Answers and Hints 99.

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1 Answers and Hints (I.) The decimal epansion of /7 = repeats after 6 digits. Since 007 = the 007 th digit is the same as the rd, which happens to be a. (I.5) Yes these are the same sets. Both sets consist of all positive real numbers: since they contain eactly the same numbers, they are the same sets. (I.6) 00 =. = + = 99 = = = 99. Similarly, 000y = 7 + y so y = In z the initial is not part of the repeating pattern, so subtract it: z = Now let w = You get 000w = w = w = w. Therefore w = From this you get z = 5 + w = = (I7.) They are the same function. Both are defined for all real numbers, and both will square whatever number you give them, so they are the same function. (I7.4) Let be any number. Then, f(), if it is defined, is the largest (I7.6) The domain of k is (0, ), and k () =. (I7.8a) False: Since arcsin is only defined if and hence not for all, it is not true that sin ( arcsin ) = for all real numbers. However, it is true that sin(arcsin ) = for all in the interval [, ]. (I7.8b) arcsin(sin ) is defined for all since sin is defined for all, and sin is always between and. However the arcsine function always returns a number (angle) between π/ and π/, so arcsin(sin ) = can t be true when > π/ or < π/. For π/ it is true that arcsin sin =. (I7.8c) Again, not true: if = π/ then tan is not defined and therefore arctan(tan ) is not defined either. Apart from that, arctan(anything) always lies between π/ and +π/, so arctan(tan ) cannot be the same as if either > π/ or < π/. (I7.8d) True. (I7.4a) Set = / in f( + ) = and you find f(0) = ( /) = 9 4. (I7.4b) Set = 0 in f( + ) = and you find f() = 0 = 0. (I7.4c) Solve + = π for : = π π. Substitute this in f( + ) = and you find f(π) = ( ) π. (I7.4d) Solve + = t for : = t. Substitute this in f( + ) = and you find f(t) = ( ) t. 75

2 76 ANSWERS AND HINTS (I7.4e) From the previous problem we know what f(t) is for any t so just substitute t = : f() = b ( ). (I7.4f) f() = ( ( )/ ) = 4. (I7.4g) f(f()) = ( f() ) { ( ) }. = (I7.5a) We know f ( ) + = for all, so if we want to know f() then we have to find an with + =. Solving + = for you find = 0. Substitute = 0 in f( ) + = and you get f() = 0 =. (I7.5b) To find f(0) you proceed as above, this time solving = 0 for. In this case there is no solution +, and therefore the equation f ( ) + = does not tell us what f(0) is. Conclusion: either 0 is not in the domain of f, or we cannot tell what f(0) is from the information provided in the problem. (I7.5c) To find f(t) you do the same as when you want to find f(). We know f ( ) + = for all, so if we want to know f(t) then we have to find an with + = t. Solving = t for you find + = t. Substitute = t in f( ) ( + = and you get f(t) = t ) = t 4. (I7.5d) f(f()) = f() 4 = f() 4 = 4. You could simplify this if you wanted to, but 4 that was not part of the question. (I7.5e) After finding f(t) = t 4 you can substitute t = and you find f() = 4. (I7.5f) f() = 4 = and therefore f(f()) = f( ) = 4 = 4. (I7.6) No. For instance if you set = you get f() = + =, and if you set = then you get f(( ) ) = ( ) +, i.e. f() = 0. But f() can t be equal to both and 0, the formula f( ) = + cannot be true for all real numbers. (I7.8) g() = ( ) = ( + ) = [ ( ) ] = ( ) +, so the range of g is (, ]. Alternatively: y = g() y = y = 0. The quadratic formula says that the solutions are = 4 ± 6 8y. 4 If 6 8y < 0 then there are no solutions and y does not belong to the range of g. If 6 8y 0 then there is at least one solution and y does belong to the range of g. Conclusion, the range of g consists of all y with 6 8y 0, i.e. all y. (II6.) (a) y = ( + ) ( + ) + [ + ] = ( ) + ( ) so that y = + (II6.4a) In this picture s(t) is on the horizontal ais and t is on the vertical ais, so horizontal and vertical have been swapped. This curve should pass the horizontal line test, which it does. (II6.4b) With a ruler I tried to draw the closest tangent lines at the four different times. Then I measured the slope of those four lines using the grid.

