• Know the definition of cos(θ) and sin(θ) as the x and y coordinates of points along the unit circle.
• Be able to easily switch between degrees and radians.
• Be able to evaluate cos(θ) and sin(θ) for special angles.
• Be able to easily recall these or similar trigonometric identities by thinking about the definition of cos(θ) and sin(θ) in terms of the unit circle.
  • cos(–θ) = cos(θ)   (thus cosine is an even function)
  • sin(–θ) = –sin(θ)   (thus sine is an odd function)
  • cos(θ + 2π)=cos(θ)
  • sin(θ + 2π)=sin(θ)
  • cos(π – θ) = –cos(θ)
  • sin(π – θ) = sin(θ)
  • cos(π/2 – θ) = sin(θ)
  • sin(π/2 – θ) = cos(θ)
  • sin²(θ) + cos²(θ) = 1
• Be able to graph y = sin(x) and y = cos(x). Know the x-intercepts along with the domain and range for both functions.
• Know the other trigonometric functions in terms of sine and cosine. That is,
  • tan(θ) = sin(θ) / cos(θ)
  • cot(θ) = cos(θ) / sin(θ)
  • sec(θ) = 1 / cos(θ)
  • csc(θ) = 1 / sin(θ)
• Be able to evaluate tan(θ), cot(θ), sec(θ) and csc(θ) for special angles by first writing everything in terms of sine and cosine.
• Be able to graph y = tan(x). I will discuss the graphs of cot(x), csc(x) and sec(x) later but try to determine the shape of these graphs yourself by rewriting everything in terms of sine and cosine.
• Know the identities tan²(θ) + 1 = sec²(θ) and cot²(θ) + 1 = csc²(θ). These are easily derived by dividing both sides of the equation sin²(θ) + cos²(θ) = 1 by either cos²(θ) or sin²(θ).

• sin(θ) = opp / hyp
• cos(θ) = adj / hyp
• tan(θ) = opp / adj
• csc(θ) = 1 / sin(θ) = hyp / opp
• sec(θ) = 1 / cos(θ) = hyp / adj
• cot(θ) = 1 / tan(θ) = adj / opp

• Given that 0 < θ < π/2 and tan(θ) = 3, determine the value of cos(θ).
• Given that 0 < θ < π/2 and sin(θ) = x/2, determine the value of tan(θ).

x	y
1	2
3	4
5	8
7	16

Quiz #1 will be given by your TA in your first discussion period next week. It will cover sections 1.1, 1.2, 1.3 and trigonometry. Homework solutions from sections 1.1 — 1.3 are now available on Illinois Compass. Solutions to the trigonometry problems are posted on the course home page. Quiz #2 on sections 1.5 and 1.6 will be given by your TA in your second discussion period next week.

In lecture I plugged some of the points from the table shown last time into the formula y = C*a^x to arrive at the formula y = sqrt(2)*sqrt(2)^x. There are other ways to write this formula.

I sketched graphs for a couple of exponential functions. In the formula y = C*a^x, we get exponential growth if a > 1 and exponential decay if 0 < a < 1.

We then discussed inverse functions. This included the important features of a table of values, a graph, and a formula for a function and its inverse. We saw multiple ways to obtain the formula for the inverse of a function. I also discussed the concept of one-to-one functions and the horizontal line test as a graphical way to see whether or not a function is one-to-one and thus has an inverse. We saw that for functions like f(x)=x^2 which are not one-to-one, we could restrict the domain so that it has an inverse.

I went over some of the basic rules for working with logarithms of any base including base e for the natural logarithm. We used these rules to solve the equation 3 = 2^{4x – 1} for x.

You should know the following identities and simplification rules.

• sin(2θ) = 2sin(θ)cos(θ)
• cos(2θ) = cos²(θ) – sin²(θ) which can also be written in the following two ways.
  • cos(2θ) = 2cos²(θ) – 1
  • cos(2θ) = 1 – 2sin²(θ)
• ln(AB) = ln(A) + ln(B)
• ln(A/B) = ln(A) – ln(B)
• ln(A^k) = k ln(A)
• e^ln(x) = x for all x > 0
• ln(e^x) = x for all x
• ln(1) = 0
• ln(e) = 1
• e⁰ = 1

1. lim_{x → 3} ( x² )
2. lim_{x → 4} ( (x² – 2x – 8) / (x – 4) )
3. lim_{x → 0} ( sin(x) / x )
4. lim_{x → π/2} ( tan(x) )
5. lim_{x → 7} ( (sqrt(x + 2) – 3)/(x – 7) )

• S ′(6) = lim_{t → 6} ( (2^t – 2⁶) / (t – 6) )

We made a table of values to approximate this limit. I mentioned that this S'(6) notation for the speed at which the tumor is growing at precisely t = 6 is called the derivative. Since some students have learned a little about derivatives before, I asked if anyone knew a formula for S'(t) so that we could simply plug t=6 into this formula. Nearly all of the students that have seen derivatives before obtained an incorrect formula for S'(t). I used this example to point out the need for understanding the method I am using with tables of values and other techniques before trying to use short-cut techniques.

In lecture today I began by proving lim_{x → 0} ( sin(x) / x ) = 1 by starting with the unit circle and drawing a small triangle, the sector of a circle, and a large triangle. We then compared areas and used The Squeeze Theorem. I then stated this theorem more formally as given on page 105.

We next evaluated the following limits.

1. lim_{x → 3⁺} ln(x – 3)
2. lim_{x → 4^–} (x + 2) / (x – 4)
3. lim_{x → 0} ( 3 + 2e^–x )
4. lim_{x → ∞} ( 3 + 2e^–x )
5. lim_{x → –∞} ( 3 + 2e^–x )
6. lim_{w → 1} (w⁴ – 1)/(w – 1)
7. lim_{x → 0} x² sin(1 / x)

Although it is good to use common sense, we saw from a few examples that what we think of as common sense may not actually be correct in a given situation. For instance, even though lim_{x → 0} ( x² ) = 0, we do not immediately know the value of lim_{x → 0} ( x² * f(x)). We looked at f(x) = 5x + 2, f(x) = 1/x⁶, and f(x) = (6x + 8) / x² to see that lim_{x → 0} ( x² * f(x)) might equal 0, ∞, 8, or some other value depending upon the choice for f(x).

