Question

The part of the lecture talking about GPS, shouldn't it be S/h to get L? Why is it L/h?

Answer 1

Can you post a link to the lecture ?

Answer 2

Here is the Link, it starts at 13:30

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-4-limits-and-continuity/

Answer 3

In the GPS example, the professor did not use S at all, so S/h is not defined.

The point of his discussion (which may be lost amongst the discussion of the GPS), is this:

Let's say you have a variable that is the function of another variable. Most often, we might say
y = f(x)

though in this case, we say L= f(h)

Now the question we are interested in, is if we put in \(x +\Delta x\) rather than x, how much will L change? (The professor associated Δx with an unknown error)
In other words, what is the difference \(\Delta L\)= f( \(x +\Delta x\)) - f(x)

He went on to say
\[ \frac{\Delta L}{\Delta x} = \frac{f(x+\Delta x)-f(x)}{\Delta x} \approx \frac{dL}{dx}\]
(only in the limit as Δx-> 0, can we equate the expressions)

He then said that the derivative is easy to find, if we have the function.
For example, if the satellite is at a known altitude A, we can use the pythagorean theorem to solve for L in terms of the fixed A and the measured h:
\[ L = \sqrt{h^2 -A^2} \]
we use calculus to find dL/dh to be (I don't think you learned this part yet)
\[\frac{dL}{dx}= \frac{h}{\sqrt{h^2 -A^2}}\]

The main point is that this expression tells us the ratio of the change in L caused by a change in x. If the change in x is "small", we can say
\[ \frac{\Delta L}{\Delta x} \approx \frac{dL}{dx} \]
and, more usefully
\[ \Delta L \approx \frac{dL}{dx} \Delta x \]
or, if we use the actual derivative
\[ \Delta L \approx \frac{h}{\sqrt{h^2 -A^2}} \Delta x \]

that is a useful expression. (though perhaps you don't yet see why)