(Part 1 – Limits & Derivatives)
    Limits
    Differential Calculus (Check out the ANIMATION!)
    Integral Calculus (updated 3/99)
    Vector Calculus

Limits (Sequences, functions, graphs)

What do the numbers 1/1, 1/2, 1/3, 1/4, 1/5, . . . get closer and closer to? It may be clear that they approach zero, so we say the limit is 0. The nth term is 1/n, so the notation is lim (n->oo) 1/n = 0. (The -> means “approaches”; the oo is a cheesy infinity symbol.)

The same idea is to consider the function f(x) = 1/x . Then as x -> oo , f(x) -> 0 , so again we say lim (x->oo) f(x) = lim (x->oo) 1/x = 0. On the graph we’d have a horizontal asymptote at y = 0 since the output values approach 0 as the graph goes off to the right. (Click here to review functions or graphing.)

Now what if x -> a and we see if f(x) approaches any specific value L. If it does, then we say that lim (x->a) f(x) = L. By definition x->a means x is near a but not equal to a.

Example 1: Let f(x) = x + 3 . Then as x -> 3, f(x) -> 3 + 3, so lim (x->3) [x+3] = 6.

Example 2: Let g(x) = (x^2 – 9) / (x – 3) . By algebra, we have

g(x) = (x + 3)(x – 3) / (x – 3) and if x =/= 3 then we can cancel, so

g(x) = x + 3 if x =/= 3. Notice that f(3) = 6 but g(3) is undefined.

The f(x) from example 1 has domain “all real numbers,” but the g(x) from example 2 has domain “all reals except 3,” so they’re different functions. But the limit as x -> 3 is the same in both cases: lim (x->3) f(x) = lim (x->3) g(x) = 6. Even though g(3) is undefined (it’d be 0 / 0) g(x) still has a limit (of 6), since x -> 3 implies x =/= 3.

Example 3: Some other interesting limits:

lim (x->0) [(sin x) / x] = 1

lim (x->0) [(1 + x)^(1/x)] = e = 2.71828 approx

lim (n->oo) [F(n+1) / F(n)] = (1 + Sqrt(5)) / 2 = 1.61805 approx.

The F(n) is the nth Fibonacci number of the sequence 1, 1, 2, 3, 5, 8, 13, . . .

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Differential Calculus

    • Tangent Lines & Derivatives
    • Differentiation Rules
    • Applications
    • Tangent Lines & Derivatives . . . . . . (top of page)

    On a straight line graph, y = m x + b, the slope is constant; it’s equal to m
    no matter where you are on the graph.
    For the slope we figure (rise)/(run) = m.

But on a parabola, like y = x^2, the direction keeps changing, so we’d expect that the “slope” doesn’t stay constant. But how do you figure out the slope of a curve?

    Example 1: The slope on the parabola y = x^2 is zero at the origin since the curve is

    horizontal there. At a different (x, y) location, what’s the slope?

Let’s take (3,9) as an example; if we join up (3,9) and a nearby point on the curve, say (3.1,9.61), and do “rise/run”: (draw yourself a picture)

. slope = (y2 – y1) / (x2 – x1) = (9.61 – 9) / (3.1 – 3) = 0.61 / 0.1 = 6.1

If we take a closer point (3.02,3.02^2) = (3.02,9.1204) then the slope is

. slope = (9.1204 – 9) / (3.02 – 3) = 0.1204 / 0.02 = 6.02

In fact for any h, the slope between (3,9) and (3+h,(3+h)^2) will be

. slope = ((3+h)^2 – 9) / (3+h – 3) = (6h + h^2) / h = 6 + h

The line joining two points on a curve is called a secant line. The slope is msec.

As h gets smaller and the points get closer together, the slope of the secant line approaches 6. The line it approaches is called the tangent line. The slope is mtan.

    Here the parabola y = x^2 has a slope of 6 at (3,9) because the limit
    (as h goes to 0) of 6 + h is 6.
    Example 2: Do the same thing at a general (x,y) point on y = f(x) = x^2:

    join up (x,x^2) and (x+h,(x+h)^2) and figure out the slope between them:

. slope = msec = ((x+h)^2 – x^2) / (x+h – x) = (2xh + h^2) / h = 2x + h.

As h goes to 0, this becomes mtan = 2x. The expression for “the slope at any x” is a new function, derived from f(x), called the derivative of f, and denoted f’(x).

Here we have: if f(x) = x^2 then f’(x) = 2x.

Checking this for our old point (3,9) we see if x = 3 then f’(x) = 2x = 2(3) = 6. Ok!

    • Related Rates
    • Max-Min Problems

    • Here’s a cool Mathematica/ QuickTime/GIFBuilder animation showing the secant lines that approach thetangent line. The tangent line is the LIMIT of the secant lines, and the slope mtan is the limit of the slopes msec.

    Note that the slope between Pand Q “settles down” to a value mtan, which is the derivative of f at x, where x = x-coord of P.
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    Differentiation Rules! . . . . .

    Another notation for f’(x) is (f(x)).
    If y = f(x) then y’ = f’(x) = (y) = dy/dx.
    Notice from above we had (x^2) = 2x . What’s the rule for the derivative of a power?
    If we had f(x) = x^3 then f’(x) = ??? Let’s use the limit definition of derivative.
    (x^3) = lim (h->0) [(x+h)^3 - x^3] / h
    = lim (h->0) [x^3 + 3 x^2 h + 3 x h^2 + h^3 - x^3] / h
    = lim (h->0) [3 x^2 h + 3 x h^2 + h^3] / h
    = lim (h->0) [3 x^2 + 3 x h + h^2]
    = 3 x^2 + 3 x(0) + 0^2
    = 3 x^2
    In general if f(x) = x^n (works for n = integer, fraction, or irrational!) then
    Power Rule : (x^n) = n x^(n-1)
    Sum Rule : [ f(x) + g(x) ] = f(x) + g(x) = f’(x) + g’(x)
    Constant Multiple Rule: [ c f(x) ] = cf(x) = c f’(x).
    Some special functions and their derivatives:

    (sin x) = cos x , (cos x) = – sin x ,
    (e^x) = e^x , (ln x) = 1 / x .
    Ex. (x^100) = 100 x^99 ;(x^2 + 5 x) = 2 x + 5 ;
    (x ) = (1/2) x ^ (-1/2) = 1 / ( 2x ) ;
    (5 sin x – 3 cos x) = 5 cos x + 3 sin x.
    We might have a product or quotient of functions:
    Product Rule :(f(x) g(x)) = f(x)g(x) + g(x)f(x)

      = f(x) g’(x) + g(x) f’(x)
    Quotient Rule :(f(x) / g(x)) = [ g(x)f(x) - f(x)g(x) ] / (g(x)^2)

      = [ g(x) f'(x) - f(x) g'(x) ] / (g(x)^2)
    Here’s what you do with a composition of functions:
    Chain Rule :(f(g(x)) = f’(g(x))g(x) = f’(g(x)) g’(x)
    Example:(sin(3x)) = cos(3x)(3x) = 3 cos(3x)

    (sin^3(x)) = [(sin x)^3]
    = 3 [ (sin x)^2 ](sin x) = 3 sin^2 x cos x
    Applications of Derivatives . (4/00)

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Related Rates

You guessed it — define some variables, try to relate their rates (of change); we get a relation between their derivatives; technically known as a “differential equation.”

Example: Spoze you blow 30 cu.in. of air per sec into a spherical balloon. How fast is the radius increasing when the balloon is 6 inches in diameter?

    Solution: Ask the important questions:
    * What are the variables involved?
    V = volume, r = radius, D = diameter, t = time.
    ** What’s the relation among them?
    V = (4/3) Pi r^3; the volume formula for a sphere.
    (Click here for lots of area and volume formulas)
    *** Now take deriv’s with respect to time t: (ok, that’s not a question)
    d/dt [V] = d/dt [(4/3) Pi r^3] = (4/3) Pi d/dt [r^3] ;
    dV/dt = (4/3) Pi r^3 * 3 r^2 dr/dt = 4 Pi r^2 dr/dt ; this is the chain rule.
    *v Then plug in what you know: D = 6 ==> r = 3, dV/dt = 30 :
    30 = 4 Pi (3^2) dr/dt ; solve for what you wanna know:
    dr/dt = 30/(36 Pi) ~=~ 0.265 in/sec.

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Max/Min Problems

The philosophy here is to optimize some quantity Q that can depend on more than one thing (variable). Find a relation (constraint) and use it to get the Q in terms of one variable; then set the first derivative equal to zero and solve!

    • (Oh, I was supposed to coax YOU into discovering all that.)

Example: Of all the rectangles of area A, which has the shortest diagonals?

    Approach: Ok, use x = width, y = height; then xy = A; we want to minimize D = -/(x^2 + y^2) ;
    we can sub in y = A/x since A is constant. Also we can minimize D^2 and D will be smallest.
    D^2 = x^2 + (A/x)^2 ; now it’s in terms of one variable; take the deriv. w.r.t. x :
    (D^2) = (x^2 + (A/x)^2) = (x^2 + (A^2) x^(-2))
    . . . . . . . . . . = 2x + A^2 (-2 x^(-3)) = 2x – (2 A^2)/(x^3) =set= 0 ;

    2x = 2 A^2 / x^3 ; x^4 = A^2 ; x = -/A ; y = A/x = -/A ;
    Analysis: The rectangle’s a square, Captain Kirk. And the -/( . ) means square root.
    Go on to Integral Calculus

    Well, that’s it for now. Check back often for new stuff!
    http://home.earthlink.net/~djbach/lessons.html
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