Derivatives can be explained in two ways. First of all, according to the Physics concept, derivatives represent the velocity or the speed of an object in its motion, which is the rate of change in distance or displacement divided by time. The other way to explain derivatives is by using a graph of a function. If the function is continuous, then the derivative of the function at a certain point will be equal to the slope of the straight line tangential to the graph. In order to find the slope of the function at point x, we can find the slope of the tangent line at the point (x, f(x)) and the other point (x_0, f(x_0)), which is close to x. When we find the slope of the tangent line, we can get the equation of the line and the derivative of the function at point x at the same time.

As we are talking about functions, we know that there are functions that are infinitely differentiable. An analytic function is a function that its Taylor series at any point x_0 in its domain converges to the function itself for x of x_0. In other words, if a function is analytic at c, then the function can be expanded to a  power series around c  has a positive radius of convergence. For example, if f(x) = e^x is analytic at x=0, then we are able to write out the power series representation ∑((x^n)/n!) as n≥0 near 0. So, the major difference between analytic functions and infinitely differentiable functions is that we can always write out a power series representation for analytic functions but not every infinitely differentiable function has a power series representation on every x in its domain.

The most difficult question on the Midterm to me is part (b) of Question 4, which is to prove the integral of f”(t)cos(t) dt ≤ f'(2π) – f'(0) from 0 to 2π. When I was doing the question, I was able to prove part (a), so I assumed that there may be some connection between the two parts. Eventually, I just could not find any tips of proving it. However, after redoing it and finding help from others, I noticed that I should use integration by parts twice in order to prove the problem. The way of solving part b is completely different from the way I solve part (a). So this part of the question is tricky and difficult to me, it teaches me not to try to solve a question in only one way.

Finally, I would like to present two study tips I recommend:

1.  First of all is to get used to all the techniques for solving problems that we learned in class because some questions on the exam may have only one way to solve. As we are familiar to all of them, there will be no worries when facing those questions.
2. Secondly, when we are studying for the exams, we should try to do the problems that we did not know how to do previously. Try to find help from others and redo those questions again and again until we get used to them. With that, the exams will not be nightmares to us.

When we are solving for more complicated integrals, we will use U-substitution to make our life easier. However, sometimes choosing the right “u” may affect our process a lot.  Here are some tips to help people choose the “u” easier:

First of all, choose the function that is not in its simplest form. For example, when we are evaluating the integral of xcos(x^2), we should choose x^2 as our “u” because du=2xdx. When we do the u-substitution, we have to represent the integral with respect to u, which means dx at this point, will be equal to du/2x. Then, with one x in each of the denominator and the numerator, we are able to cancel it. Thus, the integral will look much more simple.

Secondly, when there is a rational function,  we will try to get rid of the function at the numerator. So in this case, we will choose the function at the denominator as our “u”. For example, if we want to evaluate the integral of x/((x^2) + 3), we will choose u=(x^2) + 3 so that later on the x in the denominator and the numerator will cancel out each other.

Lastly, when we are evaluating the integrals of the product of trigonometric functions, try to think of their derivatives then decide which should be represented as “u”. For example, if we are to evaluate the integral of tan^3(x)sec^2(x), we should choose u=tan(x) because we know that the derivative of tan(x) is sec^2(x). Then the sec^2(x) will cancel out each other eventually.

If we are looking for the integral of a function, we find the area under the function within of an interval. So we will divide the graph into many small rectangles and add up all the areas of the rectangles. When there is a function that has many removable discontinuities but finite, it means that there is a limit of numbers that has a different area of a rectangle. So as we divide the function into a huge number of subintervals n, say 1000, the width of each rectangle will be relatively small. Therefore, the areas of these finite points of removable discontinuities will be very small.

In conclusion, as n we choose is very big, 1/n, which is the width of each rectangle, will be infinitely small, and it will make the areas of these rectangles infinitely small. Then they can be negligible. Thus, the area under the graph of the function will be the total of the rectangles that are made of the continuous points. So that’s why a function with finite many removable discontinuities can still be integrable.

Give a function f(x)= ln(3x^(e^x)+2). Find the derivative f'(x) of the function.

Solution:

As we use the chain rule and the power rule, we can get ln(3x^(e^x)+2) =  1/(3x^(e^x)+2) * e^x* 3x^(e^x-1)*e^x. When we simplify it, we can get f'(x) = (e^(2x))(3x^(e^x-1)/(3x^(e^x)+2).

It rains a lot during the winter in Vancouver. And there are lots of constructions on the UBC campus. Sometimes the construction workers will use traffic cones to warn the drivers that there is a construction ahead. On a rainy day, someone makes fun and puts a traffic cone at the construction near Totem Park residence upside down. So that the rain goes in the cone and start to fill it up. Assume that the bottom of the cone is a circle with a diameter of 50 cm and it has a height of 70 cm. If the height of the water is increasing at a rate of 3 cm/min, what is the rate of the rain filling up the traffic cone as it reaches a height of 40 cm?

Solution:

First of all, we know that the equation of the volume of a cone is V=(π⁄3)r^2(h). We are given that r/h= 25/70, so we can get r= (5/14)h. As we substitute (5/8)h for r and differentiate the function using power rule and chain rule, we can get dV/dt = {(25π)/196}(h^2)(dh/dt). When we plug in 40 cm for h^2 and 3 cm/min for dh/dt, we will get dV/dt = 1923 cm^3/min, which is the rate of the rain filling up the traffic cone.

Here is an example of the Intermediate Value Theorem that is more rely on real life. We claim that for every circle in the universe, there exist antipodal points of the circle that has the same temperature. We let the function be f(θ) = T(θ) −T(θ+π) which T(θ) is the temperature of any point P_θ on the circle and T(θ+π) is the temperature of the point P_θ+π, which is  the antipodal point of P_θ. So if f(x) equals to zero, that means T(θ) is equal to T(θ+π), which satisfy our claim.

In order to prove our claim, we can use the Intermediate Value Theorem, which states that for  an interval [l,r] of a continuous function, we are able to find a point c such that f(c)=L, which is between f(l) and f(r). Imagining we are flying on a plane from the Southern Hemisphere to the  Northern Hemisphere, no matter where are we taking off, we will fly across the equator for sure. Coming back to our example, when we start at the x-axis, which θ = 0, we may also get f(0) < 0. On the other hand, when we start at θ = π, we may get f(π) > 0. Since there is a temperature for every point on the circle, we are able to find a point P_a such that f(a) = 0 and f(a) equals to f(a+π). Finally, this prove rely explicitly on the concept of continuity because it is told that there is a temperature on each point.

Today, I am giving the three real-life examples of function, sequence, and series. Firstly, I am giving my function as  I have 100 dollars and I want to buy some tennis balls which is five dollars each. My purpose is to calculate how much money left is in my pocket after buying x tennis balls. Since my function is linear (f(x)=100-5x), it does not have a horizontal asymptote.

Secondly, the sports store decide to give prizes. They announce that the more we buy, the more we get! If I buy three tennis balls, I get a ping pong ball for free; if I buy four tennis balls, then I get two…etc. This is therefore a sequence a_{n} =n-n/2 for the amount of ping pong balls. However, we can buy as many tennis balls as you want so that you will also get as many ping pong balls as we can. So that the sequence diverges to positive infinity.

Lastly, when I am going to count the total of ping pong balls I get as prizes, I will have to calculate the series of a_{n}=n-n/2. Since I get more ping pong balls when I buy more tennis balls, the total of ping pong balls is going to be bigger and bigger, so the series is also diverging to positive infinity.

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