3 ANSWERS AND HINTS 77 (II6.5) At A and B the graph of f is tangent to the drawn lines, so the derivative at A is and ther derivative at B is +. (II6.6) : feet. y pounds. (II6.7) Gallons per second. y dy and are measured in pounds per feet. d (II6.8b) (a) A() is an area so it has units square inch and is measured in inches, so da is measured in d inch inch = inch. very small triangle (ignore) h thin parallelogram (b) Hint: The etra area A that you get when the side of an equilateral triangle grows from to + can be split into a thin parallelogram and a very tiny triangle. Ignore the area of the tiny triangle since the area of the parallelogram will be much larger. What is the area of this parallelogram? The area of a parallelogram is base time height so here it is h, where h is the height of the triangle. A Conclusion: h = h. The derivative is therefore the height of the triangle. (III4.) The equation (9) already contains a function f, but that is not the right function. In (9) is the variable, and g( ) = (f( + ) f())/ is the function; we want lim 0 g( ). (III4.4) δ = ε/. (III4.5) δ = min {, 6 ε} (III4.6) f() ( 7) = = 5. If you choose δ then < δ implies < <, so that 5 is at most 5 = 4. So, choosing δ we always have f() L < 4 and f() L < ε will follow from < 4 ε. Our choice is then: δ = min{, 4 ε}. (III4.7) f() =, a =, L = 7. When = one has = 7, so 7 = 0 for =. Therefore you can factor out from 7 by doing a long division. You get 7 = ( )( + + 9), and thus f() L = 7 = Never choose δ >. Then < δ will imply < < 4 and therefore = 7. So if we always choose δ, then we will always have 7 7δ for < δ. Hence, if we choose δ = min {, 7 ε} then < δ guarantees 7 < ε.

4 78 ANSWERS AND HINTS (III4.9) f() =, a = 4, L =. You have = ( )( + ) + = 4 + and therefore f() L = + 4. (0) Once again it would be nice if we could replace /( + ) by a constant, and we achieve this by always choosing δ. If we do that then for 4 < δ we always have < < 5 and hence since /( + ) increases as you decrease. <, + + So, if we always choose δ then 4 < δ guarantees f() < { which prompts us to choose δ = min, ( } + )ε. + 4, A smarter solution: We can replace /( + ) by a constant in (0), because for all in the domain of f we have 0, which implies +. Therefore 4, and we could choose δ = ε. (III4.0) Hints: so = = (III4.) We have = 4 +. If we choose δ then < δ implies < < so that Therefore 7 < we don t care 4 + < <, 5 so if we want f() < ε then we must require < 5ε. This leads us to choose δ = min {, 5ε}. (III4.6) A(, ); B( 5, ); C( 7, ); D(, 0); E( 5, ). (III4.7) False! The limit must not only eist but also be equal to f(a)! (III4.8) There are of course many eamples. Here are two: f() = / and f() = sin(π/) (see 8.) (III4.9) False! Here s an eample: f() = and g() =. Then f and g don t have limits at = 0, but f() + g() = does have a limit as 0. (III4.0) False again, as shown by the eample f() = g() =.

5 ANSWERS AND HINTS 79 (III4.a) False, for the following reason: g() is the difference of f() + g() and f(). If lim a f() eists and lim a f() + g() also eists, then { } lim g() = lim f() + g() f() a a { } = lim f() + g() lim f() a a also has to eist. (III4.b) True, as shown by the eample f() =, g() =, and a = 0. For these two functions we have lim f() = 0 (i.e. eists) 0 lim g() = does not eist 0 lim f()g() = lim = (i.e. eists) 0 0 You can make up other eamples, but to show that this statement is true you only need one eample. (III4.c) True, as shown by the same eample f() =, g() =, a = 0. This time we have lim f() = 0 (i.e. eists) 0 lim g() = does not eist 0 f() lim 0 g() = lim 0 / = lim 0 = 0 (i.e. eists) You can make up other eamples, but to show that this statement is true you only need one eample. (III4.d) False: If lim a g() and lim a f()/g() both eist then f() lim f() = lim g() a a g() ( ) lim g() a ( lim a f() ) g() and therefore lim a f() would also have to eist. (III6.) the limit is. (III6.) The limit is. Use : θ sin θ = sin θ. θ sin α cos α (III6.4) sin α = sin α cos α so the limit is lim α 0 = lim sin α α 0 cos α =. sin α sin α Other approach: sin α = α α. Take the limit and you get. sin α α α (III6.5). tan 4α tan 4α (III6.6) = sin α 4α 4α α α sin α. Take the limit and you get... = =. (III6.7) Hint: multiply top and bottom with + cos. (III6.8) Hint: substitute θ = π ϕ, and let ϕ 0. Answer: 0. (III6.9) Multiply top and bottom with + cos. The answer is. (III6.0) Substitute = u and let u 0. Answer:. (III6.) Multiply and divide by + cos. Write tan as sin cos. Answer is.

