 3, 4, ,, 9,, 30, 53, , 4, 33, 39, 6, 74 Continued on the net page Date: Monday, June 0th, 0.

2 PEYAM RYAN TABRIZIAN Section Problems 5., 5, 5, 9,, 6 5., 8,, 3, 30, 34, 36, 43, 44, 47, 5, 5, 67, , 7, 5, 7, 5, 3, 35, 4, 43, ,, 3, 5, 37, 47, 5, 58, 59, 6, , 3, 39, 46, 59, 73, 88 6., 3, 3,, 40, 4, , 6, 3, 7, 49, 5, 57, 67, , 3, 5, 9, 44, 46 Section.: Four ways to represent functions..6. Yes (by the vertical line test), Domain = [, ], Range = [, ]..8. (a) The graph of (t) should just be a line going through the origin (b) The graph of y(t) should look at first like the right half of a parabola, then should be constant for a while, and then look like the left half of a parabola (c) The graph of the horizontal velocity looks like a horizontal line (d) See announcement on bspace for a detailed solution! The picture you get is: A/Math A Summer/Solution Bank/Vertical Velocity.png

3 SOLUTION - BANK Domain = [, ], Range = [0, ], Graph is just the upper-half of the circle centered at 0 of radius f() = V () = (0 )( ) (no need to epand the answer!)..6. f is odd, g is even Section.: Mathematical models: a catalog of essential functions (a) Rational function (b) Algebraic function (c) Eponential function (d) Power function (e) Polynomial of degree 6 (f) Trigonometric function (a) G (b) f (c) F (d) g..8. (a) y = ( 3), (b) y = (a) C() = (C is the cost and is the number of chairs produced) (b) 3; Cost per chair (c) 900; Start-up cost (i.e. money needed to buy machines in order to start producing chairs) (a) y = f() + 3 (b) y = f() 3 (c) y = f( 3) (d) y = f( + 3) (e) y = f() (f) y = f( ) (g) y = 3f() (h) y = 3 f() Section.3: New functions from old functions.3.7. y = 3( + 4) ( + 4).3.4. Basically compress the graph of sin() horizontally by a factor of 3 (notice that the new period now is π 3 and then stretch the resulting graph vertically by a factor of 4 (so the new graph has range [ 4, 4] instead of [, ])

4 4 PEYAM RYAN TABRIZIAN (a) (f + g)() = 3 + (b) (f g)() = 3 + (c) (fg)() = 3 (d) ( f g ) () = 3 All of those functions have domain (, ] [, 3] EXCEPT for (d), which has domain (, ) (, 3] sin() (a) (f g)() = +sin() ; Dom = all odd multiples of π ( ) (b) (g f)() = sin ; Dom = all real numbers ecept - (c) (f f)() = = + ; Dom = all real numbers ecept (d) (g g)() = sin ( sin()); Dom = all real numbers Section.5: Eponential Functions and.5.3. Basically, the larger the base, the faster the function is increasing.5.5. Notice that ( ) 3 = 3, which means that ( 3) is the reflection of 3 across the y-ais! Similarly with The smaller the base, the faster the function is going to (a) All real numbers ; (b) All 0 real numbers.5.7. f() = f() = ( ) 3 = 3 Section.6: Inverse functions and logarithms.6.3. No; For eample, even though 6, f() = f(6) =.6.5. Yes (by the horizontal line test).6.6. (a) f (3) = 0 (b) f(f (5)) = (a) By the horizontal line test (b) Domain of f = Range of f = [, 3]; Range of f = Domain of f = [ 3, 3] (c) 0 (d).8 (.6.6. f () = ln ) ( = ln )

5 SOLUTION - BANK (a) log (8) = 3 (b) log 3 ( 9) = (a) 5 = 5 (b) ln ( (+ ) ) sin() (a) = ln(7) 3 (b) = 5 e ) (a) t = a ln ( QQ0 ; Gives the time it takes to recharge the capacitor to a given capacity Q (b) Plug in Q = 0.9Q 0 into the equation in (a), and you get t = ln(0.) 4.6 seconds (a) π 6 (b) π (a) 5 4 (b) 4 5 (use the fact that sin() = sin() cos()) If θ = sin (), then sin(θ) =, then draw a triangle with hypothenuse, and opposite side, and then the adjacent side becomes, and so our answer becomes: cos(sin ()) = cos(θ) = adjacent hypotenuse = = See the handout Proof of the derivative of arccos for a similar problem; Or look at your notes taken in section! tan(sin ) = sin(sin ()) cos(sin ()) = by the result of number 65! Section.: The limit of a function... If approaches from the left, then f() approaches 3; If approaches from the right, then f() approaches 7. No, left-hand-limits and right-hand-limits must be equal!

6 6 PEYAM RYAN TABRIZIAN..6. (a) 4 (b) 4 (c) 4 (d) Undefined (e) (f) - (g) Does not eist (left and right-side limits not equal) (h) (i) (j) Undefined (k) 3 (l) Does not eist (h does not approach one fied value as approaches 5 from the left)..8. (numerator approaches e 5 > 0 while denominator approaches ( 9 approaches 0 + and ln (0 + ) =..40. The mass blows up to ( v c goes to, so the denominator of the fraction goes to 0 +, and so the whole fraction goes to ) = 3 4 Section.3: Calculating limits using the limit laws.3.0. (a) If you plug in =, then the left hand side is not defined, but the right hand side is (b) The above equation holds if, but the point of limits is that in this case you don t care about the value at! So in this case, the equality is correct!.3.3. Does not eist (left-hand-limit is because the numerator tends to 4 and the denominator tends to 0 while the right-hand-limit is because the numerator tends to 4 and the denominator tends to 0 + ) (use the fact that 3 = ( )( + + )).3.6. (put under a common denominator t + t = t(t + ) and cancel out).3.9. (put under a common denominator and multiply by the conjugate form) (by squeeze theorem, because sin ( π ) ) (a)(i) (since = in this case) (a)(ii) (since = in this case) (b) No, since the right-hand-limit and the left-hand-limit are not equal Let a = 0 and f() = sin ( ) (or ), and g() = f().

