Supplemental Topics, Branching out more with Calculus l

Chapter 8: Supplemental Topics:

 

Identifying the graphs of f(x), f’(x) and f’’(x).

When examining the graph of f(x) to find f’(x) and examining the graph of f’(x) to find f’’(x):

• Does the function’s curve move downwards (decreasing) going right from that point to its rightmost point (in which case the slope (y-value of the derivative) would be negative) or does the curve go upwards (increasing) from that point to the rightmost point (in which case the slope (y-value of the derivative) would be positive). 

• The steeper the slope is, the greater the absolute value of the y-value of the derivative function will be. 

Limits (L’Hopital’s Rule) and Related Rates:

When we are relating the rate of two functions f(x) and g(x), we consider the following:

For indeterminate forms, recall L’Hopital’s Rule. 

L'Hopital's Rule: derive the numerator and denominator (keep it in fractional form), then plug in the value the limit approaches. If after plugging in the value, you come back to the original limit, try the rule again. If it happens to be in the exact same form again, that means that there may be no proper conclusion. 

Reaching back to F.T.O.C. and exploring the graphs of F(x), and F’(x).

Since F is just a function that refers to the area under the curve f(x), we look at the behavior of the area under the curve when we are examining F. 

Area above the x-axis within an interval is considered a positive area. F(x) is a y-value = the area from a starting point of “a” to “x”. If the area remains positive from “a” to “x”, and “x” is the x-value at which f(x) = 0, (assuming the function’s next rightmost point from “x” would be a negative y-value), then we know the area is getting bigger in value from “a” towards “x” and getting more positive and when it gets to “x”, it is at its max. 

Area below the x-axis within an interval is considered a negative area. If the area remains negative from “a” to “x”, and “x” is the x-value at which f(x) = 0, (assuming the function’s next rightmost point from “x” would be a positive y-value), then we know the area is getting smaller in value from “a” towards “x” and getting more negative and when it gets to x, it is at its min. 

Taking the derivative of F, or F’(x) we get our original curve f(x):

Discrete data: just data points, or points that don’t necessarily follow a function/ graph. 

Put them on a graph to help with visualization! This will also help with estimating different types of area. Make sure you know what the area really implies, so that you can put correct units! Find what f(t)’s units are and dt’s units are and multiply them to find the units of the area. 

Recall the ways to approximate area:

• Left/Right areas 

• Trapezoid method

• Simpson’s method

*** To find minimum and maximum units of area, use the left/right sums (identify which gives underestimates and which overestimates).***

Averages:

Velocity from any (d1, t1) to (d2, t2) = change in distance (d2 - d1) / change in time (t2 - t1)

Average velocity = change in total distance (d(t) - d(0)) / change in total time (t - 0), where “t” represents the last time value. 

Acceleration from any (v1, t1) to (v2, t2) = change in velocity (v2 - v1) / change in time (t2 - t1)

Average acceleration = change in total velocity (v(t) - v(0)) / change in total time (t - 0), where “t” represents the last time value. 

Average of value of a function:

Recall…

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Calculus l - Differential Equations

Differential equations: Equations that involve derivatives. The solution to a differential equation is not a set of values, but a function or set of functions. 

Methods to solve:

Separation of Variables method: Only works if variables are isolatable. 

Step 1.) Get all like variables on the same side. For example, all the x’s on one side, and all the y’s on the other. 

Step 2.) Integrate! (After you integrate, you’ll get a “+ C” on both sides, just join the “+ C”’s into one “+ C” on the side of the independent variable (the variable you are integrating with respect to, most commonly known as x)).

Step 4.) Isolate the dependent variable if not yet done (most commonly known as y).

Step 5.) (This step depends, you can also do this step before step 4) If you are given an initial condition, say y(1) = 2, then you can plug in 1 for x and 2 for y to solve for c. This will help us find a general equation. 

To check if a function satisfies the differential equation, we can plug it in for “y” and check. 

Tips: 

Whenever we have something multiplied by/divided by/raised to the base of/ exponent raised to the base of our constant “C”, we just convert that entire term into a “+ C”. If something is multiplied to a constant it is still just some constant.

 

If we have more than one “+ C”, then we can combine them into one “+ C” and just put it on one side.

To cancel a natural log, we can raise both sides to a base e. To cancel an absolute value, we can just write the plus or minus value of the other side. 

Growth V.S. Decay

Growth differential equation:

The actual function used to find solutions (found by separation of variables method on growth differential):

r is just k. 

Decay differential equation:

r is just k.

