Skip to main content

Section 2.1 Rates of Change and Tangents to Curves

Definition 2.1.1. Average rate of change.

The average rate of change of \(y=f(x)\) with respect to \(x\) over the interval \([x_1,x_2]\) is:

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(x_2)-f(x_1)}{x_2-x_1} \end{equation*}

Example 2.1.2. Average rate of change.

An object in free fall is described by the equation \(y=16t^2\text{,}\) where \(t\) is the time, in seconds, and \(y\) is the distance the object has fallen, in feet.

An object initially at rest falls to earth. What is the average speed of the object:

  1. over the interval from 1 second to 3 seconds?

  2. over the interval from 1 second to 1.5 seconds?

Solution.

  1. \(\displaystyle \displaystyle\frac{16\cdot 3^2 - 16\cdot 1^2}{3-1} = \frac{128}{2} = 64\)

  2. \(\displaystyle \displaystyle\frac{16\cdot (1.5)^2 - 16\cdot 1^2}{1.5-1} = \frac{20}{0.5} = 40\)

Example 2.1.3.

Find the average rate of change of the function \(g(x) = \frac{x+3}{x-1}\) over the interval \([2,5]\text{.}\)

Solution.

\(\displaystyle\frac{g(5)-g(2)}{5-2} = \frac{2-5}{5-2} = \frac{-3}{3} = -1\)

Definition 2.1.4. Secant.

The line joining \((x_1,f(x_1))\) and \((x_2,f(x_2))\) is called a secant line, and its slope is equal to the average rate of change on the interval \([x_1,x_2]\text{.}\)

Subsection 2.1.1 Instantaneous rate of change and the tangent

Goal: find the speed of the object at \(t=1\) second.

Shortening intervals: https://www.desmos.com/calculator/c9jmprwzjl 2 

Generally, we're calculating the average rate of change over \([1,1+h]\text{,}\) which produces:

\begin{equation*} \frac{\Delta y}{\Delta t} = \frac{16(1+h)^2-16(1)^2}{(1-h)-1} \end{equation*}

We can't evaluate at \(h= 0\) because that would be dividing by 0, but we can simplify:

\begin{equation*} \frac{16(1+h)^2-16(1)^2}{(1-h)-1} = \cdots = 32+16h \end{equation*}

Note that we still can't evaluate at \(h=0\text{,}\) because the average rate of change, as a function of \(h\) does not include 0 in the domain.

But we can choose \(h\) to make \(16h\) as small as we want, so we say that the limit of the average rate of change is 32 and call that the instantaneous rate of change.

Definition 2.1.6. Instantaneous rate of change.

The instantaneous rate of change of a function \(y=f(x)\) at \(x_0\) is found by simplifying

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(x_0+h)-f(x_0)}{h} \end{equation*}

Provided that \(h\) can be eliminated from the denominator, the instantaneous rate of change is the value obtained by substituting 0 for \(h\) in the simplified expression.

The tangent line to the curve \(y=f(x)\) at \(x_0\) is the line through \((x_0,f(x_0))\) whose slope is the instantaneous rate of change, which is also referred to as the slope of the tangent line.

Example 2.1.7. Equation of the tangent line.

  1. Find the slope of the curve \(y=2x^2-3x+1\) at the point \(P(5,36)\) by finding the limit of the secant slopes through \(P\text{.}\)

  2. Find an equation of the tangent line to \(y=2x^2-3x+1\) at \(P(5,36)\text{.}\)

Solution.

To find the slope at the point \(P(5,36)\text{,}\) we evaluate

\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(x_0+h)-f(x_0)}{h} \end{equation*}

for the function \(f(x)=2x^2-3x+1\text{,}\) with \(x_0=5\) and \(f(x_0)=f(5)=36\) and simplify.

\begin{align*} \frac{\Delta y}{\Delta x} \amp = \frac{f(x_0+h)-f(x_0)}{h}\\ \amp = \frac{(2(5+h)^2 - 3(5+h)+1)-36}{h}\\ \amp = \frac{2(25+10h+h^2)-3(5+h)+1-36}{h}\\ \amp = \frac{50+20h+2h^2-15-3h+1-36}{h}\\ \amp = \frac{2h^2 +17h}{h}\\ \amp = \frac{h(2h+17)}{h}\\ \amp = 2h+17 \end{align*}

Now we evaluate with \(h=0\) to obtain \(17\) as the slope of the tangent line.

With the slope determined, the equation of the tangent line is the line with slope \(17\) through the point \(P(5,36)\text{.}\) Using the point-slope form of the line, we have:

\begin{align*} y-y_0 \amp = m(x-x_0)\\ y-36 \amp = 17(x-5)\\ y \amp = 17x -49 \end{align*}
www.desmos.com/calculator/u9ntlagzlp
www.desmos.com/calculator/c9jmprwzjl