6 80 ANSWERS AND HINTS (III6.) sin( ) cos = sin( ). The answer is. cos (III6.) Substitute θ = π/ and remember that cos = cos(θ + π ) = sin θ. You get π lim π/ cos = lim θ θ 0 sin θ =. (III6.4) Similar to the previous problem, once you use tan = sin. The answer is again. cos (III6.5) /9 (III6.6) Substitute θ = π. Then lim π θ = 0, so sin lim π π = lim sin(π + θ) sin θ = lim =. θ 0 θ θ 0 θ Here you have to remember from trigonometry that sin(π + θ) = sin θ. (III6.7) Divide top and bottom by. The answer is /. (III6.8) Note that the limit is for! As goes to infinity sin oscillates up and down between and +. Dividing by then gives you a quantity which goes to zero. To give a good proof you use the Sandwich Theorem like this: Since sin for all you have sin. Since both / and / go to zero as the function in the middle must also go to zero. Hence sin lim = 0. (III6.9) zero again. (III6.) This is not a rational function, but you can use the same trick: factor out the highest power of from numerator and denominator. You get cos + = cos +. cos Using the Sandwich Theorem as in the previous problems you get lim = 0. With the limit properties you then get lim cos + = lim = = 0. cos + (III6.4). (III6.5a) tan θ sin θ lim θ 0 θ = (III6.5b) tan 0. sin 0. (0.) = , which is really a lot smaller than 0..

7 ANSWERS AND HINTS 8 (III6.6) sin 0. 0., cos 0. (0.) = 0.98, tan 0. = (sin 0.)/(cos 0.) 0.. sin ( π/ 0. ) = cos cos ( π/ + 0. ) = sin tan ( π/ 0. ) = tan = 50. (III6.7) Same approach as before, but in this problem you first have to convert 0 to radians: You get 0 = 0 60 πradians = π 8. sin 0 π 8, cos 0 π 8. You don t have a calculator, so, had this been 965, you would have enthusiastically computed these numbers by hand (to two decimals). For a really rough estimate assume π, to get 0 8 = 6 0.7, tan 0 sin 0 8 = 6 0.7, cos 0 ( ) 6 = = To find the other epressions, use sin( π + θ) = cos θ and 00 = sin 00 = cos cos 90 = cos tan 80 = ( tan 0 ) 6. (III6.9) No. As 0 the quantity sin oscillates between and + and does not converge to any particular value. Therefore, no matter how you choose k, it will never be true that lim 0 sin = k, because the limit doesn t eist. (III6.0) The function f() = (sin )/ is continuous at all 0, so we only have to check that lim 0 f() = f(0), i.e. lim 0 sin = A. This only happens if you choose A =. (III8.a) No vertical asymptote. No horizontal asymptote. If there were a slanted asymptote then m = lim = 0. But n = lim f() m = lim does not eist. (III8.5) We are given that lim f() m n = 0. Adding n to both sides gives us lim f() m = n, which is the formula for n we had to prove. To get the formula for m we multiply with lim / = 0

