Tags
calculus, comic con, integration, kinematics, math, mathematics, new york comic con, notes, nycc, study guide
Hi everyone,
I hope you are all doing well! It has been quite eventful and I am going to cover a few things in this entry.
On my adventure blog, I wrote about my experience at New York Comic Con 2015. It was my first time there and thought that it was an incredible experience. I give a lot of props to those of you who cosplay. It is hard work, and I think it’s great to see how talented these individuals are. It was also nice to see people be themselves and enjoy their time there.
If you want to read the complete entry click the picture below:
There are a lot of pictures so you can immerse yourself and see what New York Comic Con is like! You can read that first if you need a break from all of this math and science. haha
I put up the Calculus 2 handouts this past weekend and one on Monday. There is a lot of information in the handouts, but I will put up an abridged version here. You can read my complete set of notes in the link at the end of the entry. The handouts cover indefinite integrals (antiderivatives), basic integration rules, and linear motion problems.
Those of you already know my background by now and can see that I love modeling and word problems! It was fun revisiting my notes and updating them. I reworded my quiz questions and turned them into problems you can review (or will encounter in your studies). If you need a refresher on basic integration rules, read Handout 1.1A first and then proceed to 1.1 (main handout) and 1.1B (applied problems).
In this handout, I will do a quick review of basic integration, its applications to linear motion problems, and an introduction to differential equations and solving initial value problems. I will put Handout 1.1 here and a problem from 1.1B, the kinematics handout. There are a total of 3 handouts for this topic. My plan next for the Calc 2 handouts will be a review of Riemann Sums and Definite Integrals.
Happy studying and I hope you all have a great day,
Kevin MC
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Indefinite Integration (Antidifferentiation)
Let’s begin with a review of a few concepts regarding indefinite integration (or antidifferentiation.
Suppose we have the function F(x) = x^{5}. Through the Power Rule in differentiation, we find the function’s derivative: f(x) = 5x^{4}.
The function F(x) is an antiderivative of f(x) on an interval. F(x) is an antiderivative of f(x) if and only if the derivative of F(x) equals f(x). We can describe this as
if and only if
Note that “an” is highlighted above. This is because a function can have many antiderivatives. For our function f(x) = 5x^{4}, it can have antiderivatives such as
x^{5} | |
x^{5} – 0.01 | |
x^{5} + e | |
x^{5} + 10 | |
x^{5} – 473 |
etc.
As a result, we note that the function G(x) = x^{5} + C is the General Antiderivative of the function f(x) = 5x^{4}. C represents the constant of integration. The function G(x) = x^{5} + C represents the family of antiderivatives for our function f(x) = 5x^{4}. We can observe several members of this family highlighted above where C can represent -0.01, e, 10 or -473.
With this information, we can extend the definition of the indefinite integral as
We use our example of f(x) = 5x^{4} and apply it to this definition:
Differentiation and integration are inverse processes of each other. You can differentiate your answer to check your solution.
Let’s take a look at an example.
Example 1.1a
Evaluate the antiderivative (indefinite integral) and verify your solution through differentiation.
This is a polynomial function and we just need to use our integration rules to integrate it.
Step 1: Rewrite our original integral.
=
Step 2: Apply the Power Rule for Integration and the Constant Multiple Rule.
Power Rule:
≠ -1
Constant Multiple Rule:
Step 3: Simplify.
Step 4: Differentiate our result for verification.
Now we can differentiate to verify our result.
Now let’s try an example involving trigonometric functions.
Example 1.1b
Evaluate the antiderivative (indefinite integral) and verify your solution through differentiation.
Step 1: Rewrite the integrand.
In this case, we can write them as products.
Step 2: Simplify further using trigonometric identities.
From our trigonometric identities, we learned that
So we get:
Step 3: Integrate.
From our basic integration rules, we learned that
Step 4: Differentiate our result for verification.
From here, we have the simplified version of the integrand. We can expand it further to get the original integral in the problem.
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Brief Overview of Differential Equations
We’re going to briefly touch on differential equations since we will be covering them in the next section when solving initial value problems. Courses later on will go more in depth on ordinary and partial differential equations.
A differential equation is an equation involving derivatives or differentials. A few examples would be
We learned in differentiation that
denotes the derivative of the function y with respect to x.
The differential equation can also be written in differential form where
As you already noticed, the differential form is what we use throughout integration. To get , just find the derivative of the function.
For example, if we have the function y = 7x^{2} + 9x we can find the differential dy:
*If you remember what we discussed earlier where , you will also see the differential expressed as .
