Calculus 2 - Examination

Transcription

1 Calculus - Eamination Concepts that you need to know: Two methods for showing that a function is : a) Showing the function is monotonic. b) Assuming that f( ) = f( ) and showing =. Horizontal Line Test: A function is (and has an inverse) iff no horizontal line intersects the graph of f more than once. Finding inverses of functions: ) Switch s and y s. ) Solve for y. Properties of f and f : ) Domain of f = range of f. ) Range of f = domain of f. 3) f(f ()) = for all in domain of f. 4) f (f()) = for all in domain of f. 5) The graphs of f and f are symmetric about y =. 6) If g = f with f and g differentiable, then g () = f (g()). The graphs of y = e and y = ln. Properties of log : ) log(y) ( ) = log + log y ) log = log log y y 3) log p = p log NOTE: The previous three properties work for any base including ln. Change of Base Formula: log a = ln ln a

2 Properties of ln and e : ) e ln Allison = Allison. ) ln(e Allison ) = Allison. Logarithmic Differentiation: a) Take natural logs of both sides. b) Differentiate implicitly with respect to. c) Solve for y entirely in terms of. Eponential Growth and Decay: y(t) = y()e kt y(t) - Population at time t. y() - Population at time. k - Growth or decay constant. e - Rumored to be just a number..788 Inverse Trig Functions: The graphs of y = sin, y = cos, and y = tan. How to compute things similar to sin(arccos). Indeterminate Forms and l Hôpital s Rule: The two forms where l Hôpital s Rule is applicable - - and. How to manipulate the other forms,,,, and into the two forms where you can use l Hôpital s Rule. Rates of growth: How to determine if a function grows faster or slower than another function. Integration by Parts: u dv = uv v du Goal: To choose u and dv so that the product of v and du is as simple as possible to antidifferentiate. How to integrate by parts multiple times in one problem.

3 3 Trig Integrals: Knowing the three pythagorean identities: Other: sin + cos = tan + = sec + cot = csc sin = ( cos ) cos = ( + cos ) sin = sin cos Goal : To find du. Once you identify du, you can use the trig identities to substitute for whatever remains. NOTE: Usually you want to keep the trig functions raised to even powers and pull off one sin or one cos to become du. Sometimes when presented with even powers, you can try pulling off a sec or csc. Trig Substitution: NOTE: Try a non-trig substitution first... If you see something like 4 9 you should think about sin θ. If you see something like 9 4 you should think about sec θ. If there is only addition like 9 +4 you should think about tan θ. NOTE: Remember that if you are working with an indefinite integral that is in terms of, you have to label/create a TRIANGLE to get everything back in terms of at the end. Integration by Partial Fractions: Be able to rewrite a fraction using partial fractions. (Non-repeating linear factors only.) Be able to identify what A, B, C, etc. are by picking convenient values for. Be able to integrate each piece once you have decomposed the original fraction. With linear factors, each piece should integrate to ln something.

4 4 DERIVATIVES THAT YOU NEED TO KNOW!!! y = n y = e y = n n y = e y = e u y = e u du d y = a y = a ln a where a > and a y = a u y = ln y = a u ln a du d y = where a > and a y = ln u y = u du d y = log a y = ln a y = log a u y = u ln a du d y = sin y = cos y = cos y = tan y = cot y = sec y = sin y = sec y = csc y = sec tan y = csc y = sin y = y = csc cot y = sec y = y = tan y = + y = sin u y = du u d y = sec u y = u u du d y = tan u y = du + u d

5 5 INTEGRALS THAT YOU NEED TO KNOW!!! n d n+ + C provided n n + d ln + C e d a d cos d sin d sec d d e + C a + C provided a > and a ln a sin + C cos + C tan + C sin + C d + d sec + C tan + C Direct substitution is the easiest method and should be tried first!!! You should know the following forms: If you see something involving: Transform it by introducing: # # sin θ # # sec θ # + # tan θ

6 6 Sample Eam (-). Given f() =, >. Let g() = f () and H() = g 3 (). Then H (8) =. Determine the integral 4 + d. 3. Find lim ( + a) /, where a is a fied non-zero number. 4. The derivative of y = ln(sin ), where <, is 5. Solve ln = ln() ln( + 3) for. 6. (P-C) Apply logarithmic differentiation to find the derivative of f() = ( cos ). 7. (P-C) Find e 3 d. 8. (P-C) Use trignometric substitution to evaluate 9. (P-C) Find sin d. d.. (P-C) Find sec 4 d.. (P-C) After days, a sample of radon- decayed to of its 8 original amount. What is the half-life of radon- in days? Simplify as much as possible.. (P-C) Find 3 4 d. Sample Eam (3-5) 3. log 7 written in terms of natural logarithms is ( 4. Find the eact value of sin tan ). 5. If f is an increasing function, then what is true about f? The choices were: A) increasing B) decreasing C) differentiable D) continuous E) not a function 6. Evaluate lim e + e sin(). 7. Find the derivative of f() = sin (e ). 8. After trig substitution, the integral d becomes (4 ) 5/