7 SOLUTION - BANK Let a = 0 and f() = sin ( ) (or ), and g() = f() Section.4: The precise definition of a limit.4.. δ = 0.7 (remember, the smaller the δ, the better!).4.4. δ = 0. (I picked this because and.5 0., and just pick a number slightly smaller than both).4.9. δ = 5ɛ.4.3. δ = min { }, ɛ 9 { } a δ = min, ɛ ) a (+ The net two are optional, but good for practice:.4.4. δ = 4 M δ = e M (where M is negative) Section.5: Continuity (f not defined at 4; neither), (left-hand-side and right-hand-side limits not equal; continuous from the left), (ditto; continuous from the right), 4 (left-hand-side limit does not eist; continuous from the right).5.8. This is my personal opinion, you might disagree with me (a) Continuous (b) Discontinuous (because of cliffs and skyscrapers) (c) Discontinuous (you pay per mile as whole, it doesn t matter whether you ve traveled 0.9 miles or 0.99 miles) (d) Continuous.5.9. g(3) = Continuous because it s a composition of ln (continuous) and a polynomial (continuous), Dom = (, ) (, ) tan ( ) Yes, you can check that the left-hand-side-limits and the right-hand-side limits are equal! Plug in values for G, M, R if you want to, for eample G =, M = 5, R = (For etra practice) Define f() = 4 + 3, then f() = < 0, f() = 5 > 0, so by IVT, there is one number c such that f(c) = Use the fact that sin(a + h) = sin(a) cos(h) + sin(h) cos(a) Define f(t) to be the altitude of the monk on the first day, g(t) to be the altitude of the monk on the second day, and let h(t) = f(t) g(t). Then h(0) > 0, h() < 0 (where 0 means 7AM and means P M), then by IVT, there is one number c such that h(c) = 0, i.e. f(c) = g(c)

8 8 PEYAM RYAN TABRIZIAN.6.4. Section.6: Limits at Infinity; Horizontal Asymptotes (a) (b) (c) (d) (e) (f) Horizontal asymptotes: y =, y = ; Vertical asymptotes: =, = 0, = (factor out from the numerator and pull out the from inside the square root).6.6. lim ( = + = ( ( + ) ) ( ) + + ( ) ( ) )

9 SOLUTION - BANK (factor out 3 from the numerator and from the denominator) tan ( ) = π (by continuity of tan ) (by the squeeze theorem) Section.7: Derivatives and Rates of Change.7.6. y = + 4 ((y 3) = ( + ) is also acceptable).7.. (a) A runs with constant speed, while B is slow at first and then speeds up (b) 8.5 seconds (c) 9 seconds.7.7. g (0) < 0 < g (4) < g () < g ( ).7.8. (a) y = 4 3 (y + 3 = 4( 5) is also acceptable) (b) f(4) = 3, f (4) = f() = 4, a = f() = tan(), a = π F/min (slope of the red line) (a) Rate of bacterias/hour after 5 houts (b) f (0) > f (5) (basically, the more bacteria there are, the more can be produced). But if there s a limited supply of food, we get that f (0) < f (5), i.e. bacterias are dying out because of the limited supply (a) II (b) IV (c) I (d) III.8.. f (t) = 5 8t Section.8: The derivative as a function (not continuous there); (graph has a kink) (a) Acceleration (b) Velocity (c) Position.8.5. Not differentiable at the integers, because not continuous there; f () = 0 for not an integer, undefined otherwise. Graph looks like the 0-function, ecept it has holes at the integers.

10 0 PEYAM RYAN TABRIZIAN Section 3.: Derivatives of polynomials and eponential functions 3... y = V (r) = 4πr, which is the surface area of sphere! What a coincidence - or is it? ;) f (t) = t + t t = t + t y = e y = e, so y (0) =, and so the equation of tangent line is y = ( 0), i.e. y = + and equation of normal line is y = ( 0), i.e. y = + (remember that the normal line still goes through (0, ), but has slope = the negative reciprocal of the slope of the tangent line) (a) v(t) = s (t) = 3t 3; a(t) = v (t) = 6t (b) a() = (c) v(t) = 0 if t = or t =, but t > 0 (negative time doesn t make sense), so t =, and a() = First of all y = 3 and first find a point where y () = 3 (remember that two lines are parallel when their slopes are equal, and the slope of y = +3 is 3). So you want 3 = 3, so =, so = 4. Now all that you need to find out is the slope of the tangent line to the curve at 4. The equation is: y 8 = 3( 4) (because from the above calculation the slope is 3, and the tangent line goes through (4, f(4)) = (4, 8)) Section 3.: The product and quotient rules y = t(t4 3t +) t (4t 3 6t) (t 4 3t +) y = (r )e r + ( r r ) e r = ( r ) e r y () = e + e, so y (0) =, and so the tangent line has equation: y 0 = ( 0),i.e y =, and the normal line has equation: y 0 = ( 0), i.e. y = f () = (+) (+) = + ++, so f () = (+)( ++) ( +)(+) ( ++), and so f () = (+)(++) (+)(+) (++) = (4)(4) (3)(4) (4)(4) = 6 6 = 4 6 = = (a) u () = f ()g() + f()g () = ()() + ()( ) = 0 (b) v (5) = f (5)g(5) f(5)g (5) = ( 3)() (3)( 3) (g(5)) 4 = 3 4 = 8 = (900)(30593) + (96400)(400) =,345,960,000

11 SOLUTION - BANK Section 3.3: Derivatives of trigonometric functions g (t) = 3t cos(t) t 3 sin(t) y = cos() sin()() We have sin(θ) = 0, so = 0 sin(θ), so (θ) = 0 cos(θ), and ( ) π 3 = 0 cos ( ) π 3 = 0 = (multiply the fraction by 3 3 and use the fact that lim 0 sin(3) 3 = ) = 3 (multiply the numerator by 4 4 and the denominator by 6 6 sin(4) sin(6) the facts that lim 0 4 = and lim 0 6 = ) A/Math A Summer/Solution Bank/Arclength.png and use The problem is to calculate: lim θ 0 + = s d

12 PEYAM RYAN TABRIZIAN First of all, let s call the radius of the circle r. simple sub-problems: Now, let s divide this into three () Calculate s But this is not very hard! Either you know about the formula for the length of an arc, or you can easily derive it! Namely, if an angle of π radians corresponds to a length of πr (i.e. the circumference of a circle), then an angle of θ radians corresponds to a length of s. Now, because the length of an arc is proportional to the angle, we have the following equality: s θ = πr π = r So s = θr Notice that the use of radians makes this calculation particularly simple! () Calculate d This is a bit harder than the above step, but actually not that bad! Label the triangle in the figure ABC, and let H be the midpoint of BC. Then, the triangle AHB is right in A, and we can use our regular definition of sin to find a relationship between r and d, namely: sin( BAH) = BH AB But you can easily check that BAH = θ, that BH = d the midpoint of BC), and that AB = r! Hence, we get: (because H is That is, d = rsin( θ ). sin( θ ) = d r (3) Compute the limit The last thing we need to do is to calculate the required limit! But this is easy, since we know the values of s and d in terms of r: s lim θ 0 + d θ 0 + rθ r sin( θ ) θ 0 + θ sin( θ ) θ 0 + θ sin( θ ) Finally, let t = θ becomes: and notice that, as θ 0+, t 0 +, then, our limit s lim θ 0 + d t t 0 + sin(t) t 0 + sin(t) t t 0 + lim t 0 + sin(t) t = = And again, the net-to-last step is justified because both limits (numerator and denominator) eist and the limit of the denominator is nonzero! Hence, we get:

