Blog Post 10: Math is hard.

Difficulties with calculus (and example problems):

U-SUBSTITUTION

• keeping track of all of the steps and substituting correctly

 FTC PART 1 AND 2

• remembering how to do it (since it's been awhile)

 CYLINDRICAL SHELLS

• setting up subtraction correctly
• using correct variable

WASHERS

• setting up subtraction correctly
• using correct variable
• remembering to subtract inside

 AREA BETWEEN TWO CURVES

• setting up subtraction correctly

Similar problems

Blog Post 9: Washers in math?

1. Rotating about a vertical line means that the washers or disks are stacked on top of one another and the variable used should be y because the values are changing along the y axis.

Rotating about a horizontal line means the washers or disks are lined up side by side and the variable used should be x because the values are changing along the x axis.

2. If a washer is being created instead of a disk (there is open space in the middle), the volume of the inside (using the inside radius) must be subtracted from the outside volume (using the outside radius) to find the actual volume of the figure.

3. Examples:
about a vertical line:

about a horizontal line:

Blog Post 8: Antiderivatives Review

I can approximate the area under a curve via left and right endpoints for a given amount of rectangles.

• use summation equations (shown in example with variable definitions in margin) and plug in information to solve
• easy to mess up negatives, forget to add lower limit in the correct place, or use proper equation for left vs right endpoints

• plug in values for variables
• keep the delta x outside of the parentheses (more efficient) and write out the summation of the function with each i value
• add the values together and multiply by delta x to find answer

I can find the area under the curve using Riemann’s Summation.

• use a similar summation equation as with approximations, but use n rectangles and take the limit as n goes to infinity
• important to watch negatives when you break up the summation and remember to distribute any fractions that were taken out of the summation all the way through

• after plugging in values for variables, pull out delta x (just out of summation, not out of limit)
• then distribute any exponents so that the summations can be broken up
• break apart summations anywhere there is addition or subtraction to have multiple summations
• pull anything possible out of each summation until left with either a constant or i (could also be i squared)
• complete the summation, watching to distribute delta x appropriately and maintain any negatives
• multiply fractions, breaking any with addition or subtraction in the numerator apart
• take the limit as n goes to infinity to find answer

I can use U-Substitution to integrate functions.

• substitute u for a piece of the function to make integration easier, plug in given interval to find specific value rather than equation
• vital to remember to multiply any fractions back in to get the correct answer and plug the interval into whatever u equals to have the correct interval once the equation has u as the variable instead of x (if limits aren't given, it is really important to add a constant if you don't want to lose points on a quiz)

• choose a value for u that will make the equation easier to solve, in this case 2x-5 
• find du, then multiply the function by any necessary value to make sure du appears in the equation (if the function is multiplied by something, a value is needed outside of the integral to negate this change to the function: multiply the integral by one half if the function must be multiplied by two)
• plug the upper and lower limit into u to find new interval
• plug u and du into the equation
• take the antiderivative of the function
• plug the upper limit into the antiderivative and subtract the antiderivative with the lower limit plugged in to find the answer
• ****if limits are not given, plug u back in so that the original variable is back and don't forget to add a constant at the end of the antiderivative

Blog Post 7: Optimization

1. Optimization:

1. Find the first derivative of the equation
2. Set the derivative equal to zero
3. Solve for the variable

If there are two equations, you can use substitution to get it to one equation to find the derivative, then once you solve for the variable in that equation, you can plug it into the other equation to find the second variable. 

If there are multiple solutions, you can determine which is the correct one for the problem based on which is possible in the situation or which is the max/min (a problem will ask for the smallest (min) or largest (max) possible answer).

This process will allow you to use the slope to determine the best answer for the problem, either the smallest or the largest possible.

2. Find the point on the line y=2x+3 that is closest to the origin.

   

This process involves plugging the equation for y into the distance formula (after already having plugged in the coordinates of the origin). Once the equation is simplified, the power rule can be used to find the derivative, in order to find the smallest distance (minimum) from the origin. Once the derivative is set equal to zero, the value for x can be found. The value for x can then be plugged into the equation of the line to find the value for y. 

3. Based on this derivative:

Two examples of what the function could have been are:

Blog Post 6: Derivatives, but wait... there's more?

1. Derivatives do not exist:

1. f(x)=|x| 
2. cusp
3. jump
4. asymptote

(anywhere with a corner or where the limit does not exist)

2. Implicit differentiation:

Take a derivative of a function with regards to a variable that may not be in the function or is not the only variable in the function. Every time you find the derivative of a piece, you multiply by the derivative of the variable over the derivative of the variable you are finding the derivative with regards to. Example: 

3. The most important thing to remember in a related rate problem is to always multiply by the correct d(variable)/dt so that you can correctly plug in given information to find the correct answer. 

Blog Post 5: Chain Rule, mostly

1. Where f'=0 there is a maximum or minimum of f.

2. Where a function increases or decreases can be determined by plugging numbers greater than and less than the critical point (possible maximums or minimums) into f'. Where these values are positive, the function is increasing; where these values are negative, the function is decreasing.

3. The chain rule is a process for finding the derivative of a composite function. This is done by taking the derivative of the outside function, rewriting the inside function, and multiplying by the derivative of the inside function.

Example:

tangent line at x=1

4. h(x)=f(g(x))

g(-4)=5, g'(-4)=2, f'(5)=20 find h'(-4)

h'(x)=f'(g(x))g'(x)

h'(-4)=f'(g(-4))g'(-4)

h'(-4)=f'(5)2

h'(-4)=20(2)

h'(-4)=40

Blog Post 4: Derivatives, etc.

1. A function is continuous at x= a if

1. the limit exists at x = a
2. f(a) exists
3. limit at x=a equals f(a)

This function is not continuous at x=1

The limit of x=1 exists

The function exists at f(1)

The limit of 1 is 2, which does not equal f(1) which equals 0

Since the limit and the output of the function are not the same at x=1, the function is not continuous at x=1.

2. Intermediate Value Theorem:

1. solution of f(x)=3x-1 on interval [0,4]

f(0) = -1

f(4) = 11

     Since f is continuous on [0, 4] and f(0) = -1 < 0 < 11 = f(4), then there exists c in [0, 4] such that f(c) = 0.

     2.  solution of f(x)=|4x-7|+12 on interval [7,10]

f(7) = 33

f(10) = 45

     Since f is continuous on [7, 10] and f(7) = 33 > 0 < 45 = f(10), then it cannot be concluded that there exists c in [7, 10] such that f(c) = 0. 

3. The derivative of a function is the slope of the tangent line at x = a. Derivatives can be found using the limit as h approaches 0 of the difference quotient or the limit as x approaches a of the slope formula. 

derivative of f(x) = 2x-6 at x = 4

Difference quotient:

Slope formula:

The hardest part is remembering to do all of the steps.

4. Instantaneous velocity is the slope of the tangent line at one point while the average velocity is over an interval.