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Monday 10/26
Today we learned the difference between explicit functions (specifically defined functions, ie. y=x^2+3x+4) versus implicit functions (not specifically defined functions, ie. y^3x^2 + 2yx + y = 3 + 4x + yx^4). We also figured out how to find the derivative of implicit functions.

media type="custom" key="4664827"

Tuesday 10/27
We began class with the problem of the day, shown below.

First, take the deriviative of x^3 + y^3= 3xy by using the power rule and the product rule. [Hint: think of y as f(x)] Then, solve for dy/dx. The tangent line is horizontal when dy/dx=0 i.e. when y-x^2=0. Using these relationships between x and y, we can create a system to solve for x and y. We find that the tangent line is horizontal at (0,0) and (2^(1/3), 4^(1/3)). But dy/dx= (y-x^2 / y^2-x) so wouldn't the solution (0,0) make the slope undefined? Well there are two points at (0,0), which we can see by looking at the graph, and one part has an undefined slope while the other has a horizontal slope, so (0,0) works.

**REMEMBER -->**

Next, we went over a few implicit problems from last night's homework (p. 172 #1-19 odd) which are shown below.



Then, we started working on an Implicit Differentiation worksheet, which we will continue working on in class tomorrow. The first problem is shown below.

First, find the slope of the tangent line by taking the opposite reciprical of the slope of the given line. Find the derivative of the curve, which is equal to the slope of the tangent line. Then, using this equation and the equation of the curve, we can solve a system for the points.

Homework: p. 173 #25 a) and b), #26 a) and b), and #27 (for easier problems) OR work on the implicit differentiation worksheet **:D**

Wednesday 10/28




Thursday10/29
More of the implicit worksheet plus some AP Problems. Here are solutions to a few more of the implicit worksheet problems.



Friday 10/30
We started with the three problems of the day shown below. We were able to do the first two successfully but were stumped by the third because the relation was y in terms of theta but we were asked for dy/dt! Where's the t? Obviously Ms. Gentry had something in mind so after we watched the shuttle video, she explained that although the height of the shuttle above the ground could be written in terms of the angle of elevation, there were many different rates of change involved; These rates include, dy/dt which could be the speed of the shuttle, dy/dø, the rate at which the angle changes as the height changes and dø/dt, the speed at which the angle changes as the shuttle rises. We are going to have to get used to finding derivatives of variables with respect to other variables which are not in the expression!

media type="custom" key="4688925" align="left" The public observation platform from which to watch the shuttle launch is three miles from the launch pad at an estimated speed of 583 feet per second. How quickly is the angle of elevation changing three seconds after the launch?

This problem is an example of a related rates problem. As the shuttle is launched, it's distance from the ground, it's direct distance from the observation platform and the angle of elevation of the shuttle are all changing both with respect to time and with respect to each other. Assuming the speed is relatively constant for the first three seconds of flight, we can model the flight as in the the diagram below.

The problem can be broken down into the following steps:

1) What information is given? dh/dt = 583 2) What do we wish to find? dø/dt when t = 3. or when h = 3(583) = 1749 ft 3) Check the compatibility of the units: 3 miles = 3(5280) = 15840 ft. 4) Write an equation connecting the variables in the problem 5) Use the parametric chain rule to find the desired rate, dø/dt.