Does common sense help us determine lim_{x → ∞} ( 1 + 1/(2x) )^x ? Many students expect the limit to be either 1 or ∞ depending upon whether they concentrate on the term in parentheses or the exponent. We will see later that surprisingly this limit is equal to the square root of e.

Finally I defined f to be continuous at x = a if lim_{x → a} f(x) = f(a). We looked at a few graphical examples which demonstrated when f was not continuous at a particular point.

In lecture we looked carefully at why we restrict the domain of f(x) = sin(x) in order for it to have an inverse function f^-1(x) = sin^-1(x) = arcsin(x). Note that sin^-1(x) is not the same as 1/sin(x). Students should also understand how the domains of cos(x), tan(x), and sec(x) are restricted in order to have inverse functions. We then evaluated the following quantities.

• sin^-1(-1/2)
• tan^-1(sqrt(3)/2)
• sin(sin^-1(1/5))
• sin^-1(sin(3π/4)) (be very careful with this one — the answer is not 3π/4)
• cos(sin^-1(1/3))
• sec(sin^-1(x))

I introduced some of the limit laws from section 2.3 but explained why we will rarely need to use them once we understand continuity a little better. We defined f to be continuous at x=a if lim_{x → a} f(x) = f(a). Nearly every function we will use in calculus is continuous on its domain so we can often just plug in a particular x-value to determine a limit. In particular, polynomials, exponentials, logarithms, roots, trig functions, inverse trig functions and rational functions are all continuous on their domains. We can also combine continuous functions by adding, subtracting, multiplying or dividing and obtain another continuous function. We can even do composition of functions to get another continuous function. For dividing, just make sure that the denominator is not 0. For composition of functions we have to be a little bit careful but I'll explain why next time.

We then discussed the very important Intermediate Value Theorem. We used it to explore the location of the roots (i.e. x-intercepts) for f(x) = x³ – 3x² – x + 5. For homework, quizzes and tests, I expect students to very clearly show how and why they may apply theorems such as The Intermediate Value Theorem or The Squeeze Theorem.

Today we discussed limits which are of the indeterminate forms ∞ / ∞ or ∞ – ∞ and used algebra in order to evaluate the following limits.

• lim_{x → ∞} (2x² + 1) / (5x² + 2x + 3) = 2/5
• lim_{x → ∞} (6x + 1) / (5x² + 4x + 10) = 0
• lim_{x → ∞} (8x³ + x - 6) / (4x² + 2) = ∞
• lim_{x → ∞} (2x² + 5) / sqrt(9x⁴ + 2x + 6) = 2/3
• lim_{x → ∞} (sqrt(4x² + 3x) - 2x) = 3/4

sin(2*tan^-1(1/5)) = sin(2*θ) = 2*sin(θ)*cos(θ) = 2*(1/sqrt(26))*(5/sqrt(26)) = 10/26 = 5/13.

I then defined the derivative of f(x) as

f ′(x) = lim_{h → 0} (f(x+h) - f(x)) / h

I discussed the interpretation of a derivative as a rate of change as well as the slope of the tangent line. We used the limit definition of a derivative to show that f(x) = x² has derivative f ′(x) = 2x. We also used the limit definition of a derivative to investigate the form of f ′(x) given that f(x) = 2^x.

Next we used the graph of a function y = f(x) to determine a reasonable graph for the derivative function y = f ′(x).

In lecture today we saw graphically why the following three definitions of derivative are equivalent.

1. f ′(x) = lim_{h → 0} (f(x + h) – f(x)) / h
2. f ′(x) = lim_{w → x} (f(w) – f(x)) / (w – x)
3. f ′(a) = lim_{x → a} (f(x) – f(a)) / (x – a)

I again mentioned how the derivative, the rate of change, and the slope all represent the same quantity. We looked at a population P(t) = 2000 + 3t². We calculated P(5) = 2075 and P ′(5) = 30 to show that 5 years later the population is 2075 people and increasing by 30 people per year. We then looked at the height of a ball thrown upwards from an apartment window h(t) = –16t² + 96t + 160. We saw that h(0) = 160 feet is the height of the window. We graphed h(t) to see it had a slope of 0 when the ball reached its maximum height. Then we found the velocity h ′(t) = –32t + 96. We set h ′(t) = 0 to determine that the ball reached its maximum height at t = 3 seconds. We then found h(3) = 304 feet to be the maximum height. I mentioned that you could set h(t) = 0 to find when the ball fell back to the ground. We could then plug this value of t into the velocity formula to obtain the velocity at the moment the ball hit the ground.

We discussed some of the different notation to refer to the 1st derivative, 2nd derivative, 3rd derivative, etc. We used f ′(x), f ′′(x), f ′′′(x) as well as Leibniz notation dy/dx, d²y/dx², d³y/dx³. Students should be comfortable with the notation for evaluating derivatives at a point as in f ′(5) or dy/dx|_x=5. Students should also be comfortable using different variables. Using Leibniz notation the derivative of P = t³ is dP/dt = 3t² and the derivative of w = r² is dw/dr = 2r.

I showed why a function is not differentiable at any point where there is a sharp corner on a graph.

I proved the important theorem which states that if a function is differentiable at a point, then it must also be continuous at that point.

Finally we used the graph of a function y = f(x) to determine a reasonable graph for the derivative function y = f ′(x). We saw that graph of the derivative of sin(x) looks like cos(x) and the graph of the derivative of e^x looks like e^x. Later we will prove that these are actually the correct formulas for these derivatives.

In lecture we saw from the graphical interpretation of a derivative as a slope that the derivative of a constant is 0 since the graph of a constant function is a horizontal line which has slope 0. We also used this approach to see that the derivative of f(x) = mx + b is f ′(x) = m. In addition to the graphical approach, we could also have used limits to prove these derivative rules.

I then showed how limits can be used to prove the following derivative rules.

• ( xⁿ ) ′ = nx^n-1     (True for all real numbers n, our proof with limits was only for positive integers n)
• ( cf(x) ) ′ = cf ′(x)     (Do a proof with limits on your own)
• ( f(x) + g(x) ) ′ = f ′(x) + g ′(x)     (A proof with limits was given in lecture)
• ( f(x) – g(x) ) ′ = f ′(x) – g ′(x)     (Do a proof with limits on your own)
• ( f(x)g(x) ) ′ = f ′(x)g(x) + f(x)g ′(x)     (A proof with limits was given in lecture)
• ( f(x) / g(x) ) ′ = ( f ′(x)g(x) – f(x)g ′(x) ) / ( g(x) )²     (Try a proof with limits on your own.

Next week we will use limits to derive short-cut methods for finding derivatives of all the basic functions we have discussed in calculus including the composition of functions. Here are the basic derivative rules I plan to discuss. It will be helpful for you to quickly memorize (or be able to derive) these rules. I plan to give more frequent quizzes to check that you are keeping up.

• ( c ) ′ = 0     (c is any constant)
• ( xⁿ ) ′ = nx^n–1
• ( e^x ) ′ = e^x
• ( a^x ) ′ = a^xln(a)     (a > 0)
• ( ln(x) ) ′ = 1 / x
• ( sin(x) ) ′ = cos(x)
• ( cos(x) ) ′ = – sin(x)
• ( tan(x) ) ′ = sec²(x)
• ( cot(x) ) ′ = – csc²(x)
• ( sec(x) ) ′ = sec(x)tan(x)
• ( csc(x) ) ′ = – csc(x)cot(x)
• ( tan^–1(x) ) ′ = 1 / (1 + x²)
• ( sin^–1(x) ) ′ = 1 / sqrt(1 – x²)
• ( cos^–1(x) ) ′ = – 1 / sqrt(1 – x²)
• ( sec^–1(x) ) ′ = 1 / (x*sqrt(x² – 1))

In lecture today I rewrote w(x) = f(x) / g(x) as w(x)*g(x) = f(x) and took the derivative of both sides. By then solving for w ′(x) we were able to derive the quotient rule. If you search YouTube for quotient rule or quotient rule song you will likely find many different mnemonic devices or songs to remember this rule.

I used limits to find the derivative of sin(x) and mentioned the similar method for finding the derivative of cos(x). I then wrote tan(x) = sin(x) / cos(x) and used the quotient rule to obtain the derivative of tan(x). I mentioned that this same technique could be used for the other trig functions.

I also used limits to show that the derivative of f(x) = a^x is f ′(x) = f ′(0) a^x where f ′(0) = lim_{h → 0} ( (a^h – 1)/h ). This limit turns out to be ln(a) so that f(x) = a^x has derivative f ′(x) = ln(a) a^x. In particular we get that f(x) = e^x has derivative f ′(x) = e^x. The rest of the class was spent on examples and answering homework questions.

In lecture we used the product rule to show that ( f(x) g(x) h(x) ) ′ = f ′(x) g(x) h(x) + f(x) g ′(x) h(x) + f(x) g(x) h ′(x). This generalizes naturally to the product of more terms. Since ( f(x) )² = f(x) f(x), ( f(x) )³ = f(x) f(x) f(x), etc., I used this generalized product rule to obtain the following derivatives

• ( f(x)² ) ′ = 2 f(x) f ′(x)
• ( f(x)³ ) ′ = 3 ( f(x) )² f ′(x)

• ( f(x)ⁿ ) ′ = n ( f(x) )^n–1 f ′(x)

• ( f(g(x)) ) ′ = f ′(g(x)) g ′(x)

I then worked through many examples. We saw some that resulted from a chain of two functions, but others that resulted from a chain of 4 or 5 functions. We saw examples that required a combination of the product rule, quotient rule, and chain rule. We saw some complicated functions for which it was much easier to simplify before taking the derivative.

Finally we found the derivative for y = ln(x) by rewriting it as e^y = x, finding dx / dy = e^y, and then writing dy / dx = 1 / (dx / dy) = 1 / e^y = 1 / x.

We saw another way to obtain the derivative of ln(x) by writing e^ln(x) = x and taking the derivative of both sides. I then discussed the change of base formula log_b(x) = ln(x)/ln(b) to obtain that the derivative of log_b(x) is 1/(xln(b)).

We looked at explicitly defined functions versus implicitly defined functions. We found the slope of the tangent line to the curve x² + y² = 25 at the point (3,4) in two ways. First we solved for y explicitly, took the derivative, and plugged in x=3. Next we just took the derivative of both sides of the original equation to get 2x + 2y*dy/dx = 0, solved to get dy/dx = -x/y, and plugged in x=3 and y=4 to obtain the same answer. This 2nd approach is called implicit differentation and you have to think carefully to see that we used the chain rule to obtain that the derivative with respect to x of y² is 2y*dy/dx. We next found the derivative dy/dx for the following implicitly defined functions. For the first example we also plugged in (x, y) = (2, 4) to find the slope of the curve at that point.

• x³ + y³ = 9xy   (this curve is called the folium of Descartes)
• y² + x⁵y + 2x = 0
• 1 + x = sin(xy²)

I next showed how to rewrite y = sin^–1(x) as sin(y) = x and differentiate each side with respect to x in order to eventually obtain dy/dx = 1/sqrt(1 – x²). For additional homework, students should use this same approach to discover the derivative of tan^–1(x).

In lecture I showed how to obtain the derivative rule for tan^–1(x) and mentioned the other inverse trigonometric derivative rules. I did #15 and #46 from section 3.5. I then discussed logarithmic differentation and its use in finding derivatives of the following functions.

• y = ln( x⁵ e^x3 (x⁴ + 3)⁶ )   (simplify using rules of logarithms before differentiating)
• y = x³ sqrt(x⁴ + 3) / (5x + 1)⁸   (take the natural logarithm of both sides, simplify, then use implicit differentiation)
• y = x^cos(x)   (take the natural logarithm of both sides, simplify, then use implicit differentiation)

I had intended to find the derivative of y = xⁿ where n is any real number but ran out of time. The derivative rule is exactly what you would expect ( dy/dx = nx^n–1 ), but we never proved the formula to be valid when n is any real number. Interested students should prove this short-cut derivative rule by taking the natural logarithm of both sides before using implicit differentiation.

My only applications from section 3.7 are the ones concerning position, velocity and acceleration. Since I have already discussed these concepts, I will expect students to already have the tools needed to solve these homework problems. I then introduced section 3.8 by talking about differential equations. We guessed then checked potential solutions to the following differential equations and initial values.

• dy/dx = 3e^x, y(0) = 8   (solution: y = 3e^x + 5)
• dy/dx = 10x, y(2) = 30   (solution: y = 5x² + 10)
• dy/dx = 8x, y(0) = 20   (solution: y = 4x² + 20)
• dy/dx = 8y, y(0) = 20   (solution: y = 20e^8x)

In lecture I continued our analysis of differential equations with initial values such as the following.

• dP/dt = 12t, P(0) = 25   (solution: P = 6t² + 25)
• dP/dt = 12P, P(0) = 25   (solution: P = 25e^12t)

In general the differential equation dy/dx = ky where k is a constant has solution y = Ce^kx. For example dy/dx = 5y with y(0)=4 has solution y = Ce^5x. By plugging in (x, y) = (0, 4) we find that C = 4 so that our solution becomes y = 4e^5x. A second example is dy/dx = 3y with y(5) = 2 which has solution y = Ce^3x. By plugging in (x, y) = (5, 2) we find that C = 2/e¹⁵ so that our solution becomes y = (2/e¹⁵)e^3x or y = 2e^3x–15.

We looked at two populations. One population is currently 100 and increasing by 5 people per year. Another population is currently 100000 and increasing by 5000 people per year. We define relative growth rate as (dP/dt) / P. Since 5/1000 = 5000/100000 = 0.05, we see that these two populations are growing at the same relative growth rate of 5%.

If a population is currently 200 and growing at a constant relative growth rate of 3%, then this leads directly to the differential equation dP/dt = 0.03P with P(0) = 200. Solving this differential equation gives us the following formula for the population: P = 200e^0.03t.

If a quantity A is proportional to B, then this means that A = k*B where k is a constant. That is, you can translate "is proportional to" to "equals a constant times". Note then that if a population is growing at a rate which is proportional to the population size, this translates to dP/dt = k*P which has solution P = Ce^kt.

Students should also know the meaning of the term half-life and be able to use exponential functions to help solve problems involving half-lives. The example solved in class was to determine how long it takes for 100mg of caffeine in the bloodstream after a cup of coffee to be reduced to 10% of that amount. The half-life of caffeine in the bloodstream is about 4 hours for most people but closer to 10 hours for pregnant women.

I then solved the first and third problems on this related rates worksheet. Try to do the second one on your own. Solutions will be provided soon.

In lecture today I further discussed strategies for solving related rates problems. I solved problem #39 from section 3.9 and the second problem on the related rates worksheet.

We obtained a graph of the basic shape of f(x) = x⁴ – 4x³ + 16x – 16 by first looking at its derivative f ′(x) = 4x³ – 12x² + 16. Since the derivative factors as f ′(x) = 4(x + 1)(x – 2)² we can quickly see which x-values cause the derivative to be positive, negative or zero. This tells us where the graph of f(x) is increasing, decreasing or level.

We used this same approach to obtain a graph of f(x) = 5xe^–2x = 5x / e^2x. We found f ′(x) = (5 – 10x) / e^2x and noted that f ′(x) > 0 for x < 1/2, f ′(x) = 0 for x = 1/2, and f ′(x) < 0 for x > 1/2. Thus the graph of f(x) is increasing for x < 1/2, level at x = 1/2, and decreasing for x > 1/2. Even though the graph of f(x) is decreasing for x > 1/2, we see from the formula for f(x) that the y-values never become negative. We used this along with lim_{x → ∞} f(x) = lim_{x → ∞} ( 5x / e^2x ) = 0 to get a better graph for this function.

The derivative function f ′(x) tells us the shape of the graph of f(x) but not the y-values. When actually graphing a function f(x) we should plug specific x-values into f(x) to obtain the corresponding y-values. Next week we'll see how the second derivative gives us further information about the shape of a graph.

I introduced terms such as absolute maximum, absolute minimum, local maximum and local minimum. We looked at a few different graphs to determine whether or not a given function had absolute or local maxima or minima. We looked in particular at the function f(x) = tan^–1(x) along with its graph to see that it had no absolute maximum and no absolute minimum.

In lecture today we discussed the terms increasing, increasing, concave up and concave down. In particular we obtained a graph of f(x) = 2x³ + 3x² – 36x. We discussed critical points and discussed the fact that a continuous function on a closed interval always attains an absolute maximum and an absolute minimum. The closed interval method guarantees that for such continuous functions on closed intervals, we can find the absolute extreme values by plugging in the critical points (points where the derivative is 0 or undefined) and the endpoints.

We next solved the following optimization problems.

• A farmer wishes to enclose a rectangular pen with area 100 square feet next to a road. The fence along the road is to be reinforced and costs $34 per foot. Fencing that costs $16 per foot can be used for the other three sides. What dimensions for the pen will minimize the cost to the farmer? What is that minimum cost?
• A rectangle is to be inscribed in a semi-circle of radius 2. What is the largest possible area and what are the dimensions that will give this area?

• 0 ÷ 0
• ±∞ ÷ ±∞
• 0 × ±∞
• ∞ – ∞
• 1^∞
• 0⁰
• ∞⁰

We used L'Hospital's Rule to determine the following limits.

1. lim_{x → 0} ( (3x – sin(x)) / x ) = 2
2. lim_{x → 0} ( (sqrt(1 + x) – 1 – x/2) / x² ) = –1/8
3. lim_{x → 0} ( (1 – cos(x)) / (x + x²) ) = 0
4. lim_{x → ∞} ( ln(x) / (2sqrt(x)) ) = 0
5. lim_{x → ∞} ( e^2x / (x² + 3x + 5) ) = ∞
6. lim_{x → ∞} ( x² / e^x ) = 0
7. lim_{x → 0⁺} ( x ln(x) ) = 0
8. lim_{x → 0} ( 1 / sin(x) – 1 / x ) = 0
9. lim_{x → ∞} ( (1 + 1 / x)^x ) = e

(slowly)   ln(x), ..., x^1/3, x^1/2, x, x², x³, ..., e^x   (quickly)

If you need to take the limit of the ratio of two such functions, then the slowness or quickness of growth toward ∞ should be enough to tell you if the limit of the ratio is 0 or ∞. For example one should immediately see that lim_{x → ∞} ( (4x¹⁰⁰⁰ + 5x⁵⁰ + 10) / (0.001e^2x) ) = 0 since the numerator approaches ∞ slowly while the denominator approaches ∞ quickly.

I answered homework questions including #36 from section 4.7. For this problem we used similar triangles and Pythagorean's Theorem to obtain a formula for L², the square of the length of the ladder L, as a function of the distance x between the base of the ladder and the bottom of the fence. Although we could take the square root of both sides to obtain a formula for L and then take the derivative, we found it easier to immediately take the derivative of both sides with respect to x to get 2L*dL/dx on the left. Now setting dL/dx = 0 results in a simpler equation.

Use the Test 2 Notes to begin your preparation for Wednesday's test. No new homework was assigned today. Come prepared with questions on Monday.

In lecture today we used summation notation for various finite and infinite sums. I relayed the story of Gauss quickly obtaining the sum of 1 + 2 + 3 + ... + 100 as a young boy. I used what was probably his approach to find a formula the sum from k = 1 to n of k. I also gave formulas for the sum from k = 1 to n of a constant, the sum from k = 1 to n of k², the sum from k = 1 to n of k³.

We looked at area as a limit of Riemann sums. I talked about right Riemann sums, left Riemann sums, midpoint Riemann sums, and sums where an arbitrary x_k^* was chosen on each interval [ x_k–1, x_k ] in order to generate f(x_k^*). We then used this limit approach with right Riemann sums to evaluate the area between the x-axis and f(x) = 2x on the interval [1, 5]. Since the shape is just a trapezoid we also checked our answer with simple geometry.

I also defined the definite integral to be the limit of Riemann sums as shown on page 366.

In lecture today I discussed section 5.2. I wrote the definition of a definite integral as the limit given on the first page of this section. Students should be able to set up this limit for any given definite integral and should be able to evaluate the limit for some particular functions such as polynomials. By the text given just prior to theorem 4 in section 5.2 we see that we can choose whether to use a left Riemann sum, right Riemann sum, or midpoint Riemann sum in the evaluation of the limit. I suggest that you use a right Riemann sum as shown in theorem 4. It makes the calculations a bit simpler. If you are not using limits and are just approximating the value of a definite integal, then you should be prepared to compute left Riemann sums, right Riemann sums, or midpoint Riemann sums.

I used limits to solve #23. I then used the fact that the region is a semi-circle to solve #36 without limits. For #53 I used the minimum and maximum values of the function in order to approximate the definite integral of the function.

I then discussed the various properties about definite integrals found on pages 373 and 374.

Suppose a population is currently 2000 and is expected to grow by 0.3t² people per year where t represents the number of years from now. What do we expect the population to be in 10 years? We solved this in two ways.

1. We graphed 0.3t² on the interval [0, 10] and then computed a Riemann sum to approximate the change in population. To get the exact change in population, we took the limit of Riemann sums. Since by definition this limit is the definite integral, we saw that the exact change in population is the definite integral of 0.3t² as t goes from 0 to 10. Computing this definite integral as a limit of Riemann sums takes many steps but gives an answer of 100 as the change in population.
2. If we let P(t) represent the population at time t, then P ′(t) represents the rate of change of population at time t. Thus P ′(t) = 0.3t² and P(0) = 2000. By solving this differential equation we obtain the formula P(t) = 0.1t³ + 2000 as the population at time t. The exact change in population is then seen to be P(10) – P(0) = 2100 – 2000 = 100.

Our second solution to this problem demonstrates the Net Change Theorem found on page 394 which basically says that the definite integral of a rate of change gives the total change.

More generally we have the Fundamental Theorem of Calculus (part 2) (FTC) from section 5.3 which shows a wonderful shortcut for evaluating many definite integrals. We used this quick approach for many definite integrals in lecture.