Actual function used to find solutions (found by separation of variables method on decay differential):

,  k is just the negative natural log of the percent decay in decimal over the time it takes for that percentage of decay. 

Newton’s Law of Cooling

Used to answer “How long until you get the desired temperature?”. Questions relating time and temperature. 

Differential equation:

*For Newton's law of cooling, if k is not given, assume a good approximation would be 0.0005.*

Slope fields: a bunch of small dashes on points that exist on a graph that represent the slope of the function at certain points. These dashes represent the derivative of a point on the graph, but extend outwards both sides to form an actual dash or line. They are used to approximate the slope with the help of delta x. Delta x will tell the distance from one x-value to another, to approximate the slope you would connect following the path of the dashes that exist at these x-values on the graph. To better the approximation, make delta x a smaller number. 

Linearization

Local Linear Approximation or “Linearization” of f(x) at the point x = a.

y = f(a) + f’(a)(x-a)

Newton’s Method (may not work in all cases, for example, if you pick an x-value that has a y min or max, then the tangent line will have a slope of zero which is just a horizontal line that will never intersect the x-axis again and hence, won’t get you the next x-value for approximation). 

Steps to solve:

Step 1.) Get a rough sketch of the graph and identify the x-value that’s closest to the zero. Avoid picking an x-value that you know will give a max or min y-value. The x-value you pick will be your x1, (A.K.A your first guess).

Step 2.) Using the formula above, original function, function’s derivative, and x1, you can plug in x1 for xn and solve for xn+1 or x1+1 which is just x2. Repeat the process, each time replacing the xn with the value you obtained from the previous calculation, and incrementing n up by 1. For example, in the next incrementation, you would plug in the x2 that you found using the formula the first time for xn and solve for x2+1 or x3. 

Step 3.) Depending on “to __ places of accuracy”, (say, for example, 4 places of accuracy), you would keep repeating the process until you notice that the value you obtained from the previous calculation has the exact same number(s) in the same place value position as the calculation you just did.

For example x3 = 2.35443 and x4 = 2.30284. Notice how x3 and x4 seem to settle on the first two values being 2.3. Continue this process until you know for sure up to, in this example, 4 decimal places. 

The Trapezoid Rule is essentially the average of the left and right areas: 

b and a are the bounds, n is the number of intervals.

The Simpson’s Rule (most accurate approximation):

b and a are the bounds, n is the number of intervals.

 

Calculus l - Applications of Integration

Applications of Integration

Area:

Integrating using rectangles with horizontal heights:

Since the height is horizontal (parallel to the x-axis), the width must be vertical (parallel to the y-axis). The width will be a small change in y, or “dy”. This implies we are integrating with respect to y. Generally speaking, the height of a rectangle will be the right x-value minus the left x-value. Our right x-values will come from the rightmost function, and our left x-values will come from the leftmost function. When we integrate with respect to y, however, we need everything inside the integral to be in terms of y. So, we would inverse the functions.

Then we set up a definite integral because we are going to be summing up multiple of these small rectangle areas. The bound, if not specified, can be found by setting both functions to equal and finding intersection points (the y-values of those points that is).

Integrating using rectangles with vertical heights:

Since the height is vertical (parallel to the y-axis), the width must be horizontal (parallel to the x-axis). The width will be a small change in x, or “dx”. This implies we are integrating with respect to x. Generally speaking, the height of a rectangle will be the top y-value minus the bottom y-value. Our top y-values will come from the uppermost function, and our bottom y-values will come from the bottommost function. Make sure everything is in terms of x inside the integral!

Then we set up a definite integral because we are going to be summing up multiple of these small rectangle areas. The bound, if not specified, can be found by setting both functions to equal and finding intersection points (the x-values of those points that is).

Volume

Using disk/ washer method:

1. Get a sketch, it’ll help with visualization.
2. Identify the bounded area using the functions provided.
3. Rotate that area about the axis of rotation (given).

• If the axis of rotation is parallel to the y-axis, then you will be integrating with respect to y.
• If the axis of rotation is parallel to the x-axis, then you will be integrating with respect to x.
• 
  

Recall this example for visualization:

4. Create a cross section by slicing the solid perpendicular to the axis of rotation. Identify whether it is a washer or disk. (Usually, if the area is bounded between a function and the axis of rotation, it will be a disk. Conversely, if the area is bounded between functions that don’t include the axis of rotation, the result will be a washer. There are also scenarios, however, where the axis of rotation is a bound but only partially bounds a side, in cases like these, you would need to set up two different integrals to get optimal results. This is because there are two different functions that make up that side of the area on separate bounds).
5. 
   
