Math 3 Calculus Tommy Khoo Department of Mathematics Dartmouth College

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Presentation transcript:

Math 3 Calculus Tommy Khoo Department of Mathematics Dartmouth College (slides originally by Scott Pauls) Version 1, August 2016

Lesson Plan – Final Exam Cram Session 2 First 30 minutes. 1. One last thing (for real this time) about integration: ln and |x|. 2. Related rates: that classic ladder problem from our previous lecture, two ways to do that midterm 2 question. 3. That implicit differentiation question on midterm 2. 4. Some final delta-epsilon advice. Next 30 minutes is optional. Leave. Stay and work on the worksheet. Ask us questions. Finally enjoy my term break!

Final Exam Friday, Nov 18, 11. 30 am to 2. 30 pm Final Exam Friday, Nov 18, 11.30 am to 2.30 pm. Spaulding Auditorium (Hopkins Center for the Arts) Exam Accommodations Final exam for you will be in Kemeny Hall. Please email me to arrange. Office Hours 2-3 pm today. 12 – 3 pm Wednesday.

Tutor Clearing House Math 3 Review Wednesday, 3 pm to 4.30 pm, Kemeny Hall 007. Run by them and our LFs. Will be announced by Scott.

One way to start studying. Midterm 1 and 2. Worksheets. Advice given in recent lectures. 4) Examples, definitions and techniques in textbook/KA. 5) Relevant problems on past midterms and finals. 6) List of practice problems.

Integration with logarithm: one last time. We know that differentiating ln(x) gives us 1/x. What if we integrate 1/x? Reasonable to say we get ln(x) back. Since integration “reverses” differentiation. Actual answer: ln|x|. Because |x| = x when x > 0, and |x| = -x when x < 0. d ln(x)/dx = 1/x. d ln(-x)/dx = (-1)/(1/-x) = 1/x, by Chain Rule. ln(x) not defined when x < 0.

Integration with logarithm: one last time. 1/𝑥 = ln 𝑥 +𝑐. It is the same for 𝑓′(𝑥)/𝑓(𝑥) = ln 𝑓(𝑥) +𝑐. What about ln 𝑥 =𝑥 ln 𝑥 − 𝑥 ? Because we started with ln(x). So we have the condition x > 0 already. Algorithm: Whenever you start without logarithms and get ln(f(x)) while doing integration, just throw in a modulus sign to get ln|f(x)|. Say “since ln(f(x)) is undefined when f(x) ≤ 0”.

Integration by Parts Remember: you can pick anything to be your u and du. E.g. integrate ln(x). Then u = ln(x), dv = 1 dx. E.g. integrate x ln(x). Then u = ln(x), dv = x dx.  Do not have to stick to u = x, dv = ln(x).

Two definition of the derivative. What does d/dx f(x) or f’(x) mean? f’(x) = limit of [ f(x) – f(a) ] / [ x - a ], as x  a. f’(x) = limit of [ f(x+h) – f(x) ] / [ h ], as h  0. The derivative is a limit. (not every limit is a derivative) Might as well: differentiable => continuous. (but continuous does not mean differentiable)

Integration with Infinite Endpoints What does it mean to integrate from a constant number a to +infinity or –infinity? It means to take limits of the integral g x = 𝑎 𝑥 𝑓 𝑡 𝑑𝑡 . If you are confused during the exam, come ask me.

From lecture 10: that classic ladder problem. A 13 foot ladder leans against a wall so that the top of the ladder is 5 feet above the ground and the bottom of the ladder is 12 feet from the wall. At that moment, the top of the ladder is sliding down at a rate of 2 feet per second. What is the rate, 𝑑𝑥 𝑑𝑡 , that the bottom of the ladder is moving away from the bottom of the wall? 𝑑𝑥 𝑑𝑡 =− ℎ 𝑥 𝑑ℎ 𝑑𝑡 ⇝ 𝑑𝑥 𝑑𝑡 =− 5 𝑓𝑡 12 𝑓𝑡 −2 ft s = 5 6 ft s ℎ 0 𝑥 0 ℓ 𝑑ℎ 𝑑𝑡 𝑑𝑥 𝑑𝑡 =? Last, we solve this problem in a special case. Bonus question: what does this model say happens as ℎ tends to zero? Khan Academy problem sets: https://www.khanacademy.org/math/calculus-home/derivative-applications-calc/related-rates-calc/e/related-rates https://www.khanacademy.org/math/calculus-home/derivative-applications-calc/related-rates-calc/e/related-rates-multiple-rates https://www.khanacademy.org/math/calculus-home/derivative-applications-calc/related-rates-calc/e/related-rates-pythagorean-theorem https://www.khanacademy.org/math/calculus-home/derivative-applications-calc/related-rates-calc/e/related-rates-advanced

From lecture 10: related rates group work. Groups 1,2,3 Car A is traveling west at 50 mi/hr and car B is traveling north at 60 mi/hr. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection? Groups 4,5,6 Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm2? Groups 7,8,9 The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 80 mm? The instructions here are a bit different than other cases as the problems don’t sequentially build until to the last one. First task: Each group picks a problem. Draw a picture, label variables, determine known and unknown quantities. Have representatives present at the board. Second task: Find a relational equation. Third task: Do the computation – differentiate, solve for unknown rate, plug in the rest of the information.