13 SOLUTION - BANK 3 lim θ 0 + s d = Section 3.4: The Chain Rule F () = 4 ( ) 3 4 ( + 3 ) y = e k ke k f (t) = sec (e t ) + e tan(t) sec (t) y = sin( sin(tan(π))) sin(tan(π)) cos(tan(π)) sec (π)π y = αe α sin(β)+e α β cos(β); y = e α ((α β ) sin(β)+αβ cos(β)) (a) h () = f (g())g () = f ()g () = 5 6 = 30 (b) H () = g (f())f () = g (3)f () = 9 4 = (a) h () = f (f())f () = f ()f () = = (b) g () = f (4)4 = 4 = (a) f () = g( ) + g ( ) = g( ) + g ( ) (b) f () = g ( )() + 4g ( ) + g ( ) = 6g ( ) g ( ) a(t) = dv dt = dv ds ds dt = dv ds v(t) V (t) = 4 3 πr3, so dv dt = 4 dr dr 3π3r dt = 4πr dt (f()[g()] ) = f ()[g()] + f()( )[g()] g () =f ()g()[g()] f()g ()[g()] = f ()g() f()g () (g()) Section 3.5: Implicit differentiation y = y y = ey sin()+cos(y)y e y cos() cos(y) y = +

14 4 PEYAM RYAN TABRIZIAN (y ) = 3( + 3 3), or y = y = 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 ( 0 ) y 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 Whence, yy 0 b + 0 a = + (0) a =, we get:

15 SOLUTION - BANK 5 Which is what we want! y = arctan() + 0 a + y 0y b = G () = arccos() (a) f(f ()) =, let y = f (), then f(y) =, so f (y)y =, so y = (b) 3 f(y) = f(f ()) Let s denote the point of intersection between the ellipse and the tangent line by (a, b). Then, using implicit differentiation, we can show that the slope of the tangent line is a 4b Now, let K be the altitude of the lamp, our goal is to find K. Notice that the same tangent line goes through the points ( 5, 0) and (3, K), so by the slope formula, we have: Slope = K 0 3 ( 5) = K 8 In particular, since the slope is also equal to a 4b, we have: So K 8 = a 4b K = 8 a 4b = a b So all we really need to do to solve this problem is to find a b! Now we also know that the tangent line goes through the points ( 5, 0) and b 0 (a, b), so its slope is a ( 5) = b a+5, but again we know that its slope is also a 4b, and so we get: b a + 5 = a 4b So cross-multiplying, we have 4b = (a)(a + 5), that is a + 4b = 5a. HOWEVER, We also know that (a, b) is on the ellipse, so it satisfies the equation of the ellipse, and so a + 4b = 5, whence we get 5a = 5, and so a =.

16 6 PEYAM RYAN TABRIZIAN And plugging a = into a + 4(b) = 5 and assuming b > 0, we get b =, and so K = a b = =, and we re done! Section 3.6: Derivatives of logarithmic functions ( ) g () = y = log 0 ( ) + ln(0) = log 0( ) + ln(0) ( y = sin () tan 4 () ( +) cos() sin() + 4 sec () tan() ) Section 3.7: Rates of change in the natural and social sciences (a) f (t) = e t t e t = e t (b) f (3) = e 3 (c) t = (d) When t < ( ) ( ) t (e) f() f(0) + f() f(8) = e 0 + e 8e 4 = 4e 8e 4 (f) The particle is moving to the right between t = 0 and t =, and then to the left from t = to t = 8. ( ) (g) f (t) = e t t + e t ( ) f (3) = e ( ) ( = e t + t 4 ) ( = e t t 4 ) ; (h) Use a calculator (i) Speeding up when f (t) > 0 and f (t) > 0 or when f (t) < 0 and f (t) < 0. But solving those equations reveals that none of the two situations can happen! Hence the particle is constantly slowing down! (a) First solve for v(t) = 0, where v(t) = ds 80 dt = 80 3t, you get t = 3 = 5. So the maimum height s is s = s( 5 ) = = 00 (b) To find the time t when the ball is 96ft above the ground, we need to solve the equation s(t) = 96, and you get t =, 3, whence v() = 80 3 = 6 ft s and v(3) = = 6 ft s f () = 6 = linear density at. f () = 6, f () =, f (3) = 8. The density is highest at 3 and lowest at.

17 SOLUTION - BANK First of all, we know two things, namely f(0) = 0 and f (0) =. But by the chain rule: f (t) = 0.7be 0.7t a ( + be 0.7t ) = 0.7abe 0.7t ( + be 0.7t ) So from f(0) = 0, we get: And from f (0) =, we get: a + b = 0 0.7ab ( + b) = From a +b = 0, we get a = 0(+b), and plugging this into the second equation, we get: (0.7)(0)( + b)b ( + b) = 4b + b = 4b =( + b) 4b = + b b = b = 6 And so a = 0( + 6) = 0(7) = 40. Therefore, we have a = 40 and b = 6. Finally, to find out what happens in the long run, we need to calculate lim t f(t). But notice that lim t e 0.7t = 0, and so lim t f(t) = a +0 = a = (a) C () = (b) C (00) = 3; The cost of producing one more yard of a fabric once 00 yards have been produced (c) C(0) C(00) = 3.005, which is pretty close to C (00) Section 3.8: Eponential growth and decay (a) y(0) = C = 0 ( ) (b) y(t) = 0e ln(5) 6 t = 0(5) t 6 = 0 5 t (c) y(5) = (d) y (5) = Ky(5) = ln(5) (e) t = 5000 ln( 3 ) ln(5) 9.

18 8 PEYAM RYAN TABRIZIAN (a) y(t) = 00e ln( ) t 30 = 00 ( (b) y(00) = 00 ( ) 00 (c) t = 30 ln( 00 ) ln( ) ) t The problem asks about radioactive decay, so as usual, we have y = ky, so y(t) = Ce kt. Now we re given two things: First of all, the half-life is t = 5730 years, so y(5730) = y(0) = C. Moreover, we know that at a certain time t (we want to find t ), y(t ) = 0.74y(0) = 0.74C. Now even though we don t know what C is, we can still solve for t. The following calculation helps us find k: y(5730) = C Ce 5730k = C e 5730k = 5730k = ln( ) k = ln( ) 5730 Whence y(t) = Ce ln( ) 5730 t = C( ) t 5730 Now we re given that y(t ) = 0.74C, and the following calculation helps us solve for t : y(t ) =0.74C C( ) t 5730 =0.74C ( ) t 5730 =0.74 t 5730 ln( ) = ln(0.74) t =5730 ln(0.74) ln( ) t 489 So t 489 years (notice how we didn t even need info about C to figure this out!) (a) (i) 3000 ( ) ()(5) 388 (ii) 3000 ( ) ()(5) 3840 (iii) 3000 ( ) ()(5) 3850

19 SOLUTION - BANK 9 (iv) 3000 ( ) (5)(5) ) (365)(5) (v) 3000 ( (vi) 3000e 0.05(5) (b) A = 0.05A, A(0) = dh dt = 3 5π (Use V = πr h) Section 3.9: Related rates ) First of all, let s draw a picture of the situation, and remember to only label things that are constant! A/Math A Summer/Solution Bank/Street Light.png Here, is the distance between the street light and the man, and y is the distance between the man and the shadow. Also, let z = + y, the total length of the shadow. ) We want to figure out z when = 40. 3) Looking at the picture, we can use the law of similar triangles to conclude:

