or

TQ : f(Q) ~ e : f(Q+e) - f(Q)

for the function f(x), we could write this proportion like:

»

Eq. 1

Assignment:

A) Write Eq. 1 using an h in place of the e and an x in place of the Q. Does the resulting equation look familiar?

B) Use the equation from part A) with the function

, and then simplify the expression.

C) Use Fermat's Method to find the subtangent for:

at the point P

a) Use the result from part B) with x = 1.

b) As h approaches 0, the expression approaches the

length of the subtangent. What is this length?

D) Use the results of part C to determine

the slope of the tangent line at P.

(Hint: Think rise over run!)

_______________________________________

Ref: Morris Kline, Mathematical Thought

ã 1999 UHH:Lopaka

Fermat's Method for Finding the Tangent to a Curve

An Historical Perspective: Did Fermat Discover Calculus?

From Methodus ad Disquiredam Maximam et Minimam, 1637

The Basic Idea

Let PT be the desired tangent at point P on a curve. The length of the line segment TQ is called the subtangent.

Once the length of the line segment TQ is found, the tangent line is constructed by drawing the line through P and T. The slope of the tangent line is the length of PQ (the rise) divided by the length of TQ (the run), PQ/TQ.

How did Fermat calculate the length of TQ?

Fermat increased the length TQ

by a small amount e to get the point Q 1 . If the line through TP is extended to T 1 , you might notice that the triangle DTQP is similar to DPRT1 . Since these are similar

triangles, the lengths of the corresponding sides are proportional:

TQ : PQ = PR : T1 R

But since T is not known, T1 also is not known.

What do you think that Fermat does next?

Think about the effect of zooming in on the slope of the curve. As e shrinks, so does the triangle DPRT 1 .

Fermat uses the triangle formed by the points P, R and p and claims that this triangle DPRp is close enough to the triangle D PRT1 for small values of e.

If our curve is a function f , t hen the point P has coordinates (Q, f(q)) and the point p has coordinates (Q + e , f (Q + e )).

Fermat then calculates the

subtangent TQ using the

ratio of sides of similar triangles:

TQ : PQ ~ PR : Rp