8 8 ANSWERS AND HINTS and use the limit properties: f() m n lim = Work out the left hand side: This implies and thus ( lim f() m n) ( ) lim = f() 0 = lim m n. f() 0 = lim m f() lim = m. 0 0 = 0. (IV0.6d) The derivative of /( + ) is /( + ), so the derivative at = is A = 9. On the other hand /( + ) = is constant, so its derivative is B = 0. (IV0.6e) Simplicio is mistaken. The mistake is that he assumes that setting equal to some constant and then differentiating gives the same result as first differentiating w.r.t. and then setting equal to some constant. This eample shows that is not true. (IV.5) f () = tan / cos f () = / cos tan sin / cos f () = 8 sin / cos sin / cos tan / cos tan sin / cos 6. Since tan = cos one has g () = f () and g () = f (). (IV4.6a) f () = cos + sin. (IV4.6b) f () = π cos π (IV4.6c) f () = cos(cos ) ( sin ) = sin cos cos. (IV4.6d) f () = sin sin (IV4.6e) f () = 4 ( ) cos + + (IV4.6f) f () = (cos )( sin ) (cos ) (IV4.8) f () = cos π + π sin π. At C one has =, so cos π = 0 and sin π =. So at C one has f () = π. (IV4.9) v() = f(g()) = ( + 5) + = w() = g(f()) = ( + ) + 5 = + 6 p() = f()g() = ( + )( + 5) = q() = g()f() = f()g() = p(). (IV4.b) (a) If f() = sin a, then f () = a sin a, so f () = 64f() holds if a = 64, i.e. a = ±8. So sin 8 and sin( 8) = sin 8 are the two solutions you find this way. (b) a = ±8, but A and b can have any value. All functions of the form f() = A sin(8 + b) satisfy ( ). (IV4.d) (a) V = S, so the function f for which V (t) = f(s(t)) is the function f() =. (b) S (t) is the rate with which Bob s side grows with time. V (t) is the rate with which the Bob s volume grows with time.

9 ANSWERS AND HINTS 8 Quantity Units t minutes S(t) inch V (t) inch S (t) inch/minute V (t) inch /minute (c) Three versions of the same answer: V (t) = f(s(t)) so the chain rule says V (t) = f (S(t))S (t) V (t) = S(t) so the chain rule says V (t) = S(t) S (t) V = S so the chain rule says dv ds = S dt dt. (d) We are given V (t) = 8, and V (t) =. Since V = S we get S =. From (c) we know V (t) = S(t) S (t), so = S (t), whence S (t) = inch per minute. 6 (IV7.) If the distance between the bottom of the pole and the wall is (t), and if the angle between the pole and the floor is θ(t), then (t) = 0 cos θ(t). This is true at all times so we are allowed to differentiate the equation, which then leads to d dθ = 0 sin θ dt dt. When the angle is π/4 we know that d = +7 feet/sec. Therefore dt dθ dt = d 0 sin θ dt = 0 sin π 4 7 = 7 5 rad/sec. (IV7.5) This is the situation: h(t) D 4000 ft θ camera Let h(t) be the height of the rocket at time t. Then the distance between camera and rocket (let s call it D) and the angle θ satisfy h + (4000) = D h, = tan θ This is true at all times, so we can differentiate these equations, which gives us h dh dd = D dt dt, and dh 4000 dt = dθ cos θ dt. We are given that when h = 000 we have dh = 600. By Pythagoras, we then also know dt D = = 5000, and cos θ = = 4 5. Therefore and dd dt = h dh D dt = 600 = 60m/sec, 5 dθ dt = cos θ dh 4000 dt = (4/5) = rad/sec. (IV7.7) First, what is the area of an isosceles triangle whose sides and opening angle you know? This is not a formula you should memorize. Instead try figuring it out. Here is the drawing:

10 84 ANSWERS AND HINTS inch height = sin θ (inch) θ width= inch The area is width height, i.e. A(t) = Area = sin θ = sin θ in. This is true at all times, so we are allowed to differentiate, resulting in We know that da dt da dt = cos θ dθ dt. = 0.5in /sec. When θ = 60 we also know cos θ =, and hence 0.5 = dθ dθ and thus dt dt = 0.5rad/sec. (IV7.8) The drawing: ft/sec P 0ft θ If (t) is the coordinate of the point P, then 0 = tan θ(t), or, 0 = tan θ. This is true at all times, so we may differentiate both sides of the equation, resulting in dθ 0 = cos θ dt + d tan θ. dt When θ = π/ we have tan θ = so = 0/ ft. We are also given that d = ft/sec (note the dt minus, which is there because the point P is moving to the left). When θ = π/ we also have cos θ =. We are led to the following equation which implies 0 = 0 ( ) dθ dt + ( ), dθ dt = 0 4 = 9 40 rad/sec. (IV7.) The situation is thus:

