Math 3
Calculus
Tommy Khoo
Department of Mathematics
Dartmouth College
(slides originally by Scott Pauls)
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.
1. Leave.
2. Stay and work on the worksheet.
3. Ask us questions.
Finally enjoy my term break!
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.
1) Midterm 1 and 2.
2) Worksheets.
3) 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 =
𝑥
‫𝑓 𝑎׬‬
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
𝑑𝑥
second. What is the rate, , that the bottom
𝑑𝑡
of the ladder is moving away from the bottom
of the wall?
𝑑𝑥
ℎ 𝑑ℎ
𝑑𝑥
5 𝑓𝑡
ft
5 ft
=−
⇝
=−
−2
=
𝑑𝑡
𝑥 𝑑𝑡
𝑑𝑡
12 𝑓𝑡
s
6s
𝑑ℎ
𝑑𝑡
𝑑𝑥
=?
𝑑𝑡
𝑥0
ℎ0
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?
Two ways to do that midterm related rates
question.
Surface area of a cube is 𝑆 = 6𝑥 2 . Volume of a cube is V = 𝑥 3 .
Method 1: double implicit differentiation.
𝑑𝑆
𝑑𝑡
=
𝑑𝑥
12𝑥
𝑑𝑡
and
𝑑𝑉
𝑑𝑡
=
𝑑𝑥
2
3𝑥
.
𝑑𝑡
Plug in and solve.
Method 2: write one in terms of the other.
𝑆 = 6𝑥 2 = 6(𝑥 3 )2/3 = 6𝑉 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:
1. Think of 𝑦 as a function of 𝑥, 𝑦 = 𝑦 𝑥 .
2. Differentiate both sides of 𝑥 2 + 𝑦 𝑥 2 = 1:
2𝑥 + 2𝑦 𝑥 𝑦 ′ 𝑥 = 0
3. Solve for 𝑦 ′ 𝑥 :
𝑦′
2𝑥
𝑥
𝑥 =−
=−
2𝑦 𝑥
𝑦
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
lim 3𝑥 − 1 = 2 using the definition.
𝑥→1
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.
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.
1. If 0 < 𝑥 < 2, what are the maximum
and minimum values of 3𝑥 − 1 over
these values of 𝑥?
2. For 𝜖 = 1, what is a 𝛿 > 0 so that if 1 − 𝛿 < 𝑥 < 1 + 𝛿 then
1 < 3𝑥 − 1 < 3?
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 +
Previous Group Work
Let 𝑓 𝑥 = ቊ
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.
2𝑥
for 𝑥 ≥ 0
.
𝑥 + 1 for 𝑥 < 0
1. Intuitively, why doesn’t the limit exist at x = 0?
2. Let 𝜖 = 5, 𝐿 = 1. Can you find a 𝛿 > 0 that satisfies the end of the
definition?
3. Let 𝜖 = 0.5, 𝐿 = 0. Show that there is no choice of 𝛿 > 0 that satisfies the
end of the definition.
4. Explain why the work in part three demonstrates that this function
doesn’t have limit zero at 𝑥 = 0.
5. Show that lim 𝑓(𝑥) does not exist.  too tough for an exam problem
𝑥→0
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.