Differentiation, derivatives and Taylor approximations: Tangent line
Difference quotient at a point
Using the difference quotient, we can approximate the change at one point of a graph. All we have to do is fix the left point of the interval \([a,b]\) and make the length \({\vartriangle}t\) of the interval smaller and smaller.
For a smooth function \(f\) the difference quotient will then be close to a certain number, say \(m\). We call this number \(m\) the slope of the graph at the point \(\bigl(a,f(a)\bigr)\). As we will see later, this number \(m\) is the slope of the tangent to the graph of \(f\) at the point \(\bigl(a,f(a)\bigr)\). Smoothness of the function at a point \(\bigl(a,f(a)\bigr)\) is worded in mathematics as differentiability at a point, i.e. the existence of the limit \(\displaystyle \lim_{{\vartriangle}\to 0}\dfrac{f(a+{\vartriangle}t)-f(a)}{{\vartriangle}t}\).
For the quadratic polynomial function \(f(t)=3t^2-2t-1\) we calculate the difference quotient at the intervals \([1,1.1]\), \([1,1.01]\), \([1,1.001]\) and \([1,1.0001]\)
\[\begin{aligned}
{\vartriangle}t &= 0.1 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.1)-f(1)}{0.1}=\frac{0.433-0}{0.1}=4.3\\ \\{\vartriangle}t &= 0.01 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.01)-f(1)}{0.01}=\frac{0.0403-0}{0.01}=4.03\\ \\
{\vartriangle}t &= 0.001 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.001)-f(1)}{0.001}=\frac{0.004003-0}{0.001}=4.003\\ \\
{\vartriangle}t &= 0.0001 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.0001)-f(1)}{0.0001}=\frac{0.00040003-0}{0.0001}=4.0003
\end{aligned}\]
You now suspect that \(\displaystyle \frac{{\vartriangle}f}{{\vartriangle}t}\to 4\) on the interval \(1,1+{\vartriangle}t\) as \({\vartriangle}t\to 0\). We will verify this later.
You can also explore this in the interactive example below for the quadratic function.
The interactive diagram on the right-hand side shows the graph of the function \(f(t)= t^2\) with two points on the graph. The bottom point \(A=\bigl(t_A,f(t_A)\bigr)\) on the graph is free to move by dragging the magenta coloured point on the horizontal axis and the top point \(B=\bigl(t_B,f(t_B)\bigr)\) on the graph is obtained by \(t_B=t_A+{\vartriangle}t\) for a horizontal change \({\vartriangle}t\). The horizontal change \({\vartriangle}t\) is therefore: \[{\vartriangle}t =t_B-t_A\] The corresponding vertical change is: \[\begin{aligned} {\vartriangle}f&=f(t_B) -f(t_A)\\[0.2cm] &= f(t_A+{\vartriangle}t)-f(t_A)\end{aligned}\] The average rate of change \(\frac{{\vartriangle}f}{{\vartriangle}t}\) over the interval \([t_A,t_B]\) in this example does depend on the choice of the point \(A\) and the interval (or if you prefer the horizontal change \({\vartriangle}t\)); just move the slider or the point \(A\) to observe this. As you go a smaller horizontal change \({\vartriangle}t\) with a fixed choice of the point \(A\) the average rate of change \(\frac{{\vartriangle}f}{{\vartriangle}t}\) goes towards a constant.