11 ANSWERS AND HINTS 85 s(t) road car road 0 m D(t) police officer The distance from the car to the police officer is D(t). The distance along the road from the car to the nearest point to the officer on the road is s(t). The speed of the car (which we want to know) is s (t). The Doppler radar measures the rate at which D(t) is changing. The relation between D(t) and s(t) is D = s + (0). This is true at all times so we are allowed to differentiate the relation, which leads to D dd dt ds ds = s and therefore dt dt = D dd s dt. Given is that at some moment D = 0, and dd = 0 (negative because the car is moving toward the dt officer). Pythagoras says that s = D (0) = 00 = 0. Therefore ds dt = 0 0 ( 0) = 0 m/sec. Note : the phrasing of the problem is ambiguous since 0 feet away could also be taken to mean that s = 0 when the police officer measures the car s speed. In that case the above computation still is valid, ecept at the end one must substitute s = 0, D = (0) + (0) = 0 5. This leads to ds dt = 0 5 ( 0) = 5 5 m/sec. 0 (V.) At =. (V.) At = a/. (V.) At = a + a. (V.5) At = a +. (V.) False. If you try to solve f() = 0, then you get the equation + = 0. If 0 then this is the same as + = 0, which has no solutions (both terms are positive when 0). If = 0 then f() isn t even defined. So there is no solution to f() = 0. This doesn t contradict the IVT, because the function isn t continuous, in fact it isn t even defined at = 0, so the IVT doesn t have to apply. (V.6) Not necessarily true, and therefore false. Consider the eample f() = 4, and see the net problem. (V.7) An inflection point is a point on the graph of a function where the second derivative changes its sign. At such a point you must have f () = 0, but by itself that it is no enough. (V.0) The first is possible, e.g. f() = satisfies f () > 0 and f () = 0 for all. The second is impossible, since f is the derivative of f, so f () = 0 for all implies that f () = 0 for all. (V.5) y = 0 at =, 0, 0. Only sign change at =, not at = 0. = 0 loc min; = 4 loc ma; = / inflection point. No global ma or min.

12 86 ANSWERS AND HINTS (V.6) zero at = 0, 4; sign change at = 4; loc min at = 8 ; loc ma at = 0; inflection point at = 4/. No global ma or min. (V.7) sign changes at = 0, ; global min at = /4 / ; no inflection points, the graph is conve. (V.8) mirror image of previous problem. (V.9) 4 + = ( )( + ) so sign changes at = ±. Global min at = 0; graph is conve, no inflection points. (V.0) Sign changes at ±, ±; two global minima, at ± 5/; one local ma at =0; two inflection points, at = ± 5/6. (V.) Sign change at = 0; function is always increasing so no stationary points; inflection point at = 0. (V.) sign change at = 0, ±; loc ma at = /5 /4 ; loc min at = /5 /4. Inflection point at = 0. (V.) Function not defined at =. For > sign change at = 0, no stationary points, no inflection points (graph is concave). Horizontal asymptote lim f() =. For < no sign change, function is increasing and conve, horizontal asymptote with lim f() =. (V.4) global ma (min) at = (=-), inflection points at = ± ; horizontal asymptotes lim ± f() = 0. (V.5) y = 0 at = 0 but no sign changes anywhere; = 0 is a global min; there s no local or global ma; two inflection points at = ± ; horizontal asymptotes at height y =. (V.6) Not defined at =. For > the graph is conve and has a minimum at = + ; for < the graph is concave with a maimum at =. No horizontal asymptotes. (V.7) Not def d at = 0. No sign changes (ecept at = 0). For > 0 conve with minimum at =, for < 0 concave with maimum at =. (V.8) Not def d at = 0. Sign changes at = ± and also at = 0. No stationary points. Both branches ( > 0 and < 0) are increasing. Non inflection points, no horizontal asymptotes. (V.9) Zero at = 0, sign only changes at ; loc min at = ; loc ma at =. Inflection point at = /. (V.0) Changes sign at = ± and = 0; loc min at ( + 7)/, loc ma at ( 7)/; inflection point at =. (V.) Factor y = 4 = ( ). One zero is obvious, namely at = 0. For the other(s) you must solve = 0 which is beyond what s epected in this course. The derivative is y = 4. A cubic function whose coefficients add up to 0 so = is a root, and you can factor y = 4 = ( )(4 + + ) from which you see that = is the only root. So: one stationary point at =, which is a global minimum The second derivative is y = 6; there are two inflection points, at = and at = 0. (V.) Again one obvious solution to y = 0, namely = 0. The other require solving a cubic equation. The derivative is y = which is also cubic, but the coefficients add up to 0, so = is a root. You can then factor y = = ( )(4 ). There are three stationary points: local minima at =, = 4, local ma at = 4 +. One of the two loc min is a global minimum.