6. Identify the area of the face of the cross section.
7. 
   

• For washers:  pi * R^2 - pi * r^2, where “R” is the bigger radius, and “r” is the smaller radius.
• For disks:  pi * r^2
• 
  

Radius is determined by the function responsible for its change. The radius is the distance from a point on that function to the axis of rotation. The function that’s uppermost would yield a greater distance from the axis of rotation and, in turn, have a larger radius (R).

6. Last, set up an integral. The bounds, if not specified, are the outermost intersection points. Make sure to rewrite any functions in terms of whichever variable you’re integrating with respect to. Also, you can factor out a pi in washers, and then take it out of the integral or just take the pi out of the integral for disks, to look condensed like this example of integrating with respect to x:
7. 
   
8. 

Then, solve it!

Volume using shells:

If the method above doesn’t work, use the shell method… You can use the shell method when you notice that the outer function responsible for the bigger radius is the exact same function responsible for the smaller radius.

Use the formula: (circumference)(height)(thickness)

1. Follow steps from Volume using disk/ washer method up until 2.
2. Rotate the area about the axis of rotation
3. This time, instead of imagining rectangles perpendicular to the axis of rotation, our rectangles will be parallel to the axis of rotation. Connect two rectangles on opposite sides by following the path of the 3D solid. It should form a cylindrical shape:

4. When we unravel this cylinder, we will just end up getting a long rectangular box
5. 

Notice how the dimensions relate to the graph:

• Radius: see what the radius depends on.

In the first example from step 3, the radius depends on the distance from a point on sin x to the y-axis. Therefore, radius for is (the outermost function - the axis of rotation) which is (sin x - 0) or just sin x.

In the second example, the radius seems to depend on the distance from a y-value to the axis of rotation. Since both functions are continuous, both will yield the same y-value at different x-values. The distance between these x-values would be the height. Since dy is getting infinitely smaller, the position of the lengths will come increasingly closer to approach a single y-value, which is used to calculate the radius.

• Height: see what the height depends on.

In the first example, the height seems to correspond with the value of y at an x-value for sin x.

In the second example, the height is the difference between the right x-value (given by the rightmost function) and the left x-value (given by the leftmost function).

• Thickness:

Is always either dx or dy. Depends on whether the axis of rotation is parallel to the y-axis (in which case it would be dx) or parallel to the x-axis (in which case it would be dy).

First example: dx

Second example: dy

5. Set up the definite integral!

For anything else, always remember that the formula for volume is just the area of the face of a cross section multiplied by the thickness (or change in x or y). Once you find the volume for one cross section, just use a definite integral!

Work:

Hooke’s Law (springs): F = kx (Force is equal to spring constant multiplied by object’s displacement)

Basic work formula that only works when force is constant: W = Fd (Work is equal to force multiplied by displacement).

F = ma (Mass multiplied by acceleration) but acceleration in the y direction is just gravitational acceleration (9.8 m/ sec^2 only used in the metric system).

Force is also known as the weight. The work done by a variable force is the definite integral from a to b of the force function multiplied by dx.

Force isn’t constant when it requires increasing or decreasing force to lift/ drop a part of something throughout the process of work.

Units of Work, Force, and Distance:

Term	SI (metric system)	Imperial
Work	J (Joules)	Cal (Calories)
Force	N (newtons)	lb (pounds)
Distance	m (meters)	ft (feet)

 

Work required to lift/move/stretch something:

Identify how much force is required to lift/move a small amount of that thing (called dx or dy). Next, find a ratio that could relate two variables, for example, the force it takes to move an object a certain distance (you can use the spring constant here if you’re dealing with springs) and make a unit rate. Generally state the displacement from the small floating/suspended amount to the final destination as a value of x (for horizontal displacement) or y (for vertical displacement). Arrange these into a definite integral.

Work required to pump something:

Draw a diagram of the container of liquid and label the dimensions of it including the dimensions of the liquid in it. Your goal is to find how much work is required so that the liquid will come out of the container (reach the top).

Use the formula:

W = (density of liquid)*(gravitational acceleration)(definite integral from a to b)(volume of a cross section as a function of x)*(displacement) dx

Density = mass/ volume

Average value theorem:

If the function is continuous within an interval, then the function’s average value within that interval will exist as an y-value somewhere on the function in the interval. This is the formula for the average value of f(x).

Position, velocity, acceleration relations: