The solutions to Quiz #6 have been posted on the Test and Quiz Solutions page.

My blogging time has been eaten up by a variety of things lately, but I'll see if I can summarize two concepts in one entry!

We looked at the idea of using the tangent line to a function curve at a point of interest as a tool for approximating the function for points near the point of interest. The idea of the "point of interest," by the way, is that we know everything that goes on at that point exactly. Let . The line tangent to this curve at the point is given in point-slope form as:



If we solve this equation for y, we get what's called the linearization of f at c:



L(x) is an approximation to the actual value of f at a value x near a, and this gives us the linear approximation for f near a:



We did a couple of examples where we could estimate function values really easily -- yes, easily because all we were evaluating was a linear equation.

Another way to phrase the same process is through differentials. Differentials in a variable can be thought of as "a small change" in that variable. If y depends on x, what we want to know is how a small change in x affects y. We will define some actual changes first. Let be the actual change in x and be the actual change in y. If , perhaps the function f is quite hard to calculate in general. Let's assume we know the exact value of f at x, say. The actual change in y is

,

may be difficult to calculate exactly. Let's define the differential of x, . This differential is an independent variable; we can set it to any value. We will define the differential of y to be



This differential is a good approximation of and usually much easier to calculate. (See Figure 2 on p.180.) We can also use dy to come up with a good approximation for f near x, since



We did a quick example on error propagation. The basic idea is that if we make an error in measurement of a quantity (as we're likely to do because we're human), that error will go forward if we use our measurement to calculate another quantity, such as area or volume. We can estimate the maximum error in our calculated quantity by using differentials as above.

We then began looking at ways to determine various high and low points for a function. We defined absolute (global) maxima and minima and examined some graphs. Absolute extrema represented overall high and low points graphically. Not all graphs have either one, but we did determine one situation that guarantees the presence of absolute max's and min's.

Extreme Value Theorem: If f is a continuous function on a closed interval [a, b], then f attains an absolute maximum and an absolute minimum value somewhere on [a, b].

We'll look at criteria for discovering local and absolute extreme values in the cases where a graph might be difficult or impossible. See you next week.

I didn't give you any assignment on logarithmic differentiation out of 3.9. If you could try out problems 49, 51, 53, and 55 from 3.9, that'd be awesome. Thanks!

The solutions to Exam #2 are available from the Test and Quiz Solutions page.

You will get the graded exams back on Wednesday.

Midterm grades were posted late last night. The registrar is supposed to "roll" the grades this morning so that they'll be visible to you.

Included in the midterm grades were:

• All 5 quizzes you've taken.
• Exam #1 including any bonus points.
• Lab #1.

What was not included was Exam #2. I will grade that and return it to you on Wednesday. (The solutions will be posted by Tuesday at the latest.)

Monday in Calc I, we introduced our last new types of functions for which we needed to find a derivative. These were the exponential and logarithmic functions. Here's a slightly different approach from the order we used in class.

We began with the natural logarithm function. We had to go back to first principles -- definition of the derivative -- to find that



We saw several examples of how to use this new rule in conjunction with existing rules. The Chain Rule version of this was the most common variation. If u is any function of x, then we can write



If we were looking at logarithms in another base b with , the situation wasn't too difficult. By the change-of-base formula, we have that



That expression ln b is constant, so when we take the derivative we can just move it aside. Thus



To get the derivative of the exponential function, we use the inverse relationship between the exponential and natural logarithm functions, as well as a little implicit differentiation:



That is, !!! (This is the only function, aside from 0, that is its own derivative).

In its Chain Rule version, again assuming u is a function of x, we have



We looked at a couple of examples of this. For exponentials in other bases, we used the fact that to write



We then talked about logarithmic differentiation. The idea behind logarithmic differentiation is this: taking the logarithm of a quantity generally allows us to take apart the quantity, splitting it into its simple pieces. Then we take the derivative of both sides. Implicit differentiation gives us that and thus a way to get the derivative alone. The procedure is:



and since we know what y is, we have our derivative.

Wednesday and Thursday, we talked in lecture about related rates problems. The idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity (which may be more easily measured). The procedure is to find an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time.

It is impossible to cover the entire spectrum of possible problems. We need to remember (or have access to) a lot of geometry formulas to be successful at these. Many of the most common ones can be found in the stiff page near the back of your book.

A basic strategy (stolen from Stewart's 3rd edition) is the following:

1. Read the problem carefully.
2. Draw a diagram if possible.
3. Introduce notation. Assign symbols to all quantities that are functions of time.
4. Express the given information and the required rate in terms of derivatives.
5. Write an equation that relates the vatious quantities of the problem. If necessary, use the geometry of the situation to eliminate one of the variables.
6. Use the Chain Rule to differentiate both sides of the equation with respect to t.
7. Substitute the given information into the resulting equation and solve for the unknown rate.

A common error is to substitute the given information too soon. If the variable related to the information changes its value during the problem, you must wait until the end, after you take the derivative, to substitute the value.