13 ANSWERS AND HINTS 87 (V.) Global min at = 0, no other stationary points; function is conve, no inflection points. No horizontal asymptotes. (V.4) The graph is the upper half of the unit circle. (V.5) Always positive, so no sign changes; global minimum at = 0, no other stationary points; two inflection points at ±. No horizontal asymptotes since lim ± 4 + = (DNE). (V.6) Always positive hence no sign changes; global ma at = 0, no other stationary points; two inflection points at = ± 4 /5; second derivative also vanishes at = 0 but this is not an inflection point. (V.8) Zeroes at = π/4, 7π/4. Absolute ma at = π/4, abs min at = 5π/4. Inflection points and zeroes coincide. Note that sin + cos = sin( + π 4 ). (V.9) Zeroes at = 0, π, π/ but no sign change at π/. Global ma at = π/, local ma at = π/, global min at = 7π/6, π/6. (V4.) If the sides are and y, then the area is y = 00, so y = /. Therefore the height plus twice the width is f() = + y = + /. This is etremal when f () = 0, i.e. when f () = / = 0. This happens for =. (V4.) Perimeter is R + Rθ = (given), so if you choose the angle to be θ then the radius is R = /( + θ). The area is then A(θ) = θr = θ/( + θ), which is maimal when θ = (radians). The smallest area arises when you choose θ = 0. Choosing θ π doesn t make sense (why? Draw the corresponding wedge!) You could also say that for any given radius R > 0 perimeter = implies that one has θ = (/R). Hence the area will be A(R) = θr = R ( (/R) ) ) = R R. Thus the area is maimal when R =, and hence θ = radians. Again we note that this answer is reasonable because values of θ > π 4 don t make sense, but θ = does. (V4.b) (a) The intensity at is a function of. Let s call it I(). Then at the distance to the big light is, and the distance to the smaller light is 000. Therefore I() = (000 ) (b) Find the minimum of I() for 0 < < 000. I () = (000 ). I () = 0 has one solution, namely, = 000. By looking at the signs of I () you see that I() must have a minimum. If you don t like looking at signs, you could instead look at the second derivative which is always positive. I () = (000 ) 4 (V4.4a) The soup can is made of two disks (the top and bottom) and the side. Both top and bottom have area πr. The side can be made by bending a rectangular piece of metal into a cylinder. The sides of the rectangular piece of metal needed for this are h (height of the cylinder) and πr (perimeter of the cylinder). Therefore the total area of the metal that is needed to make a cylinder of height h and radius r is The volume of that cylinder is A = πr + πrh = 00in. V = area base height = πr h. We have to choose which variable we use to distinguish the different sizes of cans that we can make. The variables at hand are r and h. If we choose h then we would have to get rid of r, which involves solving the equation A = 00 for r. That equation is quadratic, and formula for r will involve square roots. The other choice is the variable r. Then we have to eliminate h, i.e. solve A = 00 for h. The solution is πr + πrh = 00 = h = 00 πr. πr

14 88 ANSWERS AND HINTS (V4.4c) (a) The optimal radius is r = 50/π, and the corresponding height is h = 00/(πr) = 00/ 50π. (V6.a) The straight line y = +, traversed from the top right to the bottom left as t increases from to +. (V6.b) The diagonal y = traversed from left to right. (V6.c) The standard parabola y =, from left to right. (V6.d) The graph = sin y. This is the usual Sine graph, but on its side. π π = sin(y) π π (V6.e) We remember that cos α = sin α, so that (t), y(t) traces out a part of the parabola y =. Looking at (t) = sin t we see the point ((t), y(t)) goes back and forth on the part of the parabola y = between = and = +. (V6.f) The unit circle, traversed clockwise, 5 times every π time units. Note that the angle θ = 5t is measured from the y-ais instead of from the -ais.