20 0 PEYAM RYAN TABRIZIAN That is: y + y = 6 5 It follows that: y = ( + y) y = 5 y = 3 z = + y = + 3 = 5 3 4) Hence z = 5 3 5) However, we know that = 5 (because the man is walking with a speed of 5 ft/s). Hence we get z = (5) = 3 So z = 5 3. Note: We did not need the fact that = 40! dd dt = 65mph (use the pythagorean theorem to conclude D = + y ) dh dt = 6 5π (use the fact that V = π h3 because h = r ) ) Again, draw a picture of the situation: Here, is the distance between P and the beam of light. ) We want to figure out d dt when = 3) Looking at the picture, because we have info about the derivative of θ (see below), we use the definition of tan(θ): tan(θ) = 3 So = 3 tan(θ) 4) Hence d dt = 3 sec (θ) dθ dt 5) First of all, we actually know what dθ dt is! Since the lighthouse makes 4 revolutions per minute and one revolution corresponds to π, we know that dθ dt = 8π (think of speed = distance/time, and here time = minute, and distance = 8π, also you put a minus-sign since is decreasing!). Moreover, by drawing the eact same picture as above, ecept with =, we can calculate sec (θ), namely: sec(θ) = hypothenuse 0 = adjacent 3 And the 0 we get from the Pythagorean theorem!

21 SOLUTION - BANK A/Math A Summer/Solution Bank/Lighthouse.png ( 0 ) It follows that sec (θ) = 3 = 0 9. Now we got all of the info we need to conclude the problem: Whence d dt = 80π 3 rad/min d dt =3 sec (θ) dθ dt d dt =3(0 9 )( 8π) d dt = 40π 9 d dt = 80π ) As usual, let s draw a picture of the situation: Here, is the distance between the runner and the friend, and s is the length of the arc corresponding to θ ) We want to figure out d dt when = 00

22 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Runners.png 3) Looking at the picture, it looks like we should use the law of cosines (because we have info about θ and about of the 3 sides of the triangle) In other words: 4) Hence d dt = (00)(00) cos(θ) = cos(θ) = sin(θ) dθ dt 5) First of all = 00 in this case. Also, by definition of a radian, we know that s = 00θ, whence ds dθ ds dt = 00 dt. But we re given that dt = 7m/s, so dθ dt = 7 00 = So all we got to figure out is sin(θ) For this, draw the same picture as above, ecept you let = 00. And in this case, we use the law of cosines again: 00 = (00)(00) cos(θ)

23 SOLUTION - BANK = cos(θ) cos(θ) = 4 Now you can use either the triangle method to figure out what sin(θ) is (all you gotta do is calculate sin(cos ( 4 ), or, even easier, notice that sin(θ) = cos (θ) (this works because sin(θ) > 0 because we assume that θ is between 0 and π. Hence sin(θ) = 6 = 5 6 = 5 4. PHEW!!! Now we have all the info we need to solve the problem: Hence d dt = d =40000 sin(θ)dθ dt dt (00) d 5 dt =40000( 4 )(0.07) 400 d dt =700 5 d dt = As is usual for related rates problems, let s draw a picture: Let θ be the angle between the hour hand and the minute hand. Now, what we want to calculate is D (θ), where the indicates differentiation with respect to the time variable. How can we relate D(θ) with what we know? This is easy! We know an angle θ and the lengths of AB and AC in the picture, so let s just use the law of cosines. We get: That is: BC = AC + AB AC AB cos(θ) Which you can write as: D(θ) = cos(θ) D(θ) = cos(θ) Now differentiate with respect to time! We get: D(θ)D (θ) = 64 sin(θ) dθ dt And now, all we need to do is to plug in everything we know!

24 4 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Clock.png First of all θ = π = π 6 (basically, the whole circle corresponds to π, and so of the circle corresponds to π ). In particular, sin(θ) = sin( π 6 ) =. Now, for d(θ), we use the law of cosines again, this time with the value θ = π 6 : D( π 6 ) = cos( pi 3 ) = = So, taking square roots, we get d( π 6 ) = = mm. Finally, we need to compute Dθ dt. But think about it! In hours, θ = π, so the speed of θ should be π = π 6 = π 6 rad/h (notice that θ is decreasing, so we wanted a negative answer!) Finally, we have all our information to get our final answer:

25 SOLUTION - BANK 5 D(θ)D (θ) = 64 sin(θ) Dθ dt (4 5 ( 3) D π ) = 64 6 π 6 (4 5 ( 3) D π ) = 3 π ( 6 6 D π ) π 6 = ( D π ) π 6 = ( D π ) π = ( 3 D π ) 8.55mm/h 6 So our final answer is D ( ) π 6 = π mm/h 8.55mm/h Section 3.0: Linear approimations and differentials L() = (because ln() = 0, and = ) (a) dy = ( sin() + cos())d ( ) (b) dy = t +t +t dt = (a) dy = 0 e 0 d (b) dy = 0 (0.) = 0.0 t +t dt (y) = y(5) y(4) = 5 4 = 0 = 0. dy = 4 () = 8 = (8.06) 3 L(8.06) = (8)( 3 )(8.06 8) = 4 + (0.06) = = Here we used the fact that f() = 3 and a = l = πr = 84, so r = 84 π = 4 π. We know dl = 0.5, so πdr = 0.5, so dr = 0.5 π = 4π (a) S = 4πr, so ds = 8πrdr = 8π 4 π 4π = 84 π. Also the relative error is ds S = 8πrdr 4πr = dr r = π π 4 = (b) V = 4 3 πr3, so dv = 4πr dr = 4π 4 π 4π = 764 π 79. Also the relative error is dv V = 4πr dr 4 = 3dr 3 πr3 r = 3 4π 4 = 3 68 = π

26 6 PEYAM RYAN TABRIZIAN df = 4kR 3 dr, so: df F = 4kR3 dr = 4kR3 dr F F And when dr R = 0.05, df F = 4(0.05) = 0. = 4kR3 dr kr 4 = 4 dr R cosh() + sinh() = e + e Section 3.: Hyperbolic functions + e e = e + e + e e = e = e 3... Just epand out the right-hand-side and use the fact that sinh() = e e, cosh(y) = ey e y, cosh() = e +e and sinh(y) = ey e y sinh() cosh() = e e e + e = 4 (e e )(e +e ) = (e e ) = e e 3... sinh() = 4 3 (use the fact that cosh () sinh () = and the fact that sinh() > 0 when > 0). Then you get tanh() = sinh() cosh() = = 4 5, sech() = cosh() = 3 5, etc. = sinh() 3..3(a). (use tanh() = e e e +e the denominator) and factor out e from the numerator and This is very similar to eample 3 on page 57. However, there is a subtle point involved, check out the document Subtle point in 3..6 for more info! 3..9(a)(b). This is also very similar to eample 4 on page 57. For (a), use the fact that cosh(cosh ()) sinh(cosh ()) = and cosh(cosh ()) =. Also, you ll need to fact that sinh(cosh ()) 0 (and this is because cosh () 0 by definition, and sinh() 0 if 0). (b) is even easier, use the fact that: tanh(tanh ()) = sech(tanh ()) and tanh(tanh ()) = f () = sinh() + cosh() sinh() = cosh() y = +tanh () (sech ()) = sech () +tanh () Section 4.: Maimum and Minimum Values Absolute maimum: Does not eist (NOT 5) - Absolute minimum: f(4) = - Local minimum: f() =, f(4) = - Local maimum: f(3) = 4, f(6) = , (g does not eist), 0 (makes g (c) = 0) (F does not eist), 4, 8 7 (makes F (c) = 0) Candidates: f( ) =, f(3) = 66 (endpoints), f(0) = 3, f( ) =, f() =. Absolute maimum: f(3) = 66, Absolute minimum: f( ) = f() =