7 7 9. Evaluate lim (e ) /.. Evaluate π cos d.. (P-C) Differentiate f() = cos.. (P-C) Evaluate sin 3 θ cos θ dθ. 3. (P-C) Evaluate 4. (P-C) Same as # 9. :) + 4 d. 5. (P-C) The half-life of radioactive cobalt is 5.7 years. a) If a sample has a mass of 4 mg, find a formula for the function y(t) which represents the mass of the sample in mg that remains after t years. b) When will the mass of the sample be reduced to mg? Additional problems from other eams/books (6-9): 6. (P-C) Evaluate sin d. 7. Evaluate lim + ( + sin ) /. 8. (P-C) Evaluate sin 3 cos 4 d. 9. Evaluate π sin cos d.

8 8 Answers:. H () = 3[g()] g () H (8) = 3[g(8)] g (8) H (8) = 3[g(8)] f (g(8)) To go any further, we need to know g(8). Since g = f, I know that to find g(8) I just have to find something that I can put into the function f that produces 8. Since f(3) = 8, I know that g(8) = 3. So H (8) = 3[3] f (3) Since f () =, we have f (3) = 6 and H (8) = 7 6 = 7 6 = 9.. The first thing that I noticed about this integral is that the denominator was something squared +. I know that u + du is tan u + C, so I want to manipulate this into that form. 4 + d Let u =. Then du =. Introduce a and a to get 4 + d = du u + = tan u + C = tan ( ) + C 3. Anytime that you have a variable in the eponent remember that you can use the properties of logarithms to pull it in front of the logarithm. With this problem, you have to identify the form first. The form is which is indeterminate. I have to set it equal to y and then take the natural log of both sides to try to epress it in a form where l Hôpital s Rule is applicable. y = lim ( + a) / ln y = lim ln( + a) / ln y = lim ln y = lim ln( + a) ln( + a)

9 9 The form is now so I can use l Hôpital s Rule. ln y = lim ln y = a y = e a a + a 4. y = 5. sin ln = ln ln( + 3) ( ) ln = ln + 3 = = + 3 = ( + 5)( ) = is the only solution as 5 is not in the domain of ln. 6. y = cos ln y = ln cos ln y = cos ln y y = (cos sin ) ln + cos y = y [(cos sin ) ln + cos ] y = cos [(cos sin ) ln + cos ] 7. e 3 d = 3 3 e 3 d 3 Let u = 3. Then du = 3 d and we have e u du = 3 eu +C = 3 e3 +C

10 8. Since I have an integral of something of the form a u, I want to try a trig substitution involving sin θ. It would be nice if = sin θ because then I would be able to replace with sin θ which is just cos θ or cos θ. That is how I choose what to substitute. Let = sin θ. Then = sin θ and d = cos θ dθ. π/ d = cos θ cos θ dθ Since this is a deinite integral, I decided to change the limits of integration. If you plug in for in the equation = sin θ, we find that θ =. Plugging in for yields that θ = π/. π/ cos θ dθ = π/ ( + cos(θ)) dθ = θ + 4 sin(θ) If this were an indefinite integral, I would have to switch everything back into by using a right triangle where sin θ =. I would also need the identity sin(θ) = sin θ cos θ. Since the integral I have is a definite integral and I have already changed the limits of integration, I do not switch back. I just plug in π/ and and subtract. ( π ) 4 + () = π 4 π/ 9. sin d There are five main methods of integration that you have learned. When I look at a problem, I try the following order:. Integration by substitution.. Integration by partial fractions. 3. Trig integral. 4. Integration by trig substitution. 5. Integration by parts. The middle three techniques should be fairly obvious. Also, the order of the middle three is not important. The important part is that you should try Integraion by substitution FIRST and Integration by parts LAST. Attacking this problem:. The only option for integration by substitution is to let u = sin. As there is no du which would be, this won t work.. There is no fraction. If there was, I would see if I could factor the denominator so that I could use integration by partial fractions.