15 ANSWERS AND HINTS 89 (V6.g) Circle with radius and center (, ) (it touches the and y aes). Traversed infinitely often in counterclockwise fashion. (V6.h) Without the this would be the standard unit circle (dashed curve below). Multiplying the component by stretches this circle to an ellipse. So ((t), y(t)) traces out an ellipse, infinitely often, counterclockwise. (V6.i) For each y = t there is eactly one t, namely, t = y /. So the curve is a graph (with as a function of y instead of the other way around). It is the graph of = y / = y. The curve is called Neil s parabola. (V6.6) Since sin t + cos t = we have y(t) = (t) on this curve. The curve is a straight line and therefore its curvature is zero. (VI9.) dy/d = e e. Local min at = ln. d y/d = e + 4e > 0 always, so the function is conve. lim ± y =, no asymptotes. (VI9.4) dy/d = e 4e. Local min at = ln 4. d y/d = 9e 4e changes sign when e = 4 9, i.e. at = ln 4 9 = ln = ln ln. Inflection point at = ln ln. lim f() = 0 so negative ais is a horizontal asymptote.

16 90 ANSWERS AND HINTS lim f() =... no asymptote there. (VI9.5a) The diagonal y = again, but since = e t can only be positive we only get the part in the first quadrant. At t = we start at the origin, as t + both and y go to +. (VI9.5b) The graph of y = ln, or = e y (same thing), traversed in the upwards direction. (VI9.5c) The part of the graph of y = / which is in the first quadrant, traversed from left to right. (VII5.b) 4 k= k = This adds up to 5, but the point of this question was to get the above answer. (VII5.c) (VII5.c) Choosing left endpoints for the c s gives you Choosing right endpoints gives f(0) + f( ) + f() + f( ) = f( ) + f() + f( ) + f() (VII5.a) F (n) F (n ) = n (n ) = n (n n + ) = n = f(n). (VII5.b) Substitute f() = F () F (0), f() = F () F (), etc... You get many terms, and almost all of them cancel: f() = F () F (0) f() = F () F () f() = F () F (). f(79) = F (79) F (78) + f(80) = F (80) F (79) So the answer is S = F (80) F (0) f() + + f(80) = F (80) F (0). In the particular eample where f(n) = n you get = (80) = 600. (VII5.4) The Riemann-sum is the total area of the rectangles, so to get the smallest Riemann-sum you must make the rectangles as small as possible. You can t change their widths, but you can change their heights by changing the c i. To get the smallest area we make the heights as small as possible. So, for the smallest possible Riemann sum choose c i to be the point in the interval [ i, i+ ] where f() attains its minimum. For the largest possible Riemann sum choose each c i to be the point in [ i, i+ ] where f() is maimal. In this eample this means you should choose: c =, c somewhere between and where f has its maimum, c =, c 4 =, c 5 = 4, and c 6 = 5. (VII0.a) The Fundamental Theorem says that F (t)dt = F () F (0), 0 so it is not right, unless F (0) = 0.

17 ANSWERS AND HINTS 9 (VII0.b) Here the integral is an indefinite integral, so F ()d stands for any function whose derivative is F (). Clearly F () is such a function, but so is F () + C for any constant C. So Simplicio is almost right. It would be more proper to say that F ()d = F () + C (VII0.7a) q ( ) ( q) = q. (VII0.7b) The answer we got in part (a) is defined for all q ±, but it is not valid for all those values of q. Here s why: The integral in part (a) is over the interval q < < q and the function that is being integrated is not defined at = ±, so those values of should not lie in the integration interval. Therefore only < q < is allowed. (VII0.9b) (a) The first derivative of erf () is, by definition erf () = π e, so you get the second derivative by differentiating this: erf () = 4 π e. This is negative when > 0, and positive when < 0 so the graph of erf () has an inflection point at = 0. (b) Wikipedia is not wrong. Let s figure out what sign erf ( ) has (for instance). By definition you have erf ( ) = e t dt. π 0 Note that in this integral the upper bound ( ) is less than the lower bound (0). To fi that we switch the upper and lower integration bounds, which introduces a minus sign: erf ( ) = 0 e t dt. π The integral we have here is positive because it s an integral of a positive function from a smaller number to a larger number, i.e. it is of the form b a f()d with f() 0 and with a < b. With the minus sign that makes erf ( ) negative. (VII0.0a) If f() = then g() = 0 ( t)tdt = ( 0 t t ) dt. Keep in mind that the integration variable is t and is, as far as the integral is concerned, just a number. So [ g() = t ] t = = t=0 6. (VIII9.) The answer is π/6. To get this using the integral you use formula (96) with f() =. You get f () = /, so + f () =. The integral of that is arcsin (+C), so the answer is arcsin arcsin 0 = π/6.