CLP-1 Differential Calculus

In this example, we recall what is meant by the slope of the straight line

• We claim that if, as we walk along this straight line, our \(x\)–coordinate changes by an amount \(\De x\text{,}\) then our \(y\)–coordinate changes by exactly \(\De y = \tfrac{1}{2}\De x\text{.}\)

• For example, in the figure on the left below, we move from the point

  \begin{gather*} (x_0,y_0)=(1\,,\,2=\tfrac{1}{2}\times 1+\tfrac{3}{2}) \end{gather*}

  on the line to the point

  \begin{gather*} (x_1,y_1)=(5\,,\,4=\tfrac{1}{2}\times 5+\tfrac{3}{2}) \end{gather*}

  on the line. In this move our \(x\)–coordinate changes by

  \begin{gather*} \De x= 5-1=4 \end{gather*}

  and our \(y\)–coordinate changes by

  \begin{gather*} \De y=4-2=2 \end{gather*}

  which is indeed \(\tfrac{1}{2}\times 4=\tfrac{1}{2}\De x\text{,}\) as claimed.

  
• In general, when we move from the point
  \begin{align*} (x_0,y_0) &= (x_0, \tfrac{1}{2}x_0+\tfrac{3}{2}) \end{align*}
  on the line to the point
  \begin{align*} (x_1,y_1) &= (x_1, \tfrac{1}{2}x_1+\tfrac{3}{2}) \end{align*}
  on the line, our \(x\)–coordinate changes by
  \begin{gather*} \De x=x_1-x_0 \end{gather*}
  and our \(y\)–coordinate changes by
  \begin{align*} \De y&=y_1-y_0\\ &=\big[\tfrac{1}{2}x_1+\tfrac{3}{2}\big] -\big[\tfrac{1}{2}x_0+\tfrac{3}{2}\big]\\ &=\tfrac{1}{2}(x_1-x_0) \end{align*}
  which is indeed \(\tfrac{1}{2}\De x\text{,}\) as claimed.
• So, for the straight line \(y=\tfrac{1}{2}x+\tfrac{3}{2}\text{,}\) the ratio \(\tfrac{\De y}{\De x} =\tfrac{y_1-y_0}{x_1-x_0}\) always takes the value \(\frac{1}{2}\text{,}\) regardless of the choice of initial point \((x_0,y_0)\) and final point \((x_1,y_1)\text{.}\) This constant ratio is the slope of the line \(y=\tfrac{1}{2}x+\tfrac{3}{2}\text{.}\)

In this example, let us fix \((x_0,y_0)\) to be the point \((2,4)\) on the parabola \(y=x^2\text{.}\) Now let \((x_1, y_1) = (x_1, x_1^2)\) be some other point on the parabola; that is, a point with \(x_1 \neq x_0\text{.}\)

• Draw the straight line through \((x_0,y_0)\) and \((x_1,y_1)\) — this is a secant line and we saw these in Chapter 1 when we discussed tangent lines ²If you do not remember this, then please revisit the first couple of sections of Chapter 1..

• The following table gives the slope, \(\tfrac{y_1-y_0}{x_1-x_0}\text{,}\) of the secant line through \((x_0,y_0)=(2,4)\) and \((x_1,y_1)\text{,}\) for various different choices of \((x_1,y_1=x_1^2)\text{.}\)

  \(x_1\)	\(1\)	\(1.5\)	\(1.9\)	\(1.99\)	\(1.999\)	\(\circ\)	\(2.001\)	\(2.01\)	\(2.1\)	\(2.5\)	\(3\)
  \(y_1=x_1^2\)	\(1\)	\(2.25\)	\(3.61\)	\(3.9601\)	\(3.9960\)	\(\circ\)	\(4.0040\)	\(4.0401\)	\(4.41\)	\(6.25\)	\(9\)
  \(\tfrac{y_1-y_0}{x_1-x_0}\)	\(3\)	\(3.5\)	\(3.9\)	\(3.99\)	\(3.999\)	\(\circ\)	\(4.001\)	\(4.01\)	\(4.1\)	\(4.5\)	\(5\)

• So now we have a big table of numbers — what do we do with them? Well, there are messages we can take away from this table.