27 SOLUTION - BANK ) Evaluate f at the endpoints 0 and π f(0) = + 0 =, f( π ) = = 0 ) Find the critical numbers of f f (t) = sin(t) + cos(t) f (t) = 0 sin(t) + cos(t) = 0 sin(t) = cos(t) Here comes the tricky part! This seems impossible to solve, but ideally we d like to write the right-hand-side just in terms of sin() in order to have a shot at solving this! Start with cos(t) = cos (t) sin (t) (the double-angle formula for cos). Moreover cos (t) = sin (t) (because cos (t) + sin (t) = ) So we get cos(t) = sin (t) sin (t) = sin (t). So our original equation becomes: sin(t) = sin (t) which you can rewrite as sin (t) + sin(t) = 0. This again looks impossible to solve, but notice that this is just a quadratic equation in sin(t)! So let X = sin(t), then we get: X + X = 0 And using the quadratic formula (or your factoring skills), we get: (X )(X + ) = 0 So X = or X =. That is, sin(t) = or sin(t) =. HOWEVER, remember that we re only focusing on [0, π ], so in particular sin(t) = has only one solution in [0, π ], namely t = π 6, and sin(t) = has NO solution in [0, π ]. It follows that the only critical number of f in [0, π ] is t = π 6 (there are no numbers where f is not differentiable, so in fact those are all the critical numbers). And we get f( π 6 ) = = 3 3 3) Compare all the candidates you have: Our candidates are f(0) =, f( π ) = 0 and f( π 6 ) =

28 8 PEYAM RYAN TABRIZIAN Hence, the absolute minimum of f on [0, π ] is f( π ) = 0 (the smallest candidate), and the absolute maimum of f on [0, π ] is f( π 6 ) = 3 3 (the largest candidate) Section 4.: The Mean Value Theorem f(0) = f(π) = 0, but f () = sec () > 0. This does not contradict Rolle s Theorem because f is not continuous on [0, π] (it is discontinuous at π f is differentiable on [0, 3]; c = ln ( e c = e Let f() = sin() ) e 6 6 (you get this by solving At least one root: f(0) = < 0 and f(π) = π > 0 and f is continuous, so by the Intermediate Value Theorem (IVT) the equation has at least one root. At most one root: Suppose there are two roots a and b. Then f(a) = f(b) = 0, so by Rolle s Theorem there is at least one c (a, b) such that f (c) = 0. But f (c) = cos(c) 0, which is a contradiction, and hence the equation has at most one root By the MVT, f(4) f() 4 = f (c) for some c in (, 4). Solving for f(4) and using f() = 0, we get f(4) = 3f (c) = This is equivalent to showing: Which is the same as: Which is the same as: sin(a) sin(b) a b sin(b) sin(a) b a sin(b) sin(a) b a But by the MVT applied to f() = sin(), we get: sin(b) sin(a) b a = cos(c) for some c in (a, b). However, cos(c), and so we re done!

29 SOLUTION - BANK Let f() = sin (), g() = cos ( ). Then f () = and: g 4 () = ( ) = = = (you get this by factoring out 4 out of the square root. This works because 0) Hence f () = g (), so f() = g() + C To get C, plug in = 0, so f(0) = g(0) + C. But f(0) = g(0) = 0, so C = 0, whence f() = g() Let f(t) be the speed at time t. By the MVT with a = : 00 and b = : 0, we get: f( : 0) f( : 00) = f (c) : 0 : 00 But : 0 : 00 = 0 minutes = 6 h, so: = f (c) Whence: f (c) = 0 for some c between : 00 pm and : 0 pm. But f (c) is the acceleration at time c, and so we re done! This is again a proof by contradiction! Suppose f has (at least) two fied points a and b. Then, by definition of a fied point, f(a) = a, and f(b) = b. However, by the Mean Value Theorem, there is a c in (a, b) such that: f(b) f(a) b a = f (c) Now using the fact that f(b) = b and f(a) = a, we get: 4 = = f () So b a b a = f (c) = f (c) That is, f (c) =. However, by assumption, f () for all, so in particular setting = c gives f (c). but since f (c) =, we get, which is a contradiction! Hence f has at most one fied point!

30 30 PEYAM RYAN TABRIZIAN Section 4.3: How derivatives affect the shape of a graph (a) f () = = 6( )( + 3); on (, 3) (, ), on ( 3, ) (b) Local ma: f( 3) = 8; Local min: f() = (c) f () = + 6; CU on (, ), CD on (, ), IP ( (a) f () = cos() sin(); on (0, π 4 ) ( 5π 4, ), on ( π 4, 5π 4 ) (b) Local ma: f( π 4 ) = ; Local min: f( 5π 4 ) = (c) f () = sin() + cos(); CU on ( 3π 4, 7π 4 ), CD on (0, 3π 4 ) ( 7π 4 ( 3π 4, 0), ( 7π 4, 0) A possible graph looks like this: A/Math A Summer/Solution Bank/Concave-Kink.png, f( 0.5) = 37 ), π), IP (a) f (θ) = sin(θ) cos(θ) sin(θ) = sin(θ)( + cos(θ)); on (π, π), on (0, π) (b) Local min: f(π) =, no local ma.

31 SOLUTION - BANK 3 (c) f () = cos(θ) + sin (θ) cos (θ) = cos(θ) + 4 cos (θ) = 4(cos (θ) cos(θ) ) = 4(cos(θ) + )(cos(θ) ); CU on ( π 3, 5π 3 ), CD on (0, π 3 ) ( 5π 3, π), IP ( π 3, 5 4 ), ( 5π 3, 5 4 ) (a) Vertical Asymptotes: =, = ; Horizontal Asymptote: y = (at ± ) (b) f () = ( ) ; on (, ) (, 0), on (0, ) (, ) (remember that f is not defined at ±) (c) Local ma: f(0) = 0; No local min (d) f () = ( ) +()()( )() ( ) = 6( )( ) ( ) ; CU on (, ) 4 (, ), CD on (, ), No inflection points (a) Vertical Asymptote: = ; Horizontal Asymptote: y = (at ± ) (b) f () = (+) e + ; on (, ) (, ) (remember f is not defined at ) (c) No local ma/min (d) f () = (+) e (+) e 4 + = (+) e 4 + ; CU on (, ) (, ), CD on (, ), IP = (, e (a) No, (b) Yes, (c) No, Section 4.4: L Hopital s Rule (a) Yes, 0 0 (b) No, 0 (c) Yes, (d) Yes, 0 (e) No, (f) Yes, e t e t lim t 0 t 3 t 0 3t e t t 0 6t e t t 0 6 = 6 lim θ π sin(θ) csc(θ) θ π cos(θ) (l Hopital s rule) cos(θ) sin (θ) θ π sin (θ)(cancelling out) =

32 3 PEYAM RYAN TABRIZIAN lim lim ln() lim = 0 tanh() lim 0 tan() sech () 0 sec () = sin () lim 0 0 = cos() sin() cos() lim 0 = 0 0 lim e = lim e = lim e = lim ln() ln() ( ) ln() + ( ) ln() ln() + + = + e = 0 + = ( ) (use conjugate form) (factor out out of the square root) ( = =, since > 0) + + ) ( = ln() lim ln() ( ) = ( 0) =

33 SOLUTION - BANK ) b ) Let y = ( + a ) ln(y) = b ln( + a ) 3) lim ln(y) b ln(+ a ) ln( + a ) ( + )( a a ) b ( )( b ) ab + a = ab ( ) 4) So lim + a b = e ab ) Let y = ) Then ln(y) = ln() 3) So lim ln(y) ln() = 0 4) Hence lim y = e 0 = (a) Here, you want to calculate lim t, so treat t as our, and leave everything else as a constant!!!!. In particular, we get: So, we get: ct lim e m = 0 because c > 0 and m > 0 t lim v = mg t c ( 0) = mg c (b) Here, we let c 0 +, so we treat c as our, and leave everything else as a constant!. Then, we have: lim v mg c 0 + c 0 + c 0 + mg ( e ct m c ( ( t m (( t ) ) m e 0 =mg (( t ) ) m =mg = mgt m =gt ) ) e ct m ) (by l Hopital s rule)

34 34 PEYAM RYAN TABRIZIAN Section 4.5: Summary of curve sketching D : R {±3} I : No intercepts, y intercept: y = 9 S : f is even A : Horizontal Asymptote y = 0 (at ± ), Vertical Asymptotes = ±3 I : f () = ( 9) ; f is increasing on (, 3) ( 3, 0) and decreasing at 0. on (0, 3) (3 ). Local maimum of 9 C : f () = 6 +3 ( 9) 3 ; f is concave up on (, 3) (3, ) and concave down on ( 3, 3); No inflection points A/Math A Summer/Solution Bank/hw0graph.png Note: First of all, f is periodic of period π, so we re only focusing on [0, π]. D : R I : intercepts: = 0, = π (basically you should get sin() = 3, which is impossible), y intercept: y = 0 S : Again, f is periodic of period π. Also, f is odd. A : No asymptotes I : f () = 3 cos() 3 cos() sin() = 3 cos()( sin ()) = 3 cos 3 (); Increasing on (0, π ) ( 3π, π); Decreasing on ( π, 3π ). Local maimum of at = π. Local minimum of at = 3π. C : f () = 9 sin() cos (); Concave down on (0, π) and Concave up on (π, π). Inflection point (π, 0)

35 SOLUTION - BANK 35 A/Math A Summer/Solution Bank/hw0graph.png D : R I : No intercepts, y intercept: y = S : No symmetries A : Horizontal Asymptotes: y = 0 (at ), y = (at ) I : f () = e (+e ) > 0, so f is decreasing on R C : f () = e e e + 3 (multiply numerator and denominator by (e ) 3 after simplifying), so f is concave down on (, 0) and concave up on (0, ). Inflection point at (0, ) A/Math A Summer/Solution Bank/hw0graph3.png

36 36 PEYAM RYAN TABRIZIAN Note: : First of all, f is periodic of period π, so from now on we may assume that [0, π] D : We want sin() > 0, so the domain is (0, π) I : No y intercepts, intercepts: Want ln(sin()) = 0, so sin() =, so = π S : Again, f is periodic of period π A : No horizontal/slant asymptotes, but lim 0 + ln(sin()) = ln(0 + ) =, so = 0 is a vertical asymptote. Also lim π ln(sin()) =, so = π is also a vertical asymptote. I : f () = cos() sin() = cot(), then f () = 0 = π, and using a sign table, we can see that f is increasing on (0, π ) and decreasing on ( π, π). Moreover, f( π ) = ln() = 0 is a local maimum of f. C : f () = csc () < 0, so f is concave down on (0, π). A/Math A Summer/Solution Bank/hw0graph.png Want to minimize + y - But y = 00, so y = 00 - Let f() = > 0 ( is positive) Section 4.7: Optimization Problems, so + y = f () = 0 00 = 0 = 00 = 0 - By FDTAEV, = 00 is the absolute minimum of f - Answer: = 00, y = = The picture is as follows:

37 SOLUTION - BANK 37 A/Math A Summer/Solution Bank/Fence.png - Want to minimize 3w + 4l - But lw =.5, so l = 0.75 w, so 3w + 4l = 3w + 3 w - Let f(w) = 3w + 3 w - w > 0 - f () = w = 0 w = w = - By FDTAEV, w = is the absolute minimum of f - Answer: w =, l = We have D = ( ) + y, so D = ( ) + y - But y = 4 4, so D = ( ) Let f() = ( ) No constraints - f () = ( ) 8 = 6 = 0 = 3 - By the FDTAEV, = 3 is the maimizer of f. - Since y = 4 4, we get y = = 3 9, so y = ± 3 9 = ± Answer: ( 3, 4 3 ) ( and 3, 4 3 ) Picture:

38 38 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/hw0opt.png We have A = y, but the trick here again is to maimize A = y (thanks for Huiling Pan for this suggestion!) - But + y = r, so y = r, so A = (r ) = r 4 - Let f() = r 4 - Constraint 0 r (look at the picture) - f () = r 4 3 = 0 = 0 or = r - By the closed interval method, = r f(0) = f(r) = 0 - Answer: = r, y = r r = r is a maimizer of f (basically - Let w be the width of the rectangle, and h the height of the rectangle. - We have A = wh + π( w ) = wh + π 4 w, but w + h + π w = 30, so h + πw + w = 30, so h = 30 (π+)w. Hence A = w( 30 (π+)w ) + π 4 w - Let f(w) = w( 30 (π+)w ) + π 4 w - Constraint: w > 0 - f (w) = 5 (π+) w = 0 w = 30 π+ - By FDTAEV, w = 30 π+ - Answer: w = 30 π+, h = 5 π+ (there s a big cancellation going on!) is the maimizer of f (a) c () = C () C(). When c is at its minimum, c () = 0, so C () C() = 0, so C () = C() = c(), so C () = c(), i.e. marginal cost equals the average cost!

