- Or create your own photobook in seconds.
- Create now!

Hello, you either have JavaScript turned off or an old version of Adobe's Flash Player.
Get the latest Flash player.

S: The Big Calculus Theory

BC: Congratulations! You have successfully explored the universe of calculus! Don't forget...the universe of calculus is forever expanding...there's still plenty to learn!

FC: The BIG Calculus Theory | By: Atima Huria, Joan Koh, Eric Zheng | Let's gp on a journey through the universe of calculus!

1: -WOW! Factor - | Our scrapbook should get creativity points because of the overall associability and incorporation of our theme into calculus. In the universe of calculus, like in any other universe, every concept builds on the another, creating an overall system: the system of calculus. This represents how there is a fundamental theorem of calculus and all the other methods are simply applications of the basic concepts. There are different parts of calculus, but in the end, it's all calculus. Similary, there are different parts to the universe, but it's all part of the universe and the basic idea overall. In addition, the universe is ever expanding like the field of calculus keeps growing and the curriculum keeps changing. Also, both our examples and explanations were integrated into the scrapbook, which really emphasized the theme.

2: -Universe of Calculus Index- | 1. Derivative Galaxy - general derivatives - Position ,Velocity,, Acceleration - Planet Related Rates - Planet Optimization 2. Integration Constellation - general integrals - Riemann Sum Stars - Area between Stars - Star Volume 3. Theorem Belt - Mean Value Theorem - Intermediate Value Theorem - Extreme Value Theorem - Average Value Theorem 4. Polar System - Polar Planets and Polar Slope - Area between Polar Planets 5. Parametric Galaxy - Parametric Slope - Velocity and Acceleration 6. Newton's Star of Cooling

3: 7. Ring length - Rectangular Ring Length - Polar Ring Length - Parametric Ring Length 8. Euler's Belt 9. Applications to Calculus - L'hopital's Moon - Partial Moons - Star Integration by Parts - Improper Integral Constellations 10. Series System

4: Derivative Galaxy | First stop, derivative galaxy!

5: This galaxy is filled with tangent lines and linearizations! It's always changing because derivatives indicate the rate of change at a specific point! However, you must beware! Not all planets in this galaxy are differentiable. Some contain holes or other discontinuities such as vertical asymptotes where you cannot differentiate. To be differentiable, a planet must be continuous at that point. Corners and cusps in the galaxy are also spots of indifferentiability. Derivatives of planets are always denoted with an apostrophe after the function, like f '(x), d/dx indicates that you are taking the derivative of a planet. The limit definition of a derivative is given by where h is equal to the distance between 2 planets on their secant line. To take the derivative of a basic function in standard form, you bring the exponent down and multiply the whole equation by it and then subtract one from the exponent. Other derivatives have different rules, such as the derivative of ln(u) is equal to 1/u multiplied by du/dx.

6: Position, Velocity, and Acceleration Often it is necessary to know the speed at which a planet or star is moving, or the velocity. Also, while acceleration is always denoted as -16t^2 on Earth, it will be different for other planets because they may not have gravity. By knowing the speed at which it is moving we can predict collisions of certain things in the universe and then protect ourselves from them. If you have a function that provides you with the position of a planet the derivative of that function with respect to time is the velocity and the second derivative is the acceleration of the planet. This can be represented with: Position: x(t) or s(t) Velocity: v(t) or x’(t) Acceleration: a(t) or x’’(t) or v’(t) The relationship between these three is that velocity is a derivative of position, since it is the rate of change between positions and acceleration is the derivative of velocity, or the 2nd derivative of position, because it is the rate of change of velocity, if the planet is not moving at a constant speed.

7: Ex) If the position of a planet at a time t is given by the equation x(t)=t^3-11t^2+24t, find the velocity and the acceleration of the particle at t=5 First, you need to take the derivative of x(t): X’(t)= 3t^2-22t+24=v(t) Second we need to plug in t=5 to find the velocity at that single point: v(t)=3(5^2)-22(5)+24=-11 Third, take the derivative of v(t) to find a(t) V’(t)=6t-22=a(t) Finally, plug in t=5 to find the acceleration that that time: a(5)=6(5)-22=8 When the velocity is negative, the planet is moving to the left. Conversely, when the velocity is positive, the planet is moving to the right. When the velocity and acceleration of the particle have the same sign, the planet’s speed is increasing. When the velocity and acceleration of the particle have opposite signs, the planet’s speed is decreasing. Finally, when the velocity is zero, the acceleration is not zero; instead, the planet is momentarily stopped and changing directions.

8: Planet Related Rates | This planet sure is adaptable! It's always changing it's size at a certain rate. Sometimes the spherical planet is getting smaller at a certain rate, and other times it is getting bigger at a certain rate. This planet is located in Derivative Galaxy because the derivative, or the instantaneous rate of change, tells us the rate of change of size of the planet with respect to the known area and lengths that are already known. The rate of change is denoted as positive if the rate of change is increasing, and negative if it is decreasing. | Ex) If planet related rates' radius is increasing at a constant rate of 2 m/s, how fast is the volume of the spherical planet increasing when the radius equals 10 m? (volume of a sphere = (4/3 * pi * r^3) d/dx(V = 4/3 * pi* r^3) = dv/dt=4 * pi * r^2 * dr/dt = dv/dt=4 * pi * (10 m)^2 * (2 m/s) = dv/dt = 800 * pi m^3/s The volume of the planet is increasing at 800(pi) meters cubed per second when the radius is equal to 10 meters.

9: Planet Optimization | Depending on the situation, this planet can be really big or really small. When it is being maximized, it is at its largest. When it is being minimized, it is at its absolute smallest. This planet is in derivative galaxy because maximums and minimums occur when the instantaneous rate of change, or the derivative, is equal to zero or undefined. Therefore, the derivative needs to be taken and set equal to zero to determine the x and y value of the possible minimum or maximum. Ex) The planet loses congruent cubes from the corners of the spherical planet optimization. The cubes are 5 m by 5 m. What dimensions give the box the largest volume? (volume of a sphere = (4/3 * pi * r^3) V = (4/3)(pi)(r - 2(5))^3 dv/dt = (4)(pi)(r - 2(5))^2 = 0 r = 10 m A radius of 10 m gives planet optimization the maximum dimensions.

10: Integration Constellation | Next stop, Integration Constellation!

11: This constellation just keeps expanding! The integration constellation keeps on gathering area and growing as the integral in the form: | To find the integral, the anti-derivative must be taken. To find the anti-derivative for basic functions, you add 1 to the exponent and then div ide by the new exponent. There are also many other types of integrations. The integral tells us the area under a curve or between two curves. It can also indicate distance, or displacement. For definite integrals, the area or distance or displacement can actually be calculated by plugging the bounds into the anti-derivative and subtracted the top from the bottom. However, if there are no definite bounds on the integral, the integral is indefinite and must have an arbitrary "+ C" placed after the anti-derivative to indicate a constant that could have been part of the anti-derivative. When solving for distance or displacement, the position of the graph above or below the x -axis must be observed. If the area is below the x-axis, it will be negative. If the area is above the x-axis, it will be positive.

12: Riemann Sum Stars | Riemann sums are approximations of the area between stars. They use shapes to approximate the area. The two different types of riemann sums include trapezoidal sums and rectangular sums. There are three different rectangular sum stars: the left hand star, the right hand star, and the midpoint star. The left hand star uses the left coordinate's height, the right hand star uses the right coordinate's height, and the midpoint star uses the point between the left and right point's height for the approximation. To find the total area, you must add up the area of the smaller rectangles or trapezoids find the approximation of the whole. The area of a rectangle is length times width and the area of a trapezoid is (base 1 + base 2) times height divided by 2. Ex) Find the area between the main galaxy and y = x^2 using the left-hand star and the right-hand star on the interval [0,3] using partition widths of 1. If the partition widths are 1, there will be 3 rectangles so the equation will be approximation = (l x w) + (l x w) + (l x w). For the left-hand approximation, the left coordinate will be used, so the height at 0, 1, and 2 will be used. Plug in the x-values to find the y-values, so the heights are 0 (0)^2, 1 (1)^2, and 4 (2)^2. left-hand approximation = (1 x 0) + (1 x 1) + (1 x 4) = 5 units. For the right-hand approximation, the right coordinate will be used, so the height at 1, 2, and 3 will be used. The heights are 1 (1)^2, 4 (2)^2, and 9 (3)^2. right-hand approximation = (1 x 1) + (1 x 4) + (1 x 9) = 14 units.

13: Area between Stars | Sometimes, the area between stars or other planets needs to be found if traveling from one to the other. This can be done by taking the integral between the equations of the two stars. To find the area between stars you use the equation: The top minus bottom can also be substitute for right minus left if the equations are in x equals form rather than y equals or if this makes the integration simpler. The bounds of the integral are x coordinates if the equations are in y equals form and y coordinates if the equations are in x equals form. Sometimes, a meteor or a comet disrupts the integral bounds and you must do the integral in two separate parts. Other times, the area is not exactly in between and the function is piecewise. Then also, you must do the integral in two separate parts. Ex) Find the area between the stars with functions f(x) = x^2 - 4x + 10 and f(x) = 4x - x^2. First graph the functions to determine which function is on top and which is on bottom and to find the bounds of integration, which can also be done by setting the equations equal to each other and solving for x. Then set up your integral. 3 1 (x^2 - 4x + 10) - (4x - x^2)dx = 16/3 units squared

14: Star Volume | To find the volume of a star through integration, the disk method or the washer method can be used. In the disk method, you add up the volumes of multiple thin slices, or disks, to get an approximation of the volume of the star. This is the definition of an integration, adding up many things to get an overall sum. The disk method is used when there is no gap between the object being rotated and the axis of rotation. When there is a gap between the object being rotated and the axis of rotation, the washer method is used to account for the gap that must be subtracted from the middle. To use the disk method, you use the equation: pi multiplied by If the equation is rotated around y = 0, you use this equation. If not, you subtract another line from it, or subtract the equation from the line. For the washer method, you use the equation: pi multiplied by Where you subtract the gap area that is not included in the volume from the actual volume area.

15: Ex) Find the volume of a star created when y = x^2, y = 0, and x = 10 is rotated around the x-axis. The lines make the bounds between 0 and 10. Since this is rotated around the x=axis, there is no need to subtract anything or have anything subtracted. So your integral setup would be pi(x^2)^2 = 157.080 units cubed Ex) Find the volume of the graph bounded by y = x^2 + 2, y = 2, and x = 8 if it is revolved around the x-axis. The lines make the bounds between 0 and 8. Your integral setup would be. (pi) (x^2 + 2)^2 - (2)^2dx = 100.531 units cubed

16: Theorem Belt | Third stop, the Theorem Belt!

17: The theorem belt is the foundation of the universe of calculus. These theorems define calculus concepts that the entire universe is built upon! Without them, there would be no integrating or deriving and the entire universe would go topsy - turvy. Some of the main theorems in the theorem belt include the mean value theorem, the intermediate value theorem, the average value theorem and the extreme value theorem.

18: Mean Value Theorem | That meteor has a bright trace that draws a continuous line through the space! The end point or the starting point of the meteor and the position the meteor is at are representing the interval the function is continuous! Also, because the meteor is always moving in a downward motion which will make the line differentiable because continuity means differentiable (of course, the interval is not inclusive for the differentiable part)! Like the Mean Value Theorem says, there will be a value, z that has the slope that is the same value as the slope that uses the two endpoints. It all works out! Ex) Let’s say that the trace of the meteor is the graph of y=(x^3)-1 with the [-1,2]. When you take the derivative of the y, then it will be dy/dx= (3x^2) and the slope of two points as part of a line will be (12-3)/(2+1)=3=dy/dx at value of z. 3=3x^2 x=1 or -1. We can only use x=1 because x=-1 is not in the interval of (-1,2). Yes! I remember that the interval is not inclusive for the derivative part. Plugging x=1 into the original y equation, we get y=(1^3)-1=0. So, in conclusion, using the mean value theorem, there will be a value that satisfying the function which is at (1,0).

19: Intermediate Value Theorem | No!! The trace of the meteorite is going away! Before the bright line of colors fade away, the Intermediate Theorem should be mentioned too! So, like I said, the trace will always draw a continuous line in the space. The interval of the line is defined the same way as we did in Mean Value Theorem. Since the meteorite is continuous, with the intermediate theorem we can say that the y values of the graph will between the y values of the two endpoint Ex) Using the same equation as in the previous example, y=(x^3)-1 with the interval of [-1,2] defines the trace of the meteor. The value when x= -1 is -2 and that of when x=2 is 7. So, using the Intermediate Theorem, we can positively conclude that in the interval of [-1,2], the graph will definitely take a value between [-2,7]. Awww! The trace of this meteor is invisible. Wait, I spot another meteor. Let’s go over there.

20: Extreme Value Theorem | Well it looks like this meteor is continuous too! That means that the extreme value theorem can be applied to this meteor. If the function of a meteor is continuous on a closed interval [a,b], then the meteor has both a maximum and a minimum on that interval [a,b]. Ex) If this meteor had an equation of f(x) = log(2x) and was continuous on the interval [2,3], I automatically know that there is both a maximum and minimum value of the meteor log(2x) on the interval [2,3].

21: Average Value Theorem | Suppose I wanted to find the average distances of meteors to earth. We already know that we find distance by integrating the absolute value of the function, but how do we find the average distance over a certain period of time? That's easy! All you have to do is multiply the distance by (1/(b-a)) or one over the difference in time. So your equation is: Ex) If I were trying to find the average distance between calculus meteor with an equation of 5x^3 between 1 and 3 seconds, what would it be? You set up your integral. (1/(3-1)) = 2 x 5x^3 dx = 50 m is the average distance.

22: Polar System | This stop is the polar system...I wonder if it will be cold....

23: This system has plenty of rose planets, loop stars, limacon planets and much more! Instead of using x's and y's in this system, r's and theta's are used. The r represents the radius of a curve and the theta represents the angle of the curve.

24: The slope of a polar planet is r=f(theta) is dy/dx. Because x and y also need to be defined, we need to derive dy/dx in terms of r and . Bear in mind that in polar system, x=rcos = f(theta )cos and y=rsin =f(theta)sin(theta). Also remember that dy/dx= (dy/dtheta )/(dx/dtheta) By substituting for x and y: ( dy/dtheta )/(dx/dtheta)=(d/dtheta[f(theta)sin(theta)])/ (d/ dtheta[f(theta)cos(theta)]) Using the product rule we are able to get dy/dx=f’(theta)sin (theta)+f(theta)cos (theta)/ f’(theta)cos (theta)-f(theta)sin (theta) Thus, this is the formula that provides for the slope of a polar planet where r=f(theta) Ex) Find the slope of the tangent line to the planet: r=2+4sin First we need to find f’(theta): f’(theta)=4cos(theta) Thus, this gives us: Dy/dx= [4cos(theta)sin(theta) + (2+4sin(theta))cos(theta)]/[4cos(theta)cos(theta)-(2+4sin(theta))sin(theta))= 2cos+8sincos/4cos^2-4(sin^2)-sin= (2cos +4sin2)/(4cos2-2sin) | Polar Planets and Polar Slope

25: Area between Polar Planets | Sometimes the area between polar planets must be found to travel around the polar system. To do this, we must find distance which we know we can get from the integration constellation. The equation is: (1/2) (r)^2 dtheta Ex) Ex: Find the area of the region in the plane enclosed by the cardiod: r=4 + 4sin(theta) Because the graph from 0 to 2pi is provided from this function, they provide our limits of integration First we need to evaluate the integral: A=1/2(the integral from 0 to 2pi of(4+4sin )^2 d(theta) Using a simple trig identity: sin^2 =1-cos2/2 we are able to rewrite the equation as: (the integral from 0 to 2pi of(12+16sin+4cos2) d(theta) After evaluating the integral, we get 24 -16+16 = 24 | Area Between Polar Planets

26: Parametric Galaxy | Parametric Galaxy...this sounds confusing

27: The parametric galaxy sure is special! Parametric planets and stars can be used to define motion that is not a function. For instance, changing the T value changes how much of the planet is actually shown. It also allows you to model planets with more than 2 variables and can answer where and when as opposed to rectangular that only shows where. To find the slope of a parametric planet, or dy/dx, you must use the equation: Ex) If planet motion is modeled by the equation x = sin(t) and y = cos(t), what is the slope of the planet in terms of t?

28: Parametric Slope | To find the second derivative of a parametric system, or d^2y/dx^2, use the equation: Ex) Find the 2nd derivative of the parametric system x = t + 3 and y = 2t +4 at t = 0.

29: Velocity and acceleration | Velocity and acceleration can easily be modeled with parametric equations. Since the initial x and y equations indicate position, the derivative of them, as we learned earlier, would be velocity. And the derivative of the velocity would be acceleration. The only difference is that instead of moving only left and right, the planets in this galaxy are able to move up and down and every which way as well! Ex) Find the velocity and acceleration equations in terms of t given the initial parametric equations x = 6t^2 and y = 7t^3. First we find the velocity by deriving both the x and y equations. x ' (t) = horizontal velocity = 12t y ' (t) = vertical velocity = 21t^2 Then we find acceleration by taking the derivative of the velocity equations. x " (t) = horizontal acceleration = 12 y '' (t) = vertical acceleration = 42t

30: Newton's Star of Cooling | Whew! This stop sure looks hot!

31: Newton's law of cooling states that a star cools at a rate proportional to the difference between its current temperature and the surrounding air. This star has been cooling for centuries! The equation process is: Where t stands for current temperature and Tm stands for surrounding temperature. To solve for a certain time or temperature , you must integrate the differential equation in terms of t.

32: Ex) Newton's star of cooling has an international temperature of 120 degrees initially, then the atmosphere cools to 72 degrees. If it is 100 degrees on the inside after 10 minutes when will it reach 80 degrees? First, let's figure out what we know and need to find out. Temperature at time of 0 = 120 Temperature of surrounding = 72 Tempeature at time of 10 min = 100 Temperature of time ? = 80 dT/dt = -k(T - Tm) dT/dt = -k(T - 72 degrees) 1/(T-72) dT = -k dt e^ln(t-72)= e^-kt+C T - 72=Ae^-kt 120 - 72 = A when t=0 A=48 T -7 2 = 48e^-kt

33: Newton's law of cooling states that a star cools at a rate proportional to the difference between its current temperature and the surrounding air. This star has been cooling for centuries! The equation process is: Where t stands for current temperature and Tm stands for surrounding temperature. To solve for a certain time or temperature , you must integrate the differential equation in terms of t. | K= -[ln(14/24)]/10 = 0.053899 T = 48e^-0.053899t +72 ln(8/48) =l n(e^-0.053899t) t=33.2429 33.2429 minutes will have passed for the star to reach 80 degrees

34: Ring Length | Whoa.

35: Ring length can be calculated for planets in the Polar System, Parametric Galaxy, and just plain rectangular planets too. Here we have to remember the skills that we learned in the integration constellation! | Rectangular Ring Length To find the arc length of a plain old rectangular planet, you must use the equation: Ex) Find the ring length of the rectangular planet's ring with equation y = (1/6)x^3 + (1/2x).

36: Polar Ring Length To find the arc length of a plain old rectangular planet, you must use the equation: Ex) Find the ring length of the polar planet's ring with equation r = e^(theta). | Whoa.

37: Parametric Arc Length To find the ringlength of a planet in the parametric galaxy, you must use the equation: Ex) Find the ring length of the parametric planet's ring with equations y = 1 - cos(t) and x = t - sin(t).

38: Euler's Belt | Oiler's Belt? Huh?

