Apr24

Trigonometric Identities Notes

Posted on by
Standard

Intro to Trigonometry:

There are 6 trigonometric functions, which take angles as their input and output to a real number. An angle is said to be in standard position: x-axis, with it vertex at the origin.  If the angle opens clockwise then the angle is considered to be negative angle. If the angle opens counter-clockwise it is considered to be a positive angle.

Two angles are called coterminal if they share a terminal side in standard position. For example, the angles 150 degrees, 510 degrees, and -210 degrees are all coterminal.

Radians are another unit of measure for angles. There are 360 degrees in a complete angle

so 360 degrees is equal to 2 pi radians; 180 degrees is equal to pi radians; 180 degrees divided by pi is equal to 1 or equal to pi divided by 180 degrees.

The unit circle is the circle centered at the origin with a radius the length of one unit.

The definition of the trig functions in the context of a point on the plane is suppose \theta  is in standard position, and p= (x,y) is a point on the terminal side of theta, and r is the distance between P and the origin. Then the six trig functions are defined as follows:

sin \theta  = \frac {y}{r}

cos \theta  = \frac {x}{r}

tan \theta  = \frac {y}{x}

csc \theta  = \frac {r}{y}

sec \theta  = \frac {r}{x}

cot \theta  = \frac {x}{y}

In the context if a point is on the unit circle. Suppose \theta is in standard position , and P= (x,y) is a point on the unit circle. Then the 6 trig functions are defined as follows:

sin \theta  =  y

cos \theta  = x

tan \theta  = \frac {y}{x}

csc \theta  = \frac {1}{y}

sec \theta  = \frac {1}{x}

cot \theta  = \frac {x}{y}

Definition of the trig functions in the context of a right triangle. Suppose \theta is an acute angle of a right triangle. Trig functions are defined as follows:

sin \theta  = \frac {opposite}{hypotenuse}

cos \theta  = \frac {adjacent}{hypotenuse}

tan \theta  = \frac {opposite}{adjacent}

csc \theta  = \frac {hypotenuse}{opposite}

sec \theta  = \frac {hypotenuse}{adjacent}

 cot \theta  = \frac {adjacent}{opposite}

Basic Identities:

sin ^2 x + cos^2 x = 1

tanx= \frac {sinx}{cosx}

sin(x + y) = sin x cos y + cos x sin y

sin (x – y) = sin x cos y – cos x sin y

cos (x + y) = cos x cos y – sin x sin y

cos (x – y) = cos x cos y + sin x sin y

The derivative of y = sin x also depends on the value of

\overset {lim}{ h \rightarrow 0} \frac {sin(x+h)-sinx}{h}

To esitmate the limit, find the qoutient \frac {sinx}{x} for various calues of x close to 0. ( make sure to set your calculator to radian mode not in degree mode)

Derivatives of Trig Functions:

Dx(sin x) = cos x

Dx(cos x) = -sin x

Dx (tan x) = sec^2 x

Dx (cot x) = -csc^2 x

Dx (sec x) = sec x tan x

Dx (csc x) = -csc x cot x

Apr17

Related Rates

Posted on by
Standard

Hey guys we are almost done!! We just took our last test of the semester and now the only thing we have is the final!! Whoooo!! We are so close! We need to not get lazy and keep up the hard work so we can pass this class with ease! I looked ahead and noticed that we are about to learn about related rates in class, so I thought I would give everyone a little insight on what to expect on the next lesson. This lesson should be fairly easy because we are just using information we already know and just adding a little extra twist in it. In these types of problems we will be given a rate of change associated with one value and then be given another rate of change related to a second value. The second value will also be related to the first value. It seems tricky, but it is really not that bad once we perform examples in class. These problems are  very similar to our optimization problems because we will have to figure out what equation we will need that includes both rates of change. When making the equation we will also have to pair up the right rate of change with the right value. These problems take a lot of because starting the problem and making the right equation is key to success.

Once you find the equations that associates the rates of change correctly, we will use the familiar chain rule to find the derivative of the equation. We will then be able to find the differentials of the equation.  We can also use the differential approximation to different rates of changes of variables, if the rate of change is not given for a certain value.