39 SOLUTION - BANK (thank you Brianna Grado-White for the solution to this problem!) The picture is as follows: A/Math A Summer/Solution Bank/hw0opt.png Here, h and h and L are fied, but varies. Now the total time taken is t = t + t = d v + d v. Now, by the Pythagorean theorem: d = + h and d = (L ) + h, so we get: And t() = + h v + (L ) + h v t L () = + v + h v (L ) + h Setting t () = 0 and cross-multiplying, we get: = v d + L v d v d (L ) = v d So, by definition of sin(θ ) and sin(θ )), we get: v v = d (L )d = d = sin(θ ) L sin(θ d ) Section 4.8: Newton s method As usual, a good picture is the key to solving the problem:

40 40 PEYAM RYAN TABRIZIAN A/Math A Summer/Solution Bank/Newton.png Note: This picture is very complete. It is meant to illustrate all the points I am making below. Here, we are given a lot of information, so let s try to tackle this problem one step at a time! First, let s calculate the equation of any tangent line to the graph of f() = sin() that goes through (0, 0). Later, we will be worrying about finding the one that has the largest slope. By definition of the derivative, the tangent line to the graph of f at c has slope f (c) = cos(c), so any such tangent line that also goes through (0, 0) has equation: y 0 = cos(c)( 0), i.e. y = cos(c). Finally, we know that the tangent line goes through (c, sin(c)) (i.e. goes through the graph of f at c), so we get: sin(c) = cos(c) c, i.e. tan(c) = c. So any tangent line at c with the above properties must solve tan(c) = c, i.e. tan(c) c = 0. So what we really need to do is to approimate the zero of the function g() = tan(). Now this looks like a Newton s method problem! But remember, that for Newton s method, we need to find a good initial guess, and here is where we use the information that the tangent line must have largest slope! First of all, notice that f() = sin() is odd, so the graph is symmetric about the origin! If you look at the picture above, you ll see that the tangent line to the graph at c is the same as the tangent line at c. This means we can restrict ourselves to the right-hand-side of the picture, i.e. c 0! (if you don t understand this argument, don t worry, it s just a simplification) And if you look at the picture again, you ll notice that if your initial guess is between 0 and π, your successive approimations will got to 0. And you don t want that because the slope of the tangent line at 0 is (which is not the greatest slope). The same problem arises with the initial guess between 3π and π (the

41 SOLUTION - BANK 4 approimations go to π) Finally, notice that when c gets larger and larger, the tangent line at c has smaller and smaller slope (see picture: The brown line has a smaller slope than the blue line, which has a smaller slope than the green line), so you d like your initial guess not to be too large. In particular, we don t want the initial guess to be larger than 3π! From this analysis, we conclude that any initial guess between π and 3π is good enough! (also look at the picture above: the green tangent line seems to be the winner here) For eample, with 0 = 4.5 (or you could try 0 = π), we get the successive approimations (remember that you are applying Newton s method to g() = tan(), NOT f()!): 0 =4.5 = = = = And so, our approimation is: c And hence the largest slope is approimately equal to cos( ) (because f (c) = cos(c)). To summarize: Draw a picture Derive the function that you want to apply Newton s method to (i.e. g() = tan() ) Argue that your intial approimation must be between π and 3π (YOU NEED TO JUSTIFY THIS PART!, maybe not as precise as I did, but there needs to be some justification Apply Newton s method to g with initial approimation between π and 3π (4 would work) Section 4.9: Antiderivatives F () = C = C f () = A, so f() = A + B f() = sin(t) + tan(t) + C, but 4 = f( π 3 ) = C = C, so f() = sin(t) + tan(t) + 4

42 4 PEYAM RYAN TABRIZIAN If f (θ) = sin(θ) + cos(θ), then f (θ) = cos(θ) + sin(θ) + C. f (0) = 4, so C = 4, so C = 5. Hence f (θ) = cos(θ) + sin(θ) + 5. Hence f(θ) = sin(θ) cos(θ) + 5θ + C. f(0) = 3, so C = 3, so C = 4. Hence f(θ) = sin(θ) cos(θ) + 5θ a(t) = 0 sin(t) + 3 cos(t), so v(t) = 0 cos(t) + 3 sin(t) + A, so s(t) = 0 sin(t) 3 cos(t) + At + B Now, s(0) = 0, but s(0) = 0(0) 3() + A(0) + B, so 3 + B = 0, so B = 3 So s(t) = 0 sin(t) 3 cos(t) + At + 3 Moreover, s(π) =, but s(π) = 0(0) 3() + A(π) + 3 = A(π), so A(π) =, so A = π = 6 π So altogether, you get: s(t) = 0 sin(t) 3 cos(t) + 6 π t First of all, the acceleration of the car is a(t) = 6, so v(t) = 6t + C. We want to find v(0) = C, so once we find C, we re done! Let t be the time when the car comes to a stop. Then v(t ) = 0, so 6t + C = 0, so C = 6t. So once we find t, we re done! Now we know that s(t ) s(0) = 00, but s(t) = 8t + Ct + C, so 00 = 8(t ) + Ct + C + 0 C(0) C = 8(t ) + 6t t = 8(t ), so 8(t ) = 00, so (t ) = 5 so t = 5 (assuming time is positive) Whence v(0) = C = 6t = (a) (i) =, so Section 5.: Areas and Distances L 6 = f(0)()+f()()+f(4)()+f(6)()+f(8)()+f(0)() = (ii) ++8 = R 6 = f()()+f(4)()+f(6)()+f(8)()+f(0)()+f()() = = (iii) M 6 = f()()+f(3)()+f(5)()+f(7)()+f(9)()+f()() = = (b) Overestimate (c) Underestimate (d) M 6 (just right, does not overshoot, like L 6, but not undershoot either, like R 6 )

43 SOLUTION - BANK (a) If n = 3, then =, and if n = 6, =, so: R 3 = f(0)() + f()() + f()() = = 8 R 6 =f( 0.5)(0.5) + f(0)(0.5) + f(0.5)(0.5) + f()(0.5) + f(.5)(0.5) + f()(0.5) =.5(0.5) + (0.5) +.5(0.5) + (0.5) + 3.5(0.5) + 5(0.5) =6.875 (b) L 3 = f( )() + f(0)() + f()() = + + = 5 L 6 =f( )(0.5) + f( 0.5)(0.5) + f(0)(0.5) + f(0.5)(0.5) + f()(0.5) + f(.5)(0.5) =(0.5) +.5(0.5) + (0.5) +.5(0.5) + (0.5) + 3.5(0.5) =5.375 (c) M 3 = f( 0.5)() + f(0.5)() + f(.5)() = 5.75 M 6 = f( 0.75)(0.5)+f( 0.5)(0.5)+f(0.5)(0.5)+f(0.75)(0.5)+f(.5)(0.5)+f(.75)(0.5) = (d) M Here n = 6 and = 0.5 L 6 =v(0)(0.5) + v(0.5)(0.5) + v()(0.5) + v(.5)(0.5) + v()(0.5) + v(.5)(0.5) =0(0.5) + 6.(0.5) + 0.8(0.5) + 4.9(0.5) + 8.(0.5) + 9.4(0.5) =34.7 R 6 =v(0.5)(0.5) + v()(0.5) + v(.5)(0.5) + v()(0.5) + v(.5)(0.5) + v(3)(0.5) =6.(0.5) + 0.8(0.5) + 4.9(0.5) + 8.(0.5) + 9.4(0.5) + 0.(0.5) = The midpoint sum seems to best approimate the area: M 6 = v(0.5)()+v(.5)()+v(.5)()+v(3.5)()+v(4.5)()+v(5.5)() = = 53ft = π n, i = πi n, so: A n i= n f( i ) n i= n ( π n ) πi πi cos( n n ) 5... The area under the curve of f() = tan() from 0 to π 4