11 3. Trig integrals consist of the si trig functions sin, cos,..., NOT their inverses. 4. There is nothing of the form a u, u a, or u + a, so integration by trig substitution is OUT. 5. I have to integrate this by parts... u = sin du = d v = v = d sin d = sin d Now, I have another integral to evaluate, so I start back at integration by substitution. Here it works. If I choose u =, then du =. By introducting a and a, we have d = d = = + C u / du = Replacing this into the earlier computation gives us sin d = sin + + C u / / =.. sec 4 d = sec sec d }{{} du so u = tan and sec = tan + = (u + ) du = u3 3 + u + C = tan3 + tan + C 3 y(t) = y()e kt y() = y()ek 8 /8 = e k ln(/8) = k (ln(/8))/ = k

12 So the original equation becomes We want the half-life, so y(t) = y()e (t ln 8 )/ y(t) = y()e ln( 8) t/ y(t) = y() ( ) t/ 8 y() = y() = ( 8 ln = t ln ln ln 8 ( 8 ) t/ ) t/ ( ) 8 = t ln 3 ln = t 3 = t. 3 4 = ( 4)( + ) I know that can be written in the form ( 4)( + ) with A and B real numbers. We have ( 4)( + ) = A 4 + B +. ( 4)( + ), we obtain = A( + ) + B( 4). A 4 + B +, By multiplying both sides by This is true for any value of. So I pick two convenient values for. Let = 4. Then we have = A(5) + so A = Let =. Then we have Then which is = + B( 5) so B = ( ( 4)( + ) d = 4 + ln 4 ln + + C ) d +

13 3 3. ln 7 = ln ln 7 ( 4. sin tan ) = sin(j) tan = J tan J = }{{} J 5. g () = f (g()) 6. f (g()) > and g must be increasing. So sin(j) = +. Since f is always increasing, f >, so lim e + e H sin = lim H = lim e e sin + cos e + e cos + cos 4 sin = + + = e 7. f () = sin (e ) + (e ) 8. Although trig substitution is not the best way to do this problem, since i have 4, I would very much like the to be 4 sin θ because then it would just be 4( sin θ) = 4 cos θ. So I set them equal to each other. = 4 sin θ = sin θ d = cos θ dθ sin θ( cos θ) dθ d = (4 ) 5/ (4 cos θ) 5/ 4 sin θ cos θ dθ = 3 cos 5 θ = 8 sin θ cos 4 θ dθ 9. y = lim (e ) / ln y = lim ln y = lim ln y = y = e ln(e ) e e

14 4. cos d u = du = d v = sin v = cos d cos d = sin sin d = sin + cos + C.. y = cos ( ) ln y = ln cos ln y = ln ln(cos ) ln y = ln ln(cos ) y y = ln + ( sin ) cos [ y = y ln + + sin ] cos y = cos [ln + + tan ] sin 3 θ cos θ dθ = = = sin θ cos ( sin θ dθ) θ }{{} du u du u ( ) u du = + u + u + C = sec θ + cos θ + C 3. ( + 4 d = ) d + 4 ( = 4 ) d = + 4 d

15 To compute this second integral, we need to recognize the need for a trig substitution. I see this because of the + 4. I would like this to become 4 tan θ + 4 = 4(tan θ + ) = 4 sec θ, so I make the following substitutions: d = = 4 tan θ = tan θ d = sec θ dθ = 4 4 sec θ sec θ dθ dθ = θ + C = tan + C Replacing this into what was above, we obtain + 4 d = tan + C 4. Same as #9. 5. y(t) = y()e kt / = e 5.7k ln(/) = 5.7k ln(/) 5.7 = k y(t) = 4e t(ln(/))/5.7 y(t) = 4e ln(/)t/5.7 y(t) = 4(/) t/ ln(/) ln(/) = (/) t/5.7 ln = = t t 5.7 ln(/)

16 6 6. sin d u = v = cos du = d v = sin d = cos cos d = cos + cos d = cos + ( sin sin d ) = cos + sin + cos + C u = du = d v = sin v = cos d y = lim + ( + sin )/ ln( + sin ) ln y = lim + cos ln y = lim + sin + ln y = y = e sin 3 cos 4 d = = = sin cos 4 ( sin d) }{{} du ( u )u 4 du (u 4 u 6 ) du = u5 5 + u7 7 + C = 5 cos5 + 7 cos7 + C 9. π/ sin cos d The first thing that you should try when integrating is direct substitution. It is the easiest of the methods that we have. For this problem, I choose u = cos. Then du is sin d By introducing two negatives, we have π/ sin cos d = du u = ln u = ln ( ln ) = ln