Find:

f(1) = undefined

x | f(x) | x | f(x) | x | f(x) |
---|---|---|---|---|---|

0.9 | -1.105 | 0.99 | -1.010 | 0.9999 | -1.001 |

1.1 | -0.91 | 1.01 | -0.990 | 1.0001 | -0.9999 |

As one can see, the values of f(x) are approaching -1 as the graph approaches the limit of x = 1.

So...

Click on the hour buttons (i.e. "One") to view information and solutions.

Why is the Second Derivative useful and important?

The First Derivative is used to find the slope of various functions (particularly those with exponents in them) at any given point along the graph that is not a limit.

The Second Derivative, as the name implies, is the "slope" of the First Derivative. In layman's terms it is the change of the change of the graph (or the slope of the slope at any given point).

One of the most common uses of the Second Derivative is its applications in Physics. In Physics, one common name for the Second Derivative is "acceleration", or the change in velocity.

Acceleration (the Second Derivative of a graph of distance traveled) aids in the understanding, conceptualizing, and caluating of problems relating to Gravity, as Gravity is constantly accelerating (changing) the velocity of any object downwards by about 10m/s.

In summation, the Second Derivative is useful for understanding what is going on with the rate of change of a graph, whether it is increasing, decreasing, or remaining constant.

Find f'(1.678) of...

To utilize the chain rule, first start with the outer most part of the equation, in this case the square root, which can be treated as taking the inner part to the power of 1/2:

Now we can use the exponent rule of multiplying the inside by the exponent, and then taking that to the power of the exponent minus one:

Finally we multiply this part by the derivative of the inner part (5z - 8), which is simply 5:

Now all we need to do is plug 1.678 into this equation...

A (fictional) professional Formula One race track can be represented with the equation (measured in feet):

When a person is at the point (180, 80), the horizontal positioning (West/East, moving towards the y-axis) is changing by 20ft/sec.

What is the rate that the vertical position is changing?

First, find the raw Derivative of the equation:

Next, we know the values of x = 180, y = 80, and that the horizontal rate of change is -20ft/sec (negative due to the fact we are approaching the y-axis).

We can substitute these into the derivative equation like so:

Then lastly, we solve for dy/dt:

Find where x is undefined in the rational function:

First, let's break up the polynomial on top into two separate monomials:

Now, seeing as the numerator contains a x - 7, along with the denominator, we can remove that from the function and use it to determine where x is undefined.

Now taking that x - 7 and setting it equal to zero we find that...

And so x is undefined at 7.

Find the average rate of change of a cube's volume on [2, 3]

Well to start we have the givens of the rate of change equation over an interval of [a, a + h], and the function of a cube's volume:

andGiven the interval [2, 3], we can plug this into the average rate of change to get ourselves started:

We can then apply the function V(s) to this to solve:

We end with the value of **nine**teen

x | f(x) | g(x) | f'(x) | g'(x) |
---|---|---|---|---|

1 | 2 | 7 | -3 | 5 |

2 | 4 | 1 | -2 | -4 |

Given the equations h(x) = f(x)g(x) and k(x) = f(x)/g(x) . . . **find:**

Using the multiplication rule "f'(x) * g(x) + f(x) * g'(x)"

h'(1) = f'(1) * g(1) + f(1) * g'(1)

And examining the data table we can replace the values like so...

h'(1) = -3 * 2 + 2 * 5

And then solve for h'(1):

-21 + 10

-11

All in all I would say that I had a fairly good time working on this portfolio. I found a way to take a manner of presentation that I enjoy working on (websites) and applying skills I learned in math class (that can be used in programming) to show what I have learned in math class. Although this website took longer than expected to create, I enjoyed a majority of the time I put into it, even when there were large errors appearing.

When it comes to what topics I selected, I generally did those that I felt I had a good grasp on, that I could convey well. I especially tried to put some focus on topics that I originally struggled a bit with, so in case someone just so happens to come across this website, it may help them with understanding some of the topics from my first semester calculus class.

The only huge problems I had was coming up with equations and solutions that could correspond to the hours on the clock, and various programming issues that came up with making the clock display the numbers and needles. It is likely that if I had the time to do this project again I would be able to complete it faster because I would know how to avoid some of the errors I came across the first time. Lastly I feel that this project is truly something that I can take a bit of pride in due to the effort and interest I showed in creating it.

- Ian Watson, Science Focus Program Student (December 2019)