44 44 PEYAM RYAN TABRIZIAN (a) We are given that the polygon is made out of n congruent triangles, so A n = n T, where T is the area of each triangle. So all we need to find is T. Here again, a picture tells a thousand words, so by drawing the picture of such a triangle, we can figure out its area: A/Math A Summer/Solution Bank/Polygon.png Using the picture, you ll notice that: T = A B = A B And we can divide the triangle into two right triangles, and hence use trigonometry to calculate A and B! Here, θ = π n, the central angle! We get: And so, we get: T = A B = r sin ( θ ) r cos cos sin ( ) θ = B r B = r cos ( ) θ = A r A = r sin ( ) θ = r sin ( ) θ ( ) θ ( ) θ cos ( ) θ = r ( sin θ ) = r sin(θ) = r sin ( ) π n

45 SOLUTION - BANK 45 Here, we used the fact that, in general, sin() cos() = sin(), so sin() cos() = sin(). And so, we have: A n = n T = n r sin ( ) π = n nr sin ( ) π n (b) Actually, the hint tells us that we don t even have to use l Hopital s rule, but rather the rule that: We have that: sin() lim = 0 ( ) lim A π n n n r n sin n = ( ) π r lim n sin n n = sin ( ) π r n lim n = r lim n n =π r lim n =πr lim 0 sin() =πr () =πr π sin ( ) π π n n sin ( ) π π n n ( = π ) n The basic idea is that, if we have an indeterminate form of the form 0, we rewrite as 0, or we rewrite 0 as. Here, for eample, we wrote n = in order to apply the hint in the problem! n And hooray, you just proved that the formula for the area of a circle of radius r is πr. But actually, you didn t, because trigonometry, which you used in (a), relies heavily on this formula! 5... = 5 = 0., so: Section 5.: The definite integral

46 46 PEYAM RYAN TABRIZIAN 0 sin( )d f(0.)(0.) + f(0.3)(0.) + f(0.5)(0.) + f(0.7)(0.) + f(0.9)(0.) = sin(0.0)(0.) + sin(0.09)(0.) + sin(0.5)(0.) + sin(0.49)(0.) + sin(0.8)(0.) = π π cos() d 5... Here = 6 n and i = + 6i n. 5 ( + 3)d n i= n i= n ( + 3 i ) n n n i= n n i= n i= n n 6 6i ( + 3( + n n )) 6 8i ( + n n ) n n + 08i n ) + n i= 08 (n) + n n 08i n ) n i) i= 08 n(n + ) + n n = + 08 = + 54 =4 And you thought 4 was not the answer to everything =)! Here = n and i = i n

47 SOLUTION - BANK 47 0 ( )d n i= n i= n ( ( i ) ) n n i= 4 n n 4i ( n n ) n 4 n 8i n 3 n 8 n n 3 i= i= 4 n n (n) 8 n(n + )(n + ) n 3 6 =4 8 6 i = = 9 n, so i = + 9i n, and so: ln()d n i= n ()( i 4 ln( i )) n n i= (( 9 + 9i ) ( 4 ln + 9i )) n n n (a) 4 (the area of the large triangle) (b) π (minus the area of the semicircle) (c) 4 π + = 9 π (the area of the large triangle minus the area of the semicircle plus the area of the small triangle) π (it s the area of a semicircle with radius ) d = 5 0 d 6 0 d = = 5 = e d = 3 e d 3 d = (e3 e) f()d+ 5 f()d f()d = 5 f()d f()d = 5 f()d m f() M, so integrating from 0 to, we get m f()d M 0 f()d = 5 f() On [0, ],, so + +, so + +, so integrating from 0 to, we get 0 + d 0 + d

48 48 PEYAM RYAN TABRIZIAN As usual, let f() be as in the problem, a = 0, b =, i = i n, and () = n. First, let s draw a picture of what s going on: A/Math A Summer/Solution Bank/Nonintegrable.png In the picture above, the green dots represent where f() = 0 and the red lines represent where f() =. This problem is unlike the problem above! In this case, the function does not blow up to infinity, but it can t make up its mind! We need to somehow use this fact in order to show that f is not integrable! But even though this problem is different, the general strategy is almost the same. f integrable means that no matter how we choose the i, we get the same answer! So to show that something is NOT integrable, we have to pick two different sets of points i and y i that give us two different answers! And here is where we use the fact that f looks the way it does. Namely, let i be your favorite rational number in [ i, i ] and yi your favorite irrational number in [ i, i ]! For eample (you don t have to write this, but it s better if you do!), you can choose: i = i = n yi = i n

49 SOLUTION - BANK 49 And you can check that i [ i, i ], and y i [ i, i ]. But the point is that i is rational, and so f ( i ) = 0 by definition of f, and thus the Riemann sum equals to: lim n i= n f ( i ) () n n i= 0 n = 0 And yi is rational, and so f (yi ) = by definition of f, and thus the Riemann sum equals to: lim n i= n f (yi ) () n n i= n n n n = n And so we get two different answers (even though we re supposed to get the same answer if f were integrable)!!! Which shows that f is not integrable on [0, ]! Let f() =, a = 0, b =. Then i = i n and () = n. How can we show that something is not integrable? The main point is: Given n we need to CHOOSE a set of points i [ i, i ] that fails (whatever that might mean). As discussed in section, the following choice is a good one: = n i = i for i This works BECAUSE [ 0, ] = [0, n ] and i [ i, i ] (for i ). Always check this on the eam! Because then, we have: n f ( i ) () f ( ) () = n = n n = n i= n Here, we use the fact that every term in the sum is positive, so the sum is greater than its first term f ( ) (). Also, () = n. And now, if we let n, the right-hand-side goes to, and so by comparison, lim n i= n f ( i ) () = So with this choice of i, things have gone awry! The Riemann sum blows up to infinity, and so f is not integrable over [0, ]. The point is: if a function is integrable, then its integral has to be finite. Other solution: Some people wrote up another solution, which is also pretty clever! Basically, let i = i = i n ( i n), which is in [ i, i ]. Then: