Chapter 15. Slope Approximation

Volume 3, Why Slopes and More Math.

Saying precisely how to compute a quantity, defines it.

Numerical methods for calculating a number or quantity exactly or in a limiting fashion are employed in mathematics as (computational) definitions. Informal non-computational descriptions of numbers and quantities have no precise role, except perhaps to supply motivation for the computational definition. And in this motivational category falls the previous description of slopes. This chapter provides an approximation and then a computational definition for slopes.

1. [Play Video] 3 minutes: Common changes of notation in the limits that yield the slope or derivative.
   

Slope Calculation

[Play Video] 4.5 minutes: Approximating Slope of a tangent line, or taking the approximation to Limit, when possible, to give a definition of the slope of a tangent. Saying how to compute or approximate a number or quantity defines.

So far the slope to a curve y = f(x) at a point (x₁,y₁) = (x₁,f(x₁)) has been physically or graphically associated with the slope of a short ski whose midpoint touches a smooth - not too bumpy - curve at the point (x₁,f(x₁)). The following diagram shows or suggests how the slope of such a ski resting on the curve at the point (x₁,y₁) could be approximated by the slope of a short chord joining (x₁,y₁) to a nearby second point (x₂,y₂) = (x₂,f(x₂)) on the curve.

h

The function $f(x)$ is assumed to be continuous at x₁ - without jumps or other discontinuity there.

Consider the following.

1. The chord or line segment joining the point $(x_1,y_1)$ to the point $(x_2,y_2)=(x_2,f(x_2)$ has slope \[m_{chord}=\frac{y_2-y_1}{x_2-x_1}=\frac{\Delta y}{\Delta x} \] and equation $y=m_{chord}(x-x_1)+ y_1$. When the ski travels between $x=x_1$ and $x=x_2$, its slope is (or should be) approximated by the slope $m_{chord}$ of the chord, alias line segment.

2. We suppose the point $(x_1,y_1)$ is fixed in place. In other words, suppose it is not moving. We further suppose the point $(x_2,y_2)= (x_2,f(x_2))$ moves along the curve $y=f(x)$ towards the point $(x_1,y_1)$. The slope $m$ of the line segment through these two points \em should \rm approach the slope $m_{ski} =f'(x_1)$ of a ski on the curve at $(x_1,y_1)$.

3. In the motion just described, as the point $(x_2,y_2)= (x_2,f(x_2))$ moves along the curve $y=f(x)$ towards the point $(x_1,y_1)$, the abscissa $x_2$ should move closer to $x_1$. The difference $\Delta x =x_2-x_1$ should also become closer and closer to zero. Thus we expect the approximation \[ m_{ski}\approx \frac{\Delta y }{\Delta x}= \frac{y_2-y_1}{x_2-x_1} =\frac{f(x_2)-f(x_1)}{x_2-x_1} \] to improve when $(x_2,y_2)= (x_2,f(x_2))$ approaches $(x_1,y_1)= (x_1,f(x_1))$ and/or as $x_2$ approaches $x_1$.

4. The continuity of $f(x)$ at $x_1$ implies the moving point $(x_2,y_2)= (x_2,f(x_2))$ will approach the non-moving, that is fixed point, $(x_1,y_1)= (x_1,f(x_1))$ when the abscissa $x_2$ approaches $x_1$ or equivalently, when $\Delta x =x_2-x_1$ approaches $0$.

5. Note the arrow $\to$ will be employed as shorthand for the phrase approaches or goes to.

If the graphical and physical expectations hold, then $m_{ski}=f'(x_1)$ should be the limiting value of $\frac{\Delta y }{\Delta x}$ as $\Delta x$ $\to$ 0. The better and better calculation of this limit should provide an arithmetic means for approximating the expected slope of the ski with greater and greater accuracy to an arbitrary number of decimal places. \em The limiting value of the segment slope should equal that of the ski. \rm This provides the computational definition and the mathematical one as well. See the next section.

Limit Definition of Slopes

The slope $m$ to a curve $y=f(x)$ at $x=x_1$ is defined by the limit calculation \[ f'(x_1)=\lim_{\mbox{ $\Delta x \to 0$}}\frac{\Delta y}{\Delta x} \] The right hand side of the above equation may be read as the limit as $\Delta x$ approaches zero of the ratio $\frac{\Delta y}{\Delta x}$.