Recall that f(x) is integrable on an interval [a, b] if the appropriate limit of the Riemann sums exists (i.e. the definite integral of f(x) from a to b exists and is finite). Theorem 3 in section 5.2 states that if f is continuous on [a, b], then f is integrable on [a, b]. For example, since functions x², e^x, cos(x) and e^x2 are all continuous on [1, 3], we know from this theorem that the limit of the appropriate Riemann sums and hence the definite integral of each of these functions from a to b exists and is finite. Note that these limits exist regardless of whether or not we are able to personally determine the limits. Prior to section 5.3, you have learned the skills necessary to evaluate the definite integral from 1 to 3 of x² as a limit. From today's discussion of the FTC we saw quick ways to evaluate the definite integral from 1 to 3 of x², e^x or cos(x) without using limits. None of this helps in determining the definite integral of e^x2 from 1 to 3. Even though e^x2 is integrable on [1, 3] our only option is to estimate the value of the definite integral using an approximation technique such as a Riemann sum with a lot of subintervals.

Note that the FTC can be used when the integrand is continuous. We saw in lecture that the FTC gave us an obviously incorrect answer when applied to the integrand x^–2 on the interval [–2, 1]. However x^–2 = 1 / x² is not continuous on [–2, 1] so we never should have tried using the FTC. We'll learn later how to deal with functions that have discontinuities.

I began lecture by answering about 4–5 questions from the homework in sections 5.3 and 5.4.

I then introduced section 5.5 where we saw that the chain rule in reverse leads to the method of substitution. In particular we looked at the following examples.

1. The indefinite integral of 6x²(x³ + 4) dx.
2. The indefinite integral of 6x²(x³ + 4)³ dx.
3. The indefinite integral of 6x²(x³ + 4)¹⁰ dx.
4. The definite integral from 1 to 2 of 6x²(x³ + 4)³ dx.
5. The indefinite integral of 1 / (3x + 10) dx.
6. The indefinite integral of x⁴cos(x⁵) dx.
7. The definite integral from –1/3 to 1/3 of tan(x) dx.
8. The indefinite integral of tan(x) dx.

In lecture today I solved #35, 36, 58 and 77 from section 5.5.

If f(x) ≥ 0 for a ≤ x ≤ b, the area of the region between the graph of y = f(x) and the x-axis for a ≤ x ≤ b is given by the definite integral from a to b of f(x) dx. Similarly if g(y) ≥ 0 for c ≤ y ≤ d, the area of the region between the graph of x = g(y) and the y-axis for c ≤ y ≤ d is given by the definite integral from c to d of g(y) dy. We saw how this comes directly from the definition of a definite integral as the limit of a Riemann sum.

We then saw how a limit of Riemann sums reveals how to find the area between curves. If f(x) ≥ g(x) for a ≤ x ≤ b, then the area of the region between these curves for a ≤ x ≤ b is the definite integral from a to b of (f(x) – g(x)) dx. When we say to find the area between curves, we usually mean the area of the finite region bounded by the curves. This often involves finding intersection points.

As examples I found the exact area for each of the following regions.

• The region between y = 1 / (x² + 1) and the x-axis for 0 ≤ x ≤ sqrt(3).
• The region bounded by f(x) = x + 3 and g(x) = 5 – x².
• The region bounded by y = 2, y = 0, x = 0, and y = ln(x).

• Area = 2 + the integral from 1 to e² of (2 – ln(x)) dx.
• Area = the integral from 0 to 2 of e^y dy.

Area = e² – 1.

I solved #20 from section 6.1 in the following two ways.

• Area = the integral from –6 to 2 of (x – (– sqrt(12 – 4x))) dx + the integral from 2 to 3 of (sqrt(12 – 4x) – (– sqrt(12 – 4x))) dx.
• Area = the integral from –6 to 2 of (3 – 0.25y² – y) dy.

I solved #23 by integrating with respect to x to obtain

• Area = the integral from 0 to π/6 of (cos(x) – sin(2x)) dx + the integral from π/6 to π/2 of (sin(2x) – cos(x)) dx.

I stated the Mean Value Theorem along with Rolle's Theorem which is a special case of the Mean Value Theorem. I proved Rolle's Theorem today and plan to quickly prove the Mean Value Theorem on Monday. Students should be able to properly state each of these theorems correctly. I drew pictures of functions which were not continuous or not differentiable to see why we need the conditions on continuity and differentiability before stating the conclusion for each theorem. Students are encouraged to think about the pictures I drew on the board which demonstrate the conclusion for each theorem. The most important aspect of these theorems is their use in proving other important theorems. In particular I will use the Mean Value Theorem to prove the Fundamental Theorem of Calculus next week.

Using Rolle's Theorem in the following way, we are able to prove that the equation x⁵ + 3x³ + 10x + 10 = 0 has exactly one real solution.

Proof: Let f(x) = x⁵ + 3x³ + 10x + 10. Since f(–1) is negative, f(0) is positive, and f is continuous everywhere, the Intermediate Value Theorem implies that there is a real root between –1 and 0. What would happen if there were two distinct real roots c₁ and c₂? Since f is continuous and differentiable everywhere with f(c₁) = f(c₂) = 0, we would apply Rolle's Theorem to find a value of c between c₁ and c₂ with f ′(c) = 0. However, this is impossible since f ′(x) = 5x⁴ + 9x² + 10 ≥ 10 for all x. Thus there cannot be two (or more) real roots. This shows that there is exactly one real solution to f(x) = 0.

We began lecture with a proof of the Mean Value Theorem. I then used the Mean Value Theorem to prove the following theorem and its corollary.

• Theorem: If f ′(x) = 0 for all x in the interval (a, b), then f(x) = C (a constant) on the interval (a, b).
• Corollary: If f ′(x) = g ′(x) for all x in the interval (a, b), then f(x) = g(x) + C on the interval (a, b) where C is a constant.

