# Strang's "Calculus Puzzle"

• January 6th 2013, 12:08 PM
zachd77
Strang's "Calculus Puzzle"
Found this "puzzle" from Strang's Calculus.

The slope-intercept form of a line (y = mx + b) requires TWO numbers, the point-slope form (y - f(a) = f'(a)(x-a)) requires THREE numbers, and the two-point form requires FOUR (a, f(a), c, f(c)). How is this possible?

(suppose f'(a) = m).

I'm curious on this one. Any ideas on this "puzzle"?
• January 6th 2013, 03:47 PM
chiro
Re: Strang's "Calculus Puzzle"
Hey zachd77.

Lets consider these one at a time.

The first one requires two numbers since m encodes information about two points (how they change over time) while b encodes information about an initial condition.

If you didn't have m and b you would have to use four data points (x0,y0,x1,y1) to get m and b.

The reason why we can reduce this is because the slope doesn't change between any two sets of points.

The four point set-up doesn't assume a particular model (in this case a linear one) while the y = mx + b does. When you can assume more information, the amount of variables you need to describe that model decreases.

For number 2 you should be aware that you can calculate f'(a) from f(a) so these quantities are not really independent in a symbolic sense (since we have information to get a derivative easily from the equation of a line).
• January 6th 2013, 11:18 PM
hollywood
Re: Strang's "Calculus Puzzle"
The combination of slope and y-intercept determine a line - two parameters are required. The three parameters in the point-slope form determine a line and one particular point on the line. The four parameters in the two-point form determine a line with two "special" points.

- Hollywood
• January 7th 2013, 04:34 AM
Deveno
Re: Strang's "Calculus Puzzle"
since lines are "linear" their derivative (slope) is constant, that is, the limit:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0} \frac{m(a+h) + b - ma - b}{h}$

$= \lim_{h \to 0} \frac{mh}{h} = m$.

so to find the "one point" f'(a) we need 4 points (3 of which can be f(a), a and x, the fourth is f(x)), so that we know m (as [f(x) - f(a)]/(x - a)).

we actually need 4 points to determine m in the y = mx + b formula. we can obtain both "m and b at once" if one of the x-coordinates is 0 (which is why it seems like we only need 2, these are "encoded" as (0,b) and (x,mx+b) which as you can clearly see is 4 data points: 0,b,x, and m).

long story short: we need two "coordinate pairs" for each form. the slope-intercept form uses a SPECIAL pair (0,b) and then some "other" pair (x,mx+b).

the derivative form uses two arbitrary pairs (a,f(a)), and (x,f(x)), the term f'(a) "hides" the fact that it is actually just [(f(x) - f(a)]/(x - a), making it look as if one data point goes away (it's just another way of writing the "two point form").

the derivative form also appears in THIS from (the "point-slope form") y - y1 = m(x - x1) with x1 playing the role of a, y[SUB]1[/SUB (= f(x1)) playing the role of f(a), and m playing the role of f'(a).