Computation of right hand side $\lim_{ \Delta x \to 0 }\frac{\Delta y}{\Delta x}$ requires the existence of a number or quantity $L$ with the following property in the absence of units.

For each whole number $k > 0$, there exist an $n>0$ such that \[\left|L-\frac{\Delta y}{\Delta x} \right| \le \frac12\cdot 10^{-k}\] if $0<|\Delta x| \le \frac12\cdot 10^{-n}$.

The value $\Delta x = 0$ is excluded as division by zero is not allowed.

Note that $\Delta y = f(x_2)-f(x_1)$. In the presence of units, the preceding requirement becomes the following.

For each whole number $k > 0$, there exist an $n>0$ such that \[\left|L-\frac{\Delta y}{\Delta x} \right| \le \frac12\cdot 10^{-k}\cdot \frac{\mbox{units of $y$}}{\mbox{units of $x$}}\] if $0<|\Delta x| \le \frac12\cdot 10^{-n} \cdot \mbox{(units of x)}$.

The Physical Limit

In the case of a slope, our initial conception was that the slope of a short ski whose midpoint was placed at a point (x₁,f(x₁)) of a curve y = f(x), would give the slope of the curve there, at least approximately. Physically, there might be some error in the placement. The ski has to be short enough so that the slope of the curve y = f(x) does not change too much. The idea of a limit can be seen here in the requirement that the ski be short enough to lie on the curve y = f(x) and not crossing several bumps or oscillations in it.

Tangent Line Revisited

The above limit definition is motivated by the graphical expectation or suggestion that the slope of the line segment joining the non-moving point (x₁,y₁) to the moving point (x₂,y₂) should approach the slope of a tangent line at the non-moving point (x₁,y₁). The tangent line should be the limiting position of the line extending this chord. See the previous diagram.

Tangent Line Equation

When a skier is located on a curve $y=f(x)$ at $(x_1,y_1)=(x_1,f(x_1))$, the slope of his or her ski is assumed to lie on a tangent line. This tangent line has - or is now assumed to have - the equation \[ y=m_{tangent}(x-x_1)+y_1 \] where $m_{tangent}=f'(x_1)$ is given by a limit $L$ discussed above. The foregoing represents the mathematical definition of the tangent line to curve y = f(x) at (x₁,y₁) = (x₁,f(x₁)).

The linear function

Linear Approximation

Suppose for a given number $k > 0$, there exist an $n>0$ such that \[\left|L-\frac{\Delta y}{\Delta x} \right| \le \epsilon = \frac12\cdot 10^{-k}\] whenever $|\Delta x| \le \delta=\frac12\cdot 10^{-n}$, then the following holds whenever the inequality $|x_2-x_1|=|\Delta x| \le \frac12\cdot 10^{-n}$ is satisfied.

1. The difference $L-\frac{\Delta y}{\Delta x}=c$ for some number $c$ with magnitude $|c|\le\epsilon= \frac12\cdot 10^{-k}$. The number $c$ will depend on $x_2$.

2. The foregoing implies \[\frac{\Delta y}{\Delta x} =L-c\] and hence that \[{\Delta y} =L{\Delta x}- c{\Delta x}\]

3. The latter in turn implies \[ f(x_2)-y_1=y_2-y_1 =L\Delta x- c{\Delta x}\] and \[ f(x_2)-y_1=L(\Delta x)- c{\Delta x}\] and hence \[ f(x_2)=y_1+L(\Delta x)+\mbox{an error}\] where the error is $-c{\Delta x}$ and its magnitude \[|c\cdot \Delta x| \le |\Delta x| \cdot \frac12\cdot 10^{-k} \le\frac12\cdot 10^{-n} \cdot \frac12\cdot 10^{-k}\]

Theorem: [Consequences of a Non-Zero Slope]

If the slope $m=f'(x_1)=L$ of $f(x)$ at $x=x_1$ is nonzero, then there exist a $\delta>0$ such that the sign of $f(x)-f(x_1)$ equals the sign of $L\cdot(x-x_1)$ whenever $|x-x_1| \le \delta$.

