Class Notes (1-18-08)
To start the class, we went over homework on anti-derivatives.
Here’s an example of a problem:
We notice that f’(x) has a coefficient of one, so we know that the anti-derivative will have a coefficient of the reciprocal of the derivative of the inside (1/4) multiplied by the reciprocal of the exponent (1/8).
How to solve the anti-derivative:
Think about the power rule, which states that you multiply the coefficient by the exponent and subtract one from the exponent to get the derivative. To get the anti-derivative, you do the opposite functions in the reverse order. You add one to the exponent, and then divide by the new exponent.
Linearization (pgs. 184-185) – Using the tangent line to find the y values of close x values. This is useful when you don’t have a calculator and if you have an equation where you can solve for certain points easily, but you cannot for other points.
In order to do this, you find the derivative of the general equation (lets say y=x^(1/2)). From here, you find the equation of the tangent line for a certain easy point. We will use the point (4,2) because it is simple. Now you will have an equation of y=0.25(x-4)+2. We can now plug in a value for x, such as 4.01, to find the y-value.
Random Notes:
In calculus, it is important to note that dy means change in the tangent line, while ∆y means change in the secant line. Here is a picture to sum that up:
Differentials:
The purpose of the differential function is to make it easier to solve for the anti-derivative. Here is how we end up with the differential equation:
See the example on page 186 to understand how to use the differential to solve for the anti-derivative (also known as an indefinite integral).
The homework for Wednesday is to do enough problems from 1-41 odd on page 193 until you are comfortable. We also have to finish the graded homework in the “Applications of the Derivative” packet.