I found some helpful strategies on the internet  that might help you, when approaching these types of problems:

  1. Draw a diagram. This is the most helpful step in related rates problems. It allows us to visualize the problem.
  2. Assign variables to each quantity in  the problem that is a function of time. Each of these values will have some rate of change over time.
  3. List all information that is given in the problem and the rate of change that we are trying to find.
  4. Write an equation that associates the variables with one another. If there are variables for which we are not given the rates of change (except for the rate of change that we are trying to determine), we must find some relation form the nature of the question that allows us to write these variables in terms of variables for which the rates of change are given. We mist then substitute these relations into the main equation.
  5. Using the chain rule, differentiate each side of the equation with respect to time
  6. Substitute all given information into the equation and solve for the required rate of change.

Like I said earlier, rates of change problems are really easy. The hardest part is getting starting and organizing your information. No need to worry! :]

Apr12

Differentials

Posted on by
Standard

In almost all of the examples and applications so far, all functions have been defined in the form y=f (x) with y given explicitly in terms of x, or as an explicit function of x, for example y = 3x – 2 is an all explicit functions of x. The equation 4xy – 3x = 6 can be expressed as an explicit function of x by solving for y. This is:

4xy – 3x= 6

4xy = 3x + 6

y = \frac {3x+6}{4x}

On the other hand, some equations in x and y cannot be readily solved for y, and some equations cannot be solved for y at all. For example, while it would be possible (but tedious) to use the formula to solve for y in the equation y^2+2yx+4x^2=0, it is not possible to solve for y in the equation y^5+by^3+6y^2x^2+2yx^3+6=0. In equations such as these last two, y is said to be given implicitly in terms of x.

In such cases, it may still be possible to find the derivative \frac {dy}{dx}   by a process called implicit differentiation. In doing so, we assume that there exists some function or functions f, which we may not be able to find, such that y= f(x) and \frac {dy}{dx} exsist. It is useful to use \frac {dy}{dx}   here rather than f'(x) to make it clear which variable is independent and which is dependent.

One notation for the derivative of y is with respect to x is \frac {dy}{dx} Notice that this similar to one way of starting slope which is \frac {\Delta y}{\Delta x} . \Delta is used to represent change. When considering \frac {dy}{dx} it is often useful to tink of the x as an infinitesimally small change. Then dy is the corresponding change in y.

REVIEW: Earlier in the year we pointed out that when y is given as a function of x, x is the independent variable. We later defined the derivative \frac {dy}{dx} when y is a function of x. in an equation such as  3xy + 4y^2  = 10 , either variable can be considered the independent variable; if it asks for \frac {dy}{dx} , consider x the independent variable ; if it ask for \frac {dx}{dy} , consider y the independent variable. A similar rule holds when other variables are used.

If y= f(x) then \frac {dy}{dx} = f”(x); dy= f'(x)dx

Steps to using Implicit Differentiation:

To find \frac {dy}{dx} for an equation containing x and y:

1. Differentiate on both sides of the equation with respect to x, keeping in mind that y is assumed to be a function of x.

2. Place all terms with \frac {dy}{dx} on one side of the equal sign, and al terms without \frac {dy}{dx} on the other side.

3. Factor out \frac {dy}{dx} , and then solve for \frac {dy}{dx} .

HINT: When an applied problem involves an equation that is not given in explicit form, implicit differentiation can be used to locate maxima and minima or to find rates of change.

 

GOOD LUCK ON THE TEST MONDAY! 😀

Apr5

Absolute Extremas to Optimization

Posted on by
Standard

So far we have only talked about local extrema (also called relative extrema). Local extrema are local maximums and minimums. These are also known as the highest or lowest points on the graph in that neighborhood.

Intuitive definition of absolute extrema:

  • An absolute maximum is the highest y value that a function achieves.
  • An absolute minimum is the lowest y value that a function achieves.

Precise definitions of absolute extrema:

  • A function f has an absolute maximum at x=c, if f(c) is greater to or equal to f(x) for all x values in the domain of f. (The absolute maximum is f(c).
  • A function f has an absolute minimum at x=c if f(c) is less than or equal to f(x) fo all x values in the domain of f. (The absolute minimum is f(c).