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Particular Solutions
We previously learned that there are many solutions to with each solution having different constants. Since it is the constants that differ for the antiderivatives of f(x), we can observe that each antiderivative is a vertical translation of each other.
For example, in our example above we can graph a few of the antiderivatives for and observe a few of its solutions. A few values of C included here are 2.5, -2.5, 5, and -5.
The general solution shows us the antiderivatives for 5x^{4} and how these antiderivatives and their different integer values for C can be a solution to the given differential equation, .
We will now learn how to find a particular solution. In order to find a particular solution, we will need to know the initial condition.
The initial condition provides the information for one specific value of x for y = F(x). In our above graph, we can observe several values for C. With the initial condition, we can find which value of C inputted into the particular solution passes through the point (x, y).
With the information from the general solution and the initial condition, we can find the particular solution. This type of problem is called an initial value problem. Let’s take a look at an example so that we can visualize this.
Example 1.1c
Find the general solution of F’(x) = 2x + 3 and determine the particular solution that satisfies the initial condition F(3) = 9.
Step 1: Find the general solution.
F’(x) = 2x + 3
We begin with F’(x) which we previously learned is the derivative of F(x). In order to find the general solution, we have to integrate F’(x) to get F(x).
F(x) = x^{2} + 3x + C
This is our general solution.
Step 2: Use the initial condition to solve for C.
We are given the initial condition F(3) = 9. We saw that a function can have many antiderivatives. However, we now know that the curve that we are looking for passes through the point (3, 9). The initial condition allows us to pinpoint and find the particular solution.
We use the information from our initial condition and apply it to our general solution to solve for C.
F(x) = x^{2} + 3x + C
F(3) = (3)^{2} + 3(3) + C
9 = 9 + 9 + C
9 = 18 + C
C = -9
Step 3: Determine the particular solution.
Now that we’ve obtained C, we can determine the particular solution that satisfies the initial condition.
The particular solution that satisfies the initial condition F(3) = 9 is F(x) = x^{2} + 3x – 9.
Example 1.1d
Solve the following initial value problems.
a)
b)
Similar to above, we have to find the general solution, utilize our initial condition, and find the particular solution that satisfies the initial condition.
For a)
Step 1: Integrate the derivative to get the general solution.
Step 2: Use the initial condition to solve for C.
f(3) = 20
Similar to our previous example, we use the information from the initial condition to solve for C.
C= -25
Step 3: Determine the particular solution.
The particular solution that satisfies the initial condition F(3) = 20 is . If you want to factor it out you can write it as .
For b)
When we solve this initial value problem, we have to perform an additional step since we are given the second derivative, (unlike in our previous example where we began with ).
Step 1: Integrate the second derivative to get .
Step 2: Use the initial condition and solve for C.
C= 1
We can now plug in C to get .
Step 3: Perform the same steps to get the general solution.
Now that we have , we can integrate it to get .
Use the initial condition and solve for C.
C = -20
The particular solution is .
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Here is one of the applied math problems from Handout 1.1B. More thorough explanations regarding position, velocity, and acceleration along with different types of problems are in the pdf version of the Handout.
Example 1
A model rocket was launched upward from an initial height of 112 feet with an initial velocity of 96 feet per second. The acceleration due to gravity is given by a(t) = -32 feet per second per second. Neglect air resistance.
- a) Find the position function s(t) expressing height s as a function of time t.
- b) How long will it take the rocket to reach the ground?
Let’s begin with part a and find the position function.
Part A
Step 1: Organize the known information.
We know that a(t) = -32 ft/s^{2}.
The initial height, s_{0} is 112 feet.
The initial velocity, v_{0} is 96 feet per second.
Step 2: Integrate the acceleration function to get the velocity function.
We learned in differentiation that acceleration is or . By integrating , we can get .
From our information, .
To get the velocity function, we integrate .
The initial velocity at t = 0 is 96 ft/s. In other words, .
We put this into the equation to find C_{1}.
C_{1} = 96
Step 3: Integrate the velocity function to find the position function.
In our previous step we found the velocity function, . We now integrate the velocity function to get the position function s(t).
Similar to the previous step, we find C_{2} at t = 0. The initial height where the rocket was launched is 112 feet.
We put C_{2} into s(t) to get the position function.
Now that we found the position function we can now calculate when the rocket hits the ground.
Part B
To find the time when the rocket hits the ground, let s(t) = 0 and solve for t.
t = -1, 7
Since time is positive (t ≥ 0) in this case, the rocket hits the ground 7 seconds after it is launched.
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Download the PDF version here in the Calculus 2 Section: Calculus 2 Review
My main math/science course site: Lecahierdekev Math and Science Review Site