Proof: In the previous discussion, choose $k$ such $\frac1210^{-k} < |L|$ and let $\delta=\frac1210^{-n}$.

This theorem implies if $m=f'(x_1)\ne0$ then no interior maximum nor minimum can occur at $x=x_1$. Finding all solutions $x=a$ of the equation $f'(x)=0$ identifies locations $x=a$ at which interior maximums and minimums might be found. The latter can also occur at points where the slope or derivative $f'(x)$ is not defined. The points $x$ where

1. $f(x)$ is undefined, and

2. $f'(x)$ is zero or undefined

1. at critical points inside that interval, and/or

2. at included endpoints.

Rules For Differentiation

A calculus course may ask for the computation of slope $m_{ski}=L$ by evaluating the limit directly. The rules and properties of limits suggest how, at least in the simpler cases. Then the course may introduce rules for differentiation. These rules for differentiation are based on or justified by the rules and properties of limits. Differentiation rules say how to compute formulas for $f'(x_1)$ in a routine mechanical manner from formulas for $f(x)$, at least when the formula for $f(x)$ is simple enough. The proof, justification and further explanation of rules for differentiation may be found in a calculus course or book.

Algebraic Evaluation of Limits

• [Play Video] 4.5 minutes: Algebraic View of Limits. Example involving sums and quotients.
• [Play Video] 2.5 minutes: Derivative of a Linear Expression cx+d via Limits.
• [Play Video] 2.3 minutes: Derivative as a Limit of a Quotient. First pass at finding the derivative or slope of f(x) = x². Algebraic View.
• [Play Video] 2.25 minutes: Second pass at finding the derivative or slope of f(x) = x² at two values of x. Numerical Examples of Limit Evaluation to suggest a pattern.
• [Play Video] 3.75 minutes: Third pass at finding the derivative or slope of f(x) = x². Back to the algebraic view and a conclusion.

Remark: Many students survive high school math courses without mastering the algebraic way of writing and thinking. But mastery of the latter is necessary for comprehension of the algebraic computations and reasoning in calculus. I have known students, who obtained excellent marks in high school mathematics, to say that the algebraic way of writing and thinking was strange to them before taking calculus. The assumption that students have mastered the algebraic thought cannot yet be made in a calculus course. Thus examples to help with this mastery may be needed.

Example 1. Let $x=2$. Then with $x_1=x=2$ and $x_2=x_1+\delta x =x +\delta x$, we have \begin{eqnarray*} \Delta y &=&f(x_2) -f(x_1) \\ &=&f(x+\Delta x) -f(x) \\ &=&f(2+\Delta x)- f(2) \\ &=&(2+\Delta x)^2 -2^2 \\ &=&(2^2+2(2)\Delta x+(\Delta x)^2) -2^2 \\ &=&2(2)\Delta x+(\Delta x)^2 \end{eqnarray*} Of course $2(2)=4$, but for the sake of pattern recognition and emphasis, we keep the arithmetic expression $2(2)$ to the end of the calculation. Now \begin{eqnarray*} \mbox{A}&=&\frac{2(2)\Delta x+(\Delta x)^2)}{\Delta x} \\ &=&2(2)+(\Delta x) \end{eqnarray*} This implies \[ \lim_{\Delta x \to 0}\mbox{A} =\lim_{\Delta x \to 0} 2(2)+(\Delta x)= 2(2) =4 \] Example 2. Let $x=3$. Then \begin{eqnarray*} \Delta y &=&f(x+\Delta x) -f(x) \\ &=&f(3+\Delta x)- f(3) \\ &=&(3+\Delta x)^2 -3^2 \\ &=&(3^2+2(3)\Delta x+(\Delta x)^2) -3^2 \\ &=&2(3)\Delta x+(\Delta x)^2 \end{eqnarray*} Therefore \begin{eqnarray*} \mbox{A}&=&\frac{2(3)\Delta x+(\Delta x)^2)}{\Delta x} \\ &=&2(3)+(\Delta x) \end{eqnarray*} This implies \[ \lim_{\Delta x \to 0}\mbox{A} =\lim_{\Delta x \to 0}2(3)+(\Delta x)= 2(3) =6 \] Example 3. Let $x=5$. Then \begin{eqnarray*} \Delta y &=&f(x+\Delta x) -f(x) \\ &=&f(5+\Delta x)- f(5) \\ &=&(5+\Delta x)^2 -5^2 \\ &=&(5^2+2(5)\Delta x+(\Delta x)^2) -5^2 \\ &=&2(5)\Delta x+(\Delta x)^2 \end{eqnarray*} Therefore we expect \begin{eqnarray*} \mbox{A}&=&\frac{2(5)\Delta x+(\Delta x)^2)}{\Delta x} \\ &=&2(5)+(\Delta x) \end{eqnarray*} The last equality suggests that \[ \lim_{\Delta x \to 0}\mbox{A} =\lim_{\Delta x \to 0}2(5)+(\Delta x) = 2(5) =10 \]