1. Using the substitution u = 2x to obtain (–1/2)*cos(2x) + C
2. Writing sin(2x) = 2sin(x)cos(x) and using the substitution u = sin(x) to obtain sin²(x) + C
3. Writing sin(2x) = 2sin(x)cos(x) and using the substitution u = cos(x) to obtain –cos²(x) + C

We saw how to calculate volume as V = lim_{n → ∞} Σ A(x_k*) Δ x where A(x_k*) is the cross-sectional area at x_k*. Since the limit of this Riemann sum is by definition a definite integral we obtain

• Volume = the definite integral of cross-sectional area

• Revolving the region R around the x-axis: We get volume = the definite integral from 1 to 8 of π (sqrt(x))² dx.
• Revolving the region R around the line y = 3: We get volume = the definite integral from 1 to 8 of (π (3)² – π (3 – sqrt(x))²) dx.
• Letting R form the base of a solid for which cross-sections perpendicular to the x-axis are squares: We get volume = the definite integral from 1 to 8 of (sqrt(x))² dx.
• Revolving the region R around the y-axis: We get volume = the definite integral from 0 to 1 of (π (8)² – π (1)²) dy + the definite integral from 1 to sqrt(8) of (π (8)² – π (y²)²) dy.

In lecture today I proved the Fundamental Theorem of Calculus (part 2) without relying on part 1. Students are not responsible for knowing the proof but I encourage math majors to make sure that they understand each step in the proof.

I looked again at the final example from the last lecture. We obtained

• volume = the definite integral from 0 to 1 of (π (8)² – π (1)²) dy + the definite integral from 1 to sqrt(8) of (π (8)² – π (y²)²) dy.

• volume = the definite integral from 1 to 8 of 2π x sqrt(x) dx.

We then found the volumes of two solids. We discussed integrating with respect to x versus integrating with respect to y – sometimes one approach will be easier than the other and sometimes one approach will be impossible. The two solids we worked with are:

• A region is bounded above by y = 9 – x² and below by the x-axis on the interval [–3, 3]. This region is revolved around the vertical line x = 5 to form a solid.
• A region is bounded above by y = 4x – x² and below by the x-axis on the interval [1, 4]. This region is revolved around the y-axis to form a solid.

In lecture today I derived the formula for the average value of a function and showed the geometric interpretation which helped to obtain an approximate value for the average before applying the formula. I then answered homework questions including #57 from section 6.2 where the cross sections are triangles.

In lecture today we sought to approximate e^0.2 without the use of a calculator. To do this we compared the function f(x) = e^x to some other functions as follows.

1. Graph f(x) = e^x and P₀(x) = 1 on the same coordinate axes. Note that f(0) = P₀(0) (i.e. same y-value). From the graphs we see that e^x ≈ 1 for x near 0. Thus e^0.2 ≈ 1.
2. Graph f(x) = e^x and P₁(x) = 1 + x on the same coordinate axes. Note that f(0) = P₁(0) and f ′(0) = P ′₁(0) (i.e. same y-value, same slope). From the graphs we see that e^x ≈ 1 + x for x near 0. Thus e^0.2 ≈ 1 + 0.2 = 1.2.
3. Graph f(x) = e^x and P₂(x) = 1 + x + 0.5x² on the same coordinate axes. Note that f(0) = P₂(0), f ′(0) = P ′₂(0) and f ′′(0) = P ′′₂(0) (i.e. same y-value, same slope, same concavity). From the graphs we see that e^x ≈ 1 + x + 0.5x² for x near 0. Thus e^0.2 ≈ 1 + 0.2 + 0.5(0.2)² = 1.22.

So for section 3.10 we will find the equation of the tangent line to the graph of f(x) at a particular point. If we call this tangent line L(x), then from the graphs of f(x) and L(x) we see that f(x) ≈ L(x) for x near the point of tangency. We used this approach to approximate the following quantities without a calculator.

• To approximate sin(0.1), we used the tangent line to the graph of sin(x) at x = 0 to obtain that sin(x) ≈ x for x near 0. Thus sin(0.1) ≈ 0.1.

• To approximate sin(315), we used the tangent line to the graph of sin(x) at x = 100π to obtain that sin(x) ≈ x – 100π for x near 100π. Thus sin(315) ≈ 315 – 100π ≈ 315 – 100(3.14159265358979323) = 0.840734641020677. Of course you don't need to use this many decimal places for π.

• To approximate sqrt(4.06), we used the tangent line to the graph of sqrt(x) at x = 4 to obtain that sqrt(x) ≈ 0.25x + 1 for x near 4. Thus sqrt(4.06) ≈ 0.25(4.06) + 1 = 2.015.

In lecture today I talked about five different methods for approximating the square root of 5.

1. Use the Intermediate Value Theorem to find the positive root of the function f(x) = x² – 5. Since f(2) = –1 < 0 and f(3) = 4 > 0 we know the root lies between 2 and 3. Now since f(2.2) = –0.16 < 0 and f(2.3) = 0.29 > 0 we know the root lies between 2.2 and 2.3. Now refine further to get the accuracy you desire.
2. Use this algorithm which was often taught to high school students 30 or more years ago.
3. Use a linear approximation. Find the tangent line to the graph of f(x) = sqrt(x) at x = 4 to be y = 0.25x + 1. Thus sqrt(x) ≈ 0.25x + 1 for x near 4. We conclude that sqrt(5) ≈ 2.25.
4. Use Newton's Method to obtain successive estimates for the positive root of the function f(x) = x² – 5. See section 4.8.
5. Use a calculator.

We ended class by approximating the x-value where the graphs of y = x³ and y = x + 1 intersect. We began by setting the equations equal to each other to get x³ = x + 1. Next we rewrote this as x³ – x – 1 = 0 in order to apply Newton's Method to the function f(x) = x³ – x – 1 with an initial estimate of x₁ = 1.

In lecture we integrated the following functions.

• sin(x) [ You should already know this. ]
• cos(x) [ You should already know this. ]
• tan(x)   [ Rewrite as sin(x) / cos(x) and use the substitution u = cos(x). ]
• cot(x)   [ Rewrite as cos(x) / sin(x) and use the substitution u = sin(x). ]
• sec(x)   [ Rewrite sec(x) = 1 / cos(x) = cos(x) / cos²(x) = cos(x) / (1 – sin²(x)) and use the substitution u = sin(x) to obtain the integral of 1 / (1 – u²) du. Since 1 / (1 – u²) = 1 / ((1 + u) (1 – u)) = 1/2 (1 / (1 + u) + 1 / (1 – u)), we get 1/2 (ln |1 + u| – ln |1 – u|) + C. Writing this in terms of x and then using a lot of simplification we obtain that the integral of sec(x) dx is ln |sec(x) + tan(x)| + C. Most people just memorize this since it takes so many steps to derive. ]
• csc(x)   [ An approach similar the one for sec(x) gives that the integral of csc(x) dx is ln |csc(x) – cot(x)|. ]
• sin³(x) [ Rewrite as sin²(x) sin(x) = (1 – cos²(x)) sin(x) and use the substitution u = cos(x). ]
• cos⁵(x) [ Rewrite as sin⁴(x) sin(x) = (1 – sin²(x))² cos(x) and use the substitution u = sin(x). ]
• sin²(x) [ Rewrite using the identity sin²(x) = 1/2 – cos(2x)/2 and then use the substitution u = 2x. ]
• sin³(x) cos⁵(x) [ Method 1: Rewrite as sin²(x) cos⁵(x) sin(x) = (1 – cos²(x)) cos⁵(x) sin(x) and use the substitution u = cos(x). Method 2: Rewrite as sin³(x) cos⁴(x) cos(x) = sin³(x) (1 – sin²(x))² cos(x) and use the substitution u = sin(x). ]
• tan(x) sec⁴(x) [ Method 1: Rewrite as tan(x) sec²(x) sec²(x) = tan(x) (tan²(x) + 1) sec²(x) and use the substitution u = tan(x). Method 2: Rewrite as sec³(x) sec(x)tan(x) and use the substitution u = sec(x). Method 3: Rewrite as sin(x) / cos⁵(x) and use the substitution u = cos(x). ]
• tan²(x) [ Rewrite as 1 – sec²(x) and integrate term by term. ]

• Fundamental Theorem of Calculus (part 1): If f is continuous on the interval [a, b], then the function defined by g(x) = the definite integral from a to x of f(t) dt for a ≤ x ≤ b is continuous on [a, b], differentiable on (a, b), and g ′(x) = f(x).

Defining a function as a definite integral with a variable as the upper limit of integration is sometimes referred to as an accumulation function. For positive integrands, as x increases you can think of this definite integral as accumulating area under a curve. Defining a function as a definite integral may seem strange at first, but it is often done in physics, chemistry, statistics and other fields. In particular we looked at the following examples.

• g(x) = the definite integral from 0 to x of t³ dt has derivative g ′(x) = x³. We see this immediately from FTC (part 1), but also checked it by first evaluating the definite integral and then differentiating.
• g(x) = the definite integral from 0 to x of (1 + t²)^1/2 dt has derivative g ′(x) = (1 + x²)^1/2. We see this immediately from FTC (part 1). I noted that it is more difficult to evaluate the definite integral and then differentiate since you may not immediately know an antiderivative for (1 + t²)^1/2. Those going on to take MATH 231 will learn to make the substitution t = tan(θ) and dt = sec²(θ) dθ followed by some more careful work to find an antiderivative.
• S(x) = the definite integral from 0 to x of sin(π t² / 2) dt has derivative S ′(x) = sin(π x² / 2). We see this immediately from FTC (part 1). It turns out that it is impossible to find an elementary function which is the antiderivative of sin(π t² / 2) so we must rely on FTC (part 1) if we need to know the derivative of this function. This particular function S(x) is called the Fresnel function which first appeared in Augustin Fresnel's theory of diffraction of light waves, but more recently has been applied to the design of highways.

• g(x) = the definite integral from 1 to x³ of cos(t) dt has derivative g ′(x) = 3x²cos(x³). We did this one by first evaluating the definite integral and then differentiating.
• y = the definite integral from 1 to x³ of e^t2 dt has derivative dy/dx = 3x²e^x6. Since it is impossible to find an elementary function which is the antiderivative of e^t2, we must rely on FTC (part 1). To do this we make the substitution u = x³ which has derivative du/dx = 3x². With this substitution, the definite integral is now y = the definite integral from 1 to u of e^t2 dt which has derivative dy/du = e^u2 by FTC (part 1). By the chain rule we now get dy/dx = (dy/du)(du/dx) = (e^u2)(3x²) = 3x²e^x6.

I introduced the following two hyperbolic functions.

• cosh(x) = (e^x + e^–x) / 2   (note that cosh(x) and cos(x) are different functions)
• sinh(x) = (e^x – e^–x) / 2   (note that sinh(x) and sin(x) are different functions)

We took derivatives to find that

• ( cosh(x) ) ′ = sinh(x)
• ( sinh(x) ) ′ = cosh(x)

The final exam is cumulative and will take place in 314 Altgeld Hall from 1:30pm to 4:30pm on Monday, May 9th. There will be 50 points of test 1 material, 50 points of test 2 material and 50 points of test 3 material. The test 2 material may include questions from section 3.11. The test 3 material may include questions from section 5.3 about the Fundamental Theorem of Calculus (part 1). You have not had quiz questions on this material so be sure to read the sections and do the assigned homework. From section 3.11, be sure that you know how sinh(x) and cosh(x) are defined in terms of e^x and e^–x. Also be sure to know the graphs and the derivatives for sinh(x) and cosh(x).

The best way to study for the final exam is to print blank copies of each test and quiz from this and past semesters. Take them in the allotted time and compare your answers to the online solutions. Additionally you should look at the notes from each lecture listed on this page, the handout of test 2 notes, and the handout of test 3 notes. For material that you struggle with, look back at the specific section in the book or stop by during office hours to see me or a TA.