You've probably already seen it in your homework problems -- "find the
equation of the line that is *tangent* to bloppity-blop curve at
the such-and-such point." And if you haven't yet, you soon shall.
It will be on the exam as well.

A line that is tangent to a curve has two properties:

- The line shares a point with the curve in question.
- At the shared point, the
*derivative of the curve is equal to the slope of the line.*

The equation of a line is, as you learned in algebra

y = mx + b eq. 4.5-1So all you have to do is find

Let's take the second property first. The slope of the line is `m`.
That value has to match the derivative of the curve at the point they
give you. So for finding `m` you need to know the derivative
of the curve. And you need to evaluate it at the point they give
in the problem. *And that value is* `m`.

Then take the first property. You need to find the `b` that
makes that first property true. And you now know what `m` is
equal to. If, for example, the point they want you to be tangent to
is `(2, 3)`` y = mx + b``2` for `x`,
`3` for `y`, whatever you came up with for `m`,
and solve for `b`. It's that easy.

Let's run through an example. Let the curve be

y(x) = xUsually it's understood that^{2}+ 1 eq. 4.5-2

Let's find the tangent at ` x = 2``x`-coordinate of the point. To find the
`y`-coordinate, simply use the equation of the curve (given in
4.5-2). That gives us ` y = 5``(2, 5)`

**Step 1: Find the derivative of the curve.** The equation of the
curve is given in 4.5-2. You know how to take its derivative. It's

y'(x) = 2x eq. 4.5-3

**Step 2: Evaluate the derivative at the point given.** The problem
says do it at ` x = 2``x`
is ` y'(2) = 4`*That is your* `m`.
Write it down. ` m = 4`

**Step 3: Solve for** `b`. Remember,
` y = mx + b``x` is,
what `y` is, and what `m` is. If you plug them
all in, you get

5 = 4×2 + b eq. 4.5-4aAnd it's trivial algebra to go from there to

b = -3 eq. 4.5-4b

**Step 4: Write the equation.** You know `m` and `b`
now. Simply substitute them into ` y = mx + b`

y = 4x - 3 eq. 4.5-5That's it. We're done. Once you know how to take the derivative, these problems are easy. Just follow the four steps here.

Figure 4.5-1 shows a graph of this problem. Observe how the tangent
line just kisses the curve at
x = 2 y'(x) = 2x = y'(2) = 4 |

Here's a slightly more complicated variant on the same problem. Find the two lines that are tangent to

y = xand pass through the point,^{2}- 2x + 1

**Step 1: Find the derivative of the curve.** In this case we
have

y' = 2x - 2We do this because we know that at the point of tangency, the derivative of the curve must be equal to the slope of the tangent line. So we need to know this derivative.

**Step 2: Write as much of the equation of the line
as you can from what you know.**
The slope, `m`, of the line is still unknown. But you know what point
it must pass through. In this case that is the point `(5,7)`. So using
the formula for the

y - 7 = m(x - 5)

**Step 3: Use the derivative to write an equation for the slope.**
We don't know the slope of the line yet, but we know that if the point,
`(x,y)`, is the point of tangency, then the slope of the line will
be equal to the derivative of the original curve at `x`. The derivative
we already determined to be ` y' = 2x - 2`

m = y' = 2x - 2

**Step 4: Substitute the slope expression into the equation for the line.**
You have ` m = 2x - 2`` y - 7 = m(x - 5)``m` in the second equation.

y - 7 = (2x - 2)(x - 5) = 2x^{2}- 12x + 10 y = 2x^{2}- 12x + 17

**Step 5: Substitute the equation for the original curve back in.**
That is, you know that ` y = x ^{2} - 2x + 1`

x^{2}- 2x + 1 = 2x^{2}- 12x + 17

**Step 6: Gather like terms and solve for **`x`.
Using simple algebra the above equation becomes the
quadratic

xYou can either use the quadratic formula, or you can factor this one in your head to get^{2}- 10x + 16 = 0

(x - 2)(x - 8) = 0So the

**Step 7: Substitute back to get **`m`.
That is, we have already seen that ` m = 2x - 2``(x, y)``x`'s are for the points of tangency, we can put those values
into the equation for `m` and get that either
` m = 2`` m = 14`

y - 7 = 2(x - 5)and

y - 7 = 14(x - 5)When you multiply those out and put them into slope-intercept form, the equations of the two lines that are tangent to

y = 2x - 3and

y = 14x - 63and you are done. The curve and the two lines are illustrated in the graph on the right here. Notice where the two lines intersect. Both points of tangency are visible on the graph, although one is very nearly at the top of the graph.

Click here to see an alternative attack on this same problem.

This is one they might throw at you just to keep you on your toes. Instead
of asking for the equation of the tangent line, they'll ask you for the
equation of the line that is *normal* to bloppity-blop curve at
such-and-such a point.

How much do you remember from algebra? Do you
remember that if two lines are *normal* to each other (that is,
they are at right angles to each other), then their slopes are
*negative reciprocals* of each other? In other words, if one
has a slope of `m`, then the other must have slope of
`-1/m`. Can you extend this definition to include a line
being normal to a curve? It's just a small variation of the rules
we had for tangent lines.

- The line shares a point with the curve in question.
- At the shared point, derivative of the curve is the negative reciprocal of the slope of the line.

Let's take our example from before where the curve is
` y = x ^{2} + 1`

The remaining steps are the same. Solve for `b` with the `x`
and `y` coordinates you have and the `m` you have just
determined. In this case we have:

5 = (-1/4)×2 + b eq. 4.5-6aand solving for

b = 5.5 eq. 4.5-6bSo the answer here is

y = (-1/4)x + 5.5 eq. 4.5-7

Figure 4.5-2 shows a plot of our curve and the line that is normal
to it at
Recall that to find where the plots of two functions cross you simply
take the difference of the two functions and solve for the y = (-1/4)x + 5.5 |

You should either have for your difference

xor the negative of that. The perscription then is to set that equal to zero and solve for^{2}+ (1/4)x - 4.5 eq. 4.5-7a

xThis is without a doubt a quadratic polynomial and we have the^{2}+ (1/4)x - 4.5 = 0 eq. 4.5-7b

_________ -1/4 ± Ö1/16 + 18 x =If you take the plus of the~~eq. 4.5-8 2~~

Let

______ y(x) = Ö1 - xThe graph of this is shown in the figure to the right. You probably recall from algebra that this function has a graph that is a semicircle. Find the equations of both the tangent line and the normal line at^{2}

**Step 1: Find the derivative of the function.** You will need it
for every step after this one. Notice that this function is a composite
of ` y(x) = f(g(x))`

_ f(x) = Öxand

_ f(x) = Öxthen

1 f'(x) =which is useful to know in finding the derivative of~~2Öx~~

**Step 2: Where is the point of tangency?** Simply substitute the
`x` given in the problem into the function to get a value for
`y(x)`. You shouldn't have to think too hard, because I gave
it away in the diagram.

**Step 3: What is the slope of the tangent line? of the normal line?**
Remember that the slope of the tangent equals the derivative of the
function at the point of tangency. So stick the value given for `x`
into the equation you have for `y'(x)` to find that value. That
is the `m` for the tangent line. It's negative reciprocal is the
`m` for the normal line.

**Step 4: Solve for **`b`** in both cases**, that is the
`b` for the tangent line and the `b` for the normal line.
Remember that you now know the `m`'s for both of them as well
as an `x` and corresponding `y`. Just substitute them
into ` y = mx + b``b`.

That completes the problem for both finding the tangent and the normal.
So write your ` y = mx + b``m` and `b` into each. Did you get
` b = 0`*any*
point on this curve, the `b` is always zero.

Simply write ` y = mx + b``m` and
substitute the expression given in the problem for `y(x)` in
for `y`. Take all the cancellations you can. You should end
up with ` b = 0`

Move on to Hilltops and Valley Floors

email me at *hahn@netsrq.com*