  • Different choices of \(x_1\) give different values for the slope, \(\tfrac{y_1-y_0}{x_1-x_0}\text{,}\) of the secant through \((x_0,y_0)\) and \((x_1,y_1)\text{.}\) This is illustrated in Figure 2.1.3 below — the slope of the secant through \((x_0,y_0)\) and \((x_1,y_1)\) is different from the slope of the secant through \((x_0,y_0)\) and \((x'_1,y'_1)\text{.}\)

    

    If the parabola were a straight line this would not be the case — the secant through any two different points on a line is always identical to the line itself and so always has exactly the same slope as the line itself, as is illustrated in Figure 2.1.4 below — the (yellow) secant through \((x_0,y_0)\) and \((x_1,y_1)\) lies exactly on top of the (red) line \(y=\tfrac{1}{2}x+\tfrac{3}{2}\text{.}\)

    

  • Now look at the columns of the table closer to the middle. As \(x_1\) gets closer and closer to \(x_0=2\text{,}\) the slope, \(\tfrac{y_1-y_0}{x_1-x_0}\text{,}\) of the secant through \((x_0,y_0)\) and \((x_1,y_1)\) appears to get closer and closer to the value \(4\text{.}\)

It is very easy to generalise what is happening in Example 2.1.2.

• Fix any point \((x_0,y_0)\) on the parabola \(y=x^2\text{.}\) If \((x_1,y_1)\) is any other point on the parabola \(y=x^2\text{,}\) then \(y_1=x_1^2\) and the slope of the secant through \((x_0,y_0)\) and \((x_1,y_1)\) is
  \begin{align*} \text{slope} &= \frac{y_1-y_0}{x_1-x_0} =\frac{x_1^2-x_0^2}{x_1-x_0} && \text{since }y=x^2\\ & =\frac{(x_1-x_0)(x_1+x_0)}{x_1-x_0} && \text{remember } a^2-b^2 = (a-b)(a+b)\\ & =x_1+x_0 \end{align*}
  You should check the values given in the table of Example 2.1.2 above to convince yourself that the slope \(\tfrac{y_1-y_0}{x_1-x_0}\) of the secant line really is \(x_0+x_1 = 2+x_1\) (since we set \(x_0=2)\text{.}\)

• Now as we move \(x_1\) closer and closer to \(x_0\text{,}\) the slope should move closer and closer to \(2x_0\text{.}\) Indeed if we compute the limit carefully — we now have the technology to do this — we see that in the limit as \(x_1 \to x_0\) the slope becomes \(2x_0\text{.}\) That is

  \begin{align*} \lim_{x_1 \to x_0} \frac{y_1-y_0}{x_1-x_0} &= \lim_{x_1 \to x_0} (x_1+x_0) \qquad \text{by the work we did just above}\\ &= 2x_0 \end{align*}

  Taking this limit gives us our first derivative. Of course we haven't yet given the definition of a derivative, so we perhaps wouldn't recognise it yet. We rectify this in the next section.

  

• So it is reasonable to say “as \(x_1\) approaches \(x_0\text{,}\) the secant through \((x_0,y_0)\) and \((x_1,y_1)\) approaches the tangent line to the parabola \(y=x^2\) at \((x_0,y_0)\)”. This is what we did back in Section 1.1.

  The figure above shows four different secants through \((x_0,y_0)\) for the curve \(y=x^2\text{.}\) The four hollow circles are four different choices of \((x_1,y_1)\text{.}\) As \((x_1,y_1)\) approaches \((x_0,y_0)\text{,}\) the corresponding secant does indeed approach the tangent to \(y=x^2\) at \((x_0,y_0)\text{,}\) which is the heavy (red) straight line in the figure.

  Using limits we determined the slope of the tangent line to \(y=x^2\) at \(x_0\) to be \(2x_0\text{.}\) Often we will be a little sloppy with our language and instead say “the slope of the parabola \(y=x^2\) at \((x_0,y_0)\) is \(2x_0\)” — where we really mean the slope of the line tangent to the parabola at \(x_0\text{.}\)