Class Notes 070827

The function

T (L) = .2 sqrt(L)

gives us the instantaneous rate of change of period with

respect to length.

The function

T ' (L) = .1 / sqrt(L)

gives us the instantaneous rate of change of period with

respect to length.

This function is derived by using a little moderately tricky

algebra to rearrange the expression

( T(L + `dL) - T(L) ) / `dL

and determine its limiting value as `dL approaches 0.

We can relate this to graphs and to the algebra we did last

time.

The function T ' (L) is called the derivative of T with

respect to L.

If A is a function of B, then the derivative function A '

(B) represents the derivative of A with respect to B. This function gives
the instantaneous rate of change of A with respect to B.

Most of first-semester calculus is concerned with

finding formulas for calculating derivatives, and applying that knowledge to
rate-of-change situations.

The main functions we will analyze this semester are

combinations of linear, quadratic, power, trigonometric and exponential
functions, and their inverse functions.

Linear functions:

y = m x + b is standard form, where m is slope and b is

y-intercept.

If we know two distinct points on a straight line there is

exactly one straight line containing those two points.

Suppose the points are (x1, y1) and (x2, y2).

What is the slope between these points?

What is the equation of the straight line between these
points?

Answers:

The slope is rise / run. The rise is y2 - y1 and
the run is x2 - x1 (you should be able to draw the picture for this
situation). So the slope is (y2 - y1) / (x2 - x1).

Let (x, y) be any point. Then the slope from (x1,
y1) to (x, y) is (y - y1) / (x - x1).

(x, y) is on the line if and only if the two slopes are
equal; i.e.,

(x, y) is on the line if, and only if,

(y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).

Note that this formula can be rearranged (just multiply
both sides by x - x1) to the form

y - y1 = (y2 - y1) / (x2 - x1) * (x - x1),

which you probably memorized as the two-point form of the
equation of a straight line.

Derivative of linear function:

Write the function in function notation as y(x) = m x +
b.

Ave roc of y with respect to x is `dy / `dx.

Between x and x + `dx we have `dy = y(x+`dx) - y(x).

y(x + `dx) = m ( x + `dx ) + b so

`dy = y(x + `dx) - y(x) = m ( x + `dx) + b - ( m x + b) =
mx + m `dx + b - mx - b = m `dx.

Thus `dy / `dx = m `dx / `dx = m.

So `dy / `dx = m, which doesn't change as `dx shrinks to 0.

Thus

The derivative function for y(x) = m x + b is just y '
(x) = m.

Quadratic functions:

The general form of a quadratic function is y(x) = a x^2 + b x

+ c.

Let's go ahead and calculate the derivative function.

Just as before, the derivative is the limiting value of `dy
/ `dx.

Just as before, `dy = y(x + `dx) - y(x).

In this case, y(x + `dx) = a ( x + `dx)^2 + b ( x + `dx)
+ c, so

`dy = y(x + `dx) - y(x) = a ( x + `dx)^2 + b ( x + `dx) +
c - (a x^2 + bx + c).

This simplifies easily enough. Note that (x + `dx)^2
= x^2 + 2 x `dx + `dx^2 and see if you agree with the following:

`dy = (a x^2 + 2 a x `dx + a `dx^2 + b x + b `dx + c) - (a
x^2 + b x + c)

= a x^2 + 2 a x `dx + a `dx^2 + b x + b `dx + c - a x^2 -
b x - c

= 2 a x `dx + a `dx^2 + b `dx.

Using this expression for `dy we find that

`dy / `dx = (2 a x `dx + a `dx^2 + b `dx) / `dx.

Being careful to avoid incorrect 'cancellation' we factor
`dx out of the numerator to get

`dy / `dx = (2 a x + a `dx + b) * `dx / `dx, which
simplifies to just

`dy / `dx = 2 a x + a `dx + b.

To get the derivative we have to consider what happens as
| `dx | shrinks to 0.

As | `dx | approaches 0, the term a `dx must also approach
zero (since a just a fixed number, we can make a `dx as small as we like by
making `dx sufficiently small). So

as | `dx | -> 0, we have `dy / `dx -> 2 a x + b

and our derivative is therefore

y ' (x) = 2 a x + b.

For the quadratic function

y(x) = a x^2 + b x + c, the derivative function is

y ' (x) = 2 a x + b.

This derivative function tells us the instantaneous rate at

which y is changing with respect to x for any value of x. The value of
this derivative function is equal to the corresponding slope of the graph of y
vs. x.

Class Notes 070827

The function

T (L) = .2 sqrt(L)

gives us the instantaneous rate of change of period with

respect to length.

The function

T ' (L) = .1 / sqrt(L)

gives us the instantaneous rate of change of period with

respect to length.

This function is derived by using a little moderately tricky

algebra to rearrange the expression

( T(L + `dL) - T(L) ) / `dL

and determine its limiting value as `dL approaches 0.

We can relate this to graphs and to the algebra we did last

time.

The function T ' (L) is called the derivative of T with

respect to L.

If A is a function of B, then the derivative function A '

(B) represents the derivative of A with respect to B. This function gives
the instantaneous rate of change of A with respect to B.

Most of first-semester calculus is concerned with

finding formulas for calculating derivatives, and applying that knowledge to
rate-of-change situations.

The main functions we will analyze this semester are

combinations of linear, quadratic, power, trigonometric and exponential
functions, and their inverse functions.

Linear functions:

y = m x + b is standard form, where m is slope and b is

y-intercept.

If we know two distinct points on a straight line there is

exactly one straight line containing those two points.

Suppose the points are (x1, y1) and (x2, y2).

What is the slope between these points?

What is the equation of the straight line between these
points?

Answers:

The slope is rise / run. The rise is y2 - y1 and
the run is x2 - x1 (you should be able to draw the picture for this
situation). So the slope is (y2 - y1) / (x2 - x1).

Let (x, y) be any point. Then the slope from (x1,
y1) to (x, y) is (y - y1) / (x - x1).

(x, y) is on the line if and only if the two slopes are
equal; i.e.,

(x, y) is on the line if, and only if,

(y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).

Note that this formula can be rearranged (just multiply
both sides by x - x1) to the form

y - y1 = (y2 - y1) / (x2 - x1) * (x - x1),

which you probably memorized as the two-point form of the
equation of a straight line.

Derivative of linear function:

Write the function in function notation as y(x) = m x +
b.

Ave roc of y with respect to x is `dy / `dx.

Between x and x + `dx we have `dy = y(x+`dx) - y(x).

y(x + `dx) = m ( x + `dx ) + b so

`dy = y(x + `dx) - y(x) = m ( x + `dx) + b - ( m x + b) =
mx + m `dx + b - mx - b = m `dx.

Thus `dy / `dx = m `dx / `dx = m.

So `dy / `dx = m, which doesn't change as `dx shrinks to 0.

Thus

The derivative function for y(x) = m x + b is just y '
(x) = m.

Quadratic functions:

The general form of a quadratic function is y(x) = a x^2 + b x

+ c.

Let's go ahead and calculate the derivative function.

Just as before, the derivative is the limiting value of `dy
/ `dx.

Just as before, `dy = y(x + `dx) - y(x).

In this case, y(x + `dx) = a ( x + `dx)^2 + b ( x + `dx)
+ c, so

`dy = y(x + `dx) - y(x) = a ( x + `dx)^2 + b ( x + `dx) +
c - (a x^2 + bx + c).

This simplifies easily enough. Note that (x + `dx)^2
= x^2 + 2 x `dx + `dx^2 and see if you agree with the following:

`dy = (a x^2 + 2 a x `dx + a `dx^2 + b x + b `dx + c) - (a
x^2 + b x + c)

= a x^2 + 2 a x `dx + a `dx^2 + b x + b `dx + c - a x^2 -
b x - c

= 2 a x `dx + a `dx^2 + b `dx.

Using this expression for `dy we find that

`dy / `dx = (2 a x `dx + a `dx^2 + b `dx) / `dx.

Being careful to avoid incorrect 'cancellation' we factor
`dx out of the numerator to get

`dy / `dx = (2 a x + a `dx + b) * `dx / `dx, which
simplifies to just

`dy / `dx = 2 a x + a `dx + b.

To get the derivative we have to consider what happens as
| `dx | shrinks to 0.

As | `dx | approaches 0, the term a `dx must also approach
zero (since a just a fixed number, we can make a `dx as small as we like by
making `dx sufficiently small). So

as | `dx | -> 0, we have `dy / `dx -> 2 a x + b

and our derivative is therefore

y ' (x) = 2 a x + b.

For the quadratic function

y(x) = a x^2 + b x + c, the derivative function is

y ' (x) = 2 a x + b.

This derivative function tells us the instantaneous rate at

which y is changing with respect to x for any value of x. The value of
this derivative function is equal to the corresponding slope of the graph of y
vs. x.