39: Euler’s Belt is used throughout the solar system to estimate the approximate answer to a differential equation. Because there are limits on the capabilities of the human race in terms of technology, we often have to use approximations of values. Here is what you need to know about Euler's belt:

40: Let’s use an example as we continue our journey through the universe of calculus Ex) Use Euler’s Method with change in x = 0.2, to estimate y(1) if y’ = y - 2 and y(0) = 4 Because we are given that the curve goes through the coordinate (0,4), we are able to use x0 = 0 and y0 = 4. The slope is found by y0 = 4 plugged into the y’= y - 2, so that we are able to have an initial slope of 2. Now we need to find the other sets of points: Increase x0 by 0.2 to get x1: x1 = 0.2 Multiply 0.2 by y’0 and add to y0 to get y1: y1 = 4 + 0.2 (2) = 4.4 Then find y'1 by plugging y1 into the equation for y’: y'1 = 4.4 - 2 = 2.4

41: Repeat this process into you have reached the desired point of x=1. Your final answer should be: y5 = 6.1472 + 0.2(4.1472) = 6.97664

42: Applications of Calculus | Whew, I sure am tired!

43: The universe of calculus can seem pointless at times, but it really is useful in solving logical problems. Here are some mathematical applications of calculus theorems.

44: Wow, this moon is sure hard to understand sometimes. Often times, it seems that things are not possible. While sometimes it seems that it is impossible to understand or discover in the solar system, sometimes you just need to take things from another perspective. The L'Hopital rule states that if the limit of a function gives us an undefined expression, we can take the derivative of the top and bottom and see if we are able to get a determinate equation. If this is not conclusive yet again, we are able to repeat the process again. Ex) | L'hopital's Moon

45: L'hopital's House | On L'hopital's moon is the very famous L'hopital's house. L'hopital's house demonstrates the pathway to finding the limit of a function. At first, the limit is indeterminate. But after the derivative is taken at L'hopital's house, the limit goes to infinity as x approaches infinity. To view the limit of the function, you must look inside the window of L'hopital's house. The mirrors inside the house represent the infinity. Since the limit goes to infinity, it increases forever. Likewise, the image of infinity in the mirrors can be seen forever.

46: We use partial moons to evaluate types of integrals that contain rational expressions. However, it is first important to understand the Algebra behind the method: For instance, if we wanted to add the expressions: 3/(x-1) and 5/(x+2) we would have to find a least common denominator and then simply add the fractions together. This would result in: (3(x+2)+5(x-1))/((x-1)(x+2)) = (8x+1)/(x-1(x+2)) Now instead, imagine if you were only given the final result and had to find out what two fractions were added together to give you the final result. What constants would be in use? Let's do an example to figure it out. Ex) | Partial Moons

47: Star Integration by Parts | When we were in the integration constellation, we saw stars being integrated to create large constellations. However, some of those were not easily integrated and needed a special method to be integrated. Integration by parts is used when you have two stars being multiplied together that can not be integrated normally. You must pick which star is u and which is dv. Pick the star with an exponent that will eventually equal 0 for u, or a natural log. You must use the equation: Ex)

48: Improper integrals in star occur when either one of the bounds is infinity, or when the distance between the constellations has a discontinuity. Therefore, the integrals have two options. They will either diverge and grow infinitely, meaning that the stars will never collide, or they will converge and collide eventually. To solve an improper integral you must turn the integral into a limit problem. If the bound has a discontinuity, break up the limit at the discontinuity. Then evaluate the integral. If you get a definite number, the integral converges. If you get infinity, then the integral diverges. | Improper Integral Constellations

49: Ex)

50: Series System | Woo-hoo! Last stop, Series System!

51: The planets in the series system really sum it all up! A series is a sum of terms. The sum either diverges and grows infinitely, or converges to a specific number that cannot always be determined. These planets undergo many series tests each day to determine whether they will eventually converge with another planet, or if they will diverge from them forever! On the next page, there is a list of all the tests that the planets in the series system go through and the rules for convergence and divergence.

**By:**EricZheng S.**Joined:**over 5 years ago**Published Mixbooks:**2

No contributors

**Title:**The Big Calculus Theory!- Calculus Scrapbook
**Tags:**None**Published:**over 5 years ago

Get up to **50**% off your first order!

or via e-mail

By clicking on the "Create your account" button, you agree to Mixbook's
Terms of Service.

Welcome back! Go ahead and Log In

or via e-mail

Your first order