An absolute extrema will always be located at either a local extrema or endpoints of the interval. Since local extrema occur at critical values, we can also say absolute extrema occur at either critical values or endpoints of the interval.

Extreme Value Theorem

If f is continuous on the interval [a,b], then there will be absolute extrema that exist on that interval. If f is not continuous on the interval (a,b), then there could possibly be no extrema in the interval, if there is a extrema that exists it will not be an endpoint because the endpoints in the interval are not included. ( You could technically get closer and closer to endpoints of an interval)

Now that we know all about what occurs at extremas and endpoints, we can move onto optimization in calculus.

Optimization in calculus is the process of finding the greatest or least value of a function for some constraint, which must be true regardless of the solution. This part of calculus can me very confusing for some students because you have to read problems carefully to find what they are optimizing and understand what the constraints of the problem are.

When approaching an optimization problem, students need to read the problem carefully and throughly. Setting up the problem is the HARDEST part. When reading the problem, write down what the problem is asking for. Also, write down any constraints of the problem. The constraints are also known as the intervals given in the problem. For example, x values can not be less than zero.

Next you need to find the function of the problem using the constraint  . For example, if we are trying to find maximum area of rectangle we would use the function: Length x Width. Make sure to add the constraint to the function so that the values that are found are true values to the problem.

Third step is to find the derivative of the function which contains the constraint. This will give you true values to the function. Once you find your true values, plug them into the ORGIONAL function ( without constraint). Note: most fo the time you have to use the constraint to find pairs of numbers, for example length and width. Make sure to test endpoints of the interval (only if included) and all true values (critical values). These numbers will be the solution to where your maximum or minimum optimization is occurring.

Take your time and read these problem slowly. They can be tricky!

Mar27

Curve- Sketching

Posted on by
Standard

This is the process of sketching a curve, based on information about the function. We look at different parts of the function such as  its derivative and second derivative.

Step 1. Consider any information given about the type of function. For example if the graph is polynomial or  rational. Whatever type of function it is will determine its domain. The domain is very important when sketching the graph.

Step 2. Mark any asymptotes, which are straight lines approached by a given curve as one of the variables in the equation of the curve approaches infinity.

  • If  the limit of f(x) as x approaches negative infinity or positive infinity is equal to k, then there is a horizontal asymptote of y=k
  • If the limit of f(x) as x approaches c is negative infinity or positive infinity, then there is a vertical asymptote of x=c.

Step 3.  Plot all given points

  • These are any known points on the graph such as x-intercept and y-intercept.  

Step 4. Consider first derivative sign chart. The first derivative sign chart tells you where the graph is increasing or decreasing at a certain point. The points where the graph changes from increasing to decreasing is considered a critical value point on the graph.The critical values are where the second derivative is equal to zero. If the first derivative is positive at a point then it is increasing at that point and if the first derivative is negative at a point it is decreasing at that point.

  • Mark horizontal tangent lines whenever f'(x)=0
  • Mark f'(x) is not defined at a point, determine whether it has a discontinuity, vertical tangent line, or sharp bend and mark according
  • Mark all local minimums and maximums. (remember to pay attention to whether it’s a curvy max/min or sharp bend.

Step 5. Consider the second derivative sign chart. The second derivative tells you if the graph is concave up or concave down at a certain point. The second derivative can tell us the inflection points of the graph which are points where the graph changes concavity. If the second derivative is negative at a point it is concave down at that certain point and if the second derivative is positive at a point it is concave up at that certain point.  Sketch in the curve one piece at a time taking the intervals from the second derivative sign chart, paying attention to the concavity and also pay attention to the notes you have already marked on the graph.

Step 6. You should be able to sketch the graph.

Special Notes:

When the derivative does not exist:

There are 3 reasons the derivative may not exist:

1. Different types of discontinuities

  • Point Discontinuities

  • Jump Discontinuities

  • Vertical Asymptotes

2. Vertical Tangent Lines

3. Sharp Bend

Once you make the sign charts for the first derivative and the second derivative, sketching the graph is easy. You have to sketch the graph one piece at a time. It is important to focus on key points of the graph and not try to drawl the whole graph at once. This will lead to errors and simple mistakes. Sketching graphs is not a speedy process. It takes time, patience, and diligence.