[Play Video] 2.25 minutes: Derivative as a Limit of a Quotient. First pass at finding the derivative or slope of f(x) = x². Algebraic View.

[Play Video] 2.25 minutes: Second pass at finding the derivative or slope of f(x) = x² at two values of x. Numerical Examples of Limit Evaluation to suggest a pattern.

[Play Video] 3.75 minutes: Third pass at finding the derivative or slope of f(x) = x². Back to the algebraic view and a conclusion.

The Common Algebraic Pattern

The three examples follow the same pattern. We will rewrite the above calculations with the letter $a$ replacing the numbers $2$, $3$ and/or $5$ above, to emphasize the pattern. In the rewrite below, note that the role of $a$ below could be played or assumed by each of the numbers $2$, $3$ or $5$ above, another number or another letter!

Example n. Let $x=a$. Then as before \begin{eqnarray*} \Delta y &=&f(x+\Delta x) -f(x) \\ &=&f(a+\Delta x)- f(a) \\ &=&(a+\Delta x)^2 -a^2 \\ &=&(a^2+2a\Delta x+(\Delta x)^2) -a^2 \\ &=&2a\Delta x+(\Delta x)^2 \end{eqnarray*} Therefore \begin{eqnarray*} \mbox{A}&=&\frac{2a\Delta x+(\Delta x)^2)}{\Delta x} \\ &=&2(a)+(\Delta x) \end{eqnarray*} This implies \[ \lim_{\Delta x \to 0}\mbox{A}=\lim_{\Delta x \to 0}2(a)+(\Delta x) = 2a \] Note that in the limit calculation, the variable $a$ is held constant while $\Delta x \to 0$.

Now we can replace $a$ in the above pattern by $x$. This yields \[ f'(x)=\lim_{\Delta x \to 0}\mbox{A} = \cdots = \lim_{\Delta x\to0} 2x+(\Delta x) =2x \] The $\cdots$ indicates reasoning similar or identical to that has gone before.

Remark - technical. The ratio $\mbox{A}=\frac{f(x_1+\Delta x)-f(x_1)}{\Delta x}$ is not defined at $\Delta x =0$ as division by zero is not defined. But the algebraic manipulations above shows that $\lim_{\Delta x \to 0} $A does exist - at least for the simple case treated.

[Play Video] 2.25 minutes: Derivative of x³ algebraically via Limits.

Recapitulation

In the absence of units, the slope of the ski could be the finite number L with the following arithmetic property:

For each whole number $k > 0$, there exist an $n>0$ such that \[\left|L-\frac{\Delta y}{\Delta x} \right| \le \frac12\cdot 10^{-k}\] IF $|\Delta x| \le \frac12\cdot 10^{-n}$.

The above limit L, if such a finite limit exists, is taken as the definition of the slope to curve at or above the point x = x₁. This mathematical definition or convention provides the computational answer to the two questions:

• what is the slope of a curve $y=f(x)$ at a point $(x_1,y_1)= (x_1,f(x_1))$?

• what is the slope of a curve $y=f(x)$ at the point on it determined by x=x₁$?

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
3. Prime Factors - quickly
4. Fractions + Ratios
5. Arith with units - science