Two ways to do that midterm related rates question. Surface area of a cube is 𝑆=6 𝑥 2 . Volume of a cube is V= 𝑥 3 . Method 1: double implicit differentiation. 𝑑𝑆 𝑑𝑡 =12𝑥 𝑑𝑥 𝑑𝑡 and 𝑑𝑉 𝑑𝑡 =3 𝑥 2 𝑑𝑥 𝑑𝑡 . Plug in and solve. Method 2: write one in terms of the other. 𝑆= 6𝑥 2 = 6(𝑥 3 ) 2/3 =6 𝑉 2/3 . Implicit differentiate, plug in, and solve.

From lecture 9: finding the slope to a curve using implicit differentiation. Motivating example: 𝑥 2 + 𝑦 2 =1 vs. 𝑦=± 1− 𝑥 2 Technique summary: Think of 𝑦 as a function of 𝑥, 𝑦=𝑦 𝑥 . Differentiate both sides of 𝑥 2 +𝑦 𝑥 2 =1: 2𝑥+2𝑦 𝑥 𝑦 ′ 𝑥 =0 Solve for 𝑦 ′ 𝑥 : 𝑦 ′ 𝑥 =− 2𝑥 2𝑦 𝑥 =− 𝑥 𝑦 For the review and continuation in the beginning of class 9, it is a good idea to do several different examples as this technique is one students generally have a difficult time with. It is also a great place to introduce this technique as a way to compute derivatives of inverse functions, with the application of the arc-trigonometric functions as well as the logarithm (given the derivative of the exponential function). Possibilities: Find 𝑑𝑦 𝑑𝑥 if 𝑥 3 + 𝑦 4 =4𝑥𝑦. Find the derivative of sin −1 (𝑥) using the fact that if 𝑦= sin −1 (𝑥) then 𝑥=sin(𝑦). Reading: Stewart 3.5

Implicit Differentiation Questions If the question ask you to find the tangent line to a point like in midterm 2, you can leave dy/dx or y’ in terms of both x and y. Since you will plug in a point (x,y) = (a,b) anyway. If the question just ask you to find dy/dx or y’, try to get the answer in terms of just x. If you cannot, leave it in terms of both x and y. It is not wrong. Old stuff but make sure you know the equation of the tangent line to a point.

One last thing on Delta Epsilon I emailed out a delta epsilon package earlier.

Previous Group Work We’re going to show that The limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is equal to 𝐿 if, for every 𝜖>0, there exists a 𝛿>0 so that whenever 𝑎−𝛿<𝑥<𝑎+𝛿, and 𝑥≠𝑎 we have 𝐿−𝜖<𝑓 𝑥 <𝐿+𝜖. We’re going to show that lim 𝑥→1 3𝑥−1=2 using the definition. If 0<𝑥<2, what are the maximum and minimum values of 3𝑥−1 over these values of 𝑥? For 𝜖=1, what is a 𝛿>0 so that if 1−𝛿<𝑥<1+𝛿 then 1<3𝑥−1<3? Now let 𝜖 be a fixed positive number. What are the maximum and minimum values of ( 3𝑥−1 −2) on the interval 1−𝛿,1+𝛿 1−𝛿,1+𝛿 ? How can we pick a 𝛿>0 that satisfies the definition? The limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is equal to 𝐿 if, for every 𝜖>0, there exists a 𝛿>0 so that whenever −𝛿<𝑥−𝑎<𝛿 and 𝑥≠𝑎 , we have −𝜖<𝑓 𝑥 −𝐿<𝜖. Why are we doing this? Isn't the answer obvious? Well, maybe so, but think about the example with sin 𝜋 𝑥 . There, the answer isn't obvious and the function is a difficult one to deal with. By choosing a simple function to start with, we can understand the definition and how to use it in a situation where the rest of the math - the algebraic manipulations - are simpler. Setting the stage for class 4, indicate that we will be doing an honest proof of a claim – showing a particular function has a particular limit. Tell them about the difficult elements – complex algebra, abstract reasoning, and abstract computation. Encourage them to think of the last question here in terms of the first 3.

Previous Group Work Let 𝑓 𝑥 = 2𝑥 for 𝑥≥0 𝑥+1 for 𝑥<0 . The limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is equal to 𝐿 if, for every 𝜖>0, there exists a 𝛿>0 so that whenever |𝑥−𝑎|<𝛿, we have |𝑓 𝑥 −𝐿|<𝜖. Let 𝑓 𝑥 = 2𝑥 for 𝑥≥0 𝑥+1 for 𝑥<0 . Intuitively, why doesn’t the limit exist at x = 0? Let 𝜖=5, 𝐿=1. Can you find a 𝛿>0 that satisfies the end of the definition? Let 𝜖=0.5,𝐿=0. Show that there is no choice of 𝛿>0 that satisfies the end of the definition. Explain why the work in part three demonstrates that this function doesn’t have limit zero at 𝑥=0. Show that lim 𝑥→0 𝑓(𝑥) does not exist.  too tough for an exam problem In the second half class 4, return to the proof but emphasize the places where it might fail. The purpose of this set of problems is to explore when the definition fails to be true and, consequently, that the limit doesn’t exist.

Optional 30 minutes + Worksheet The next 30 minutes and the worksheet is optional. Do question 1 a), question 2 a) and 2 b). Feel free to leave. Or stay and work on it. Or ask us questions. Good luck with the finals! Feel free to email me any time if you have questions.