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I was given the point P (4,2) and Q $(x,\sqrt{x})$ and was told to find the secant line. I forgot what the difference between secant and tangent was but I thought it might have something to do with that formula I was suppose to memorize but forgot. As far as I could make sense of this the formula should be something like $\frac{(a-h)+a}{h}$ so I used that and I got 1 which I think is wrong so I wrote 2. I was then given points like 3.5 3.9 3.99 and such and told to find the tangent lines I think and then estimate the tangent I think. I didn't know how to do this because I got -1 and 1 which can't be right so I made up numbers getting closer to two. I was then told to estimate the secant lines and I guessed 2 because I don't know how to do that.

I tried to take the derivative of it which gave me $\frac{1}{2}x^\frac{-3}{4}$ which gave me .56 something or so so I just made up an answer and guessed 2 because 4/2=2.

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Do you have some sort of notes with definitions you could consult? – anon Jan 19 '12 at 2:11
I have Stewarts calculus book but it isn't very useful for learning calculus. – user138246 Jan 19 '12 at 2:19

Secant lines go through two points, while tangent lines meet smoothly with the curve at one point:

$\hskip 2in$ pic

$\hskip 1.5in$ pic2

A secant line for two points $(x_A,y_A)$ and $(x_B,y_B)$ is given by a rise-over-run formula:

$$\frac{y-y_A}{x-x_A}=\frac{y_B-y_A}{x_B-x_A}. \tag{S}$$

This of course might look awfully daunting to remember. How to memorize it? The key is to see what both sides of the equation stand for: the left side is the slope $m$ (rise/run) of the line between the points $(x,y)$ and $(x_A,y_A)$ while the right side is the slope between $(x_A,y_A)$ and $(x_B,y_B)$. So if you want, you can use this diagram I made up as a mnemonic to be unpackaged whenever:

$\hskip 1.75in$ pic3

(This says the slope of $\overline{AB}$ equals that of $\overline{BC}$. Here we relabel $(x,y)$ as $C$.) This of course makes sense: any point $(x,y)$ that is off this line will not form a line with $A$ having the same slope as $\overline{AB}$.

Now for a curve given by $(x,f(x))$ in coordinates (i.e. $y=f(x)$), secant lines will indeed have slope given by $$m=\frac{f(b)-f(a)}{b-a},$$ which is similar to what you remember if we let $x=a$ and $b=x+h$.

Tangent lines are trickier. We must let the two points $A$ and $B$ approach each other indefinitely, and in the limit we get a derivative in the expression:


These two formulas for secant and tangent lines, $(S)$ and $(T)$ above, are what you need to (1) remember consistently and (2) know how to apply in actual problems.

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I don't understand this at all, how does that formula work? Why are the secant lines equal with slope? Why can't they cross the same line twice and have different slopes? I can think of many examples where that works. – user138246 Jan 20 '12 at 16:17

It's a trick question, sort of. A secant line is any line that touches the graph in (at least) two places; the tangent line is what you get when you let the two places get infinitely close together. The secant line between $P$ and $Q$ is just the line between those two points; for points $(a,b)$ and $(c,d)$, the line will have the expression

$$y = \frac{d-b}{c-a}(x-a) + b$$

For the specific points they give you, the line will be a function of, let's say, $s$, where $Q$ is the point $(s,\sqrt{s})$ (I would use $x$ to match your notation, but then it conflicts with the convention of using $x$ and $y$ for a line.)

For the rest of your question, are you sure you're remembering the problem correctly? I think they might be asking you to use the secant lines to estimate the tangent line (to the function $f(x) = \sqrt{x}$) at the point $(4,2)$.

You can do this by noting that the slop of the secant line between $P(4,2)$ and $Q(x,\sqrt{x})$ is $(\sqrt{x}-2)/(x-4)$ and then plugging in the numbers they gave you for $x$; since the $x$-values are increasingly close to $4$, the slopes of the secants will be close to the slope of the tangent line.

Taking the derivative is a good way to check your work, since you then have the exact value you're supposed to be approximating. In this case, $\frac{d}{dx}\sqrt{x}$ is $$\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{1/2-1}=\frac{1}{2\sqrt{x}}$$ so that the derivative at $x=4$ is $1/4$.

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