Derivative

Derivative indicates the change rate of a function at a given moment. The function has to be continuous which means a smooth line without any break points, gaps and jumps. The graph of the function must be able to be draw by only one motion. If the derivative is positive, the function is increasing. When the derivative is negative, the function is decearsing. When the derivative is 0, the function is not changing at this moment. There is no derivative in the discontinuous part of a function. It is a very basic idea in calculus, which can be used in Calculus, Physics and more scientific subjects.

A function is analytic at a point c means the function have a convergent power series at point c and the function is infinitely differentiable at point c. For instance, if power series of the function converges to a constant. There is an example f(x)=|x|. It is not differentiable at x=0. It cannot even have MacLaurin series, which does not have a convergent power series at x=0.

It is useful, because it indicate us if a function-related question is able to solve by using the power series of it. As well, if a function is analytic at a point c, it is infinitely differentiable at point c; however, if a function is infinitely differentiable at point c, it may not be analytic at point c.

I think the part (f) in the first question is the most difficult one. Firstly it is hard to start since a part of the question looks like the derivative of arctan or arcsin, which is confusing. Actually the technique of integral by substitution should be used for the first step. After t is substituted by tan(x), the result is still complex, but it seems can be solved immediately. A while later, it is realized maybe another integral by substitution should be done, but it need to be revised first. Sec(x) and tan(x) should be written by 1/cos(x) and sin(x)/cos(x). Simplify the equation, finally we have a simple equation which is able to apply integral by substitution easily.

Tips:
1. For the first question which is basic stuff, it is definitely a good way to try three techniques for solving integrals.
2. Also, because of basic question, calm down and be patient. Trust yourself you can do it after a lot of practices. Of course, do practices.
3. Review the basic stuff like integral techniques first. It is the core techniques for the whole course.
4. Don’t forget the specific parts, such as absolute value and the thing like dt, dx or dh.

1. Take the derivative of each complex part in the question. For example, the question 2 in assignment 3, the derivative of cos(t) is -sin(t). There is a sin(t) over cos^2(t), which seems to substitute cos(t). As well, if there is cos(x^3), it is reasonable to take the derivative of x^3 first.

2. Be sensitive when you see a number or expression. In Wednesday class, there is a question which ask to evaluate the integral of x^3cos(x^2)dx from 0 to square root of pi. Except the first tip, there is a square root of pi, which is a little bit complicated. It is easiest way to simplify it to square it, which turn it to pi. In this case, it seems to substitute x^2.

3. If there is a question asking to calculate the integral of x^2*sin(x^2+4), always substitute x^2+4 which is a part of sin. In this way, the derivative of x^2 and x^2+4 is identical, but it is much easier if we substitute x^2+4.

4. Memorize the formulae. If a question is same as the pattern of a formula, try to transform it to the formula by substitution, just like the question 2b in assignment 3.

In order to judge whether a function is integrable or not, firstly it must be defined for each points on the graph. As the question stated, the functions with finitely many removable discontinuities are defined everywhere. Then let the functions are partitioned into n subintervals. Consider the area in each subinterval as a regular rectangle with the width of each subintervals. Then choose a sample point for each subinterval as the length of the rectangle. If it is integrable, whatever the sample is chosen, the area should always exist and equal. For a function with finitely many removable discontinuities, when the continuous parts are chosen or the discontinuous parts are chosen, one overestimate the area, the other underestimate the area. If the results are same in these two situation, the integral should be same as the results. If the discontinuities are finite, when it multiply with the width which is extremely small, the result will be negligible. The main part are always the infinite many continuities times the length. Finitely many removable discontinuities is not powerful enough to change the answer, cause as the n increase, the width is extremely small. Functions with finitely many removable discontinuities are integrable.

There is a clock with a minute hand of 20cm and a hour hand of 10cm. At what rate is the distance between the two tips changing at 4:30?

Let the distance between two tips be c and the length of two hands be a and b. And degree of the angle between two hand is i. The two hands and the distance between tips bounded a triangle. The length of two hands are known, I need to computer the distance between two tips. c^2=a^2+b^2-2abcos(i) In order to find the rate of change, we need to take the first derivative. 2c*dc/dt=2absin(i)*di/dt. At 4:30, i=pi/4, c^2=20^2+10^2-2*20*10*sin(pi/4). Then I get c=14.74. di/dt is same as the difference between the rate of change of two hands. The rate of change of minute hand is 6 degree/minute=pi/30. The rate of change of hour hand is 1/2 degree/minute=pi/360. di/dt=pi/30-pi/360=11pi/360. Now, I get all the value I need, just plug in. 2*14.74*dc/dt=2*10*20*sin(pi/4)*11pi/360, dc/dt=0.92cm/minute.

A student made his Frisbee on roof of the common block in Walter Gage Residence. He finds 20-feet-long a ladder which leans against wall in order to climb on the roof. He does not set the ladder well, the bottom of the ladder slides away from the building horizontally at a rate of 3 ft/sec. So how fast is the ladder sliding down the house when the top of the ladder is 16 feet from the ground?

When the ladder leans against wall, the ladder, the wall and the ground between them form a right triangle. According the Pythagorean theorem, square of the height of the wall(h) and square of the length of the ground(l) between the ladder and the wall equals square of the length of the ladder, which is h^2+l^2=20^2. When the top of the ladder is 8 feet from the ground, 16^2+l^2=20^2, so we solved for l, the answer is 12. Then take the derivative of the equation for both sides, 2h*h’+2l*l’=0. As we known, h=16, l=12 and h’=-3 because it slides down, 2*16*(-3)+2*12*l’=0. We can get l’=4 ft/sec.

Let the difference of temperature between angle θ and angle θ+π be f(θ). If f(θ)=0 when θ=0, we have already found two antipodal points which have a same temperature. If f(0) is not 0, then f(0)>0 f(π)<0 or f(0)<0 f(π)>0. The circle is a closed figure, so the graph which show the relationship between θ and f(θ) is continuous, no matter what the graph looks like. The change of θ is similar with a running race. When you are the NO.10 initially and your aim is NO.1, you need to exceed everyone in front of you. The first step is to exceed the one who is right in front of you and become NO.9. Repeat this process several times, you will be the NO.1. When we review the whole process, you have at least once to be in each position on the court, because you keep running during the race. As same as the graph, when you start at f(0) and try to reach f(π), you have to experience each point between f(0) and f(π) at least once. As we known, f(0) and f(π) have different sign, so 0 must between them. That is why there exist a number a in (0,π) such that f(a)=0.

A Function:
Bob has worked for A company for 2 years. He got $100000 for his salary. Now his employer decides to raise his salary because of his ability. After arising, company pays him $5000 per month.

It is the function between the time(t) and money (f) he owns. f(t)=100000+5000t

A sequence:
A biology student is observing cell division. There is only one bacterium cell in a petri dish. Each second, one cell division occurs. After first division, two bacterium cells appear in the dish.
The sequence is n=1*2^(t-1)

A series:
A boss invests a really effective and risky task which monthly benefit is 1%. He spends $100,000,000 to invest initially. Because of the risk, after each month he takes 20% out of his investment(not include the benefit) to minimize the risk.

There is a series which is between time(t) and benefit(y).

Does the graph of your function have a horizontal asymptote?
No, because it is a linear function.
Does your sequence converge?
No, since it is a geometric sequence which has a common ratio of 2 being greater than 1.
Does your series converge?
Yes, it is a geometric series and the absolute value of r is smaller than 1.

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