When you first learned about slope you learned the mnemonic device “rise over run” to help you remember that to calculate the slope between two points you use the following formula:

\begin{align*}m=\frac{y_2-y_1}{x_2-x_1}\end{align*}

In Calculus, you will learn that for curved functions, it makes more sense to discuss the slope at one precise point rather than between two points. The slope at one point is called the slope of the tangent line and the slope between two separate points is called a secant line.

Consider a car driving down the highway and think about its speed. You are probably thinking about speed in terms of going a given distance in a given amount of time. The units could be miles per hour or feet per second, but the units always have time in the denominator. What happens when you consider the instantaneous speed of the car at one instant of time? Wouldn’t the denominator be zero?

### Instantaneous Rate of Change

The slope at a point \begin{align*}P\end{align*}**instantaneous rate of change** at that point. The slope at a point \begin{align*}P\end{align*}**secant line** is a line that passes through two distinct points on a function. A **tangent line** to a function at a given point is the straight line that just touches the curve at that point.

Because you are interested in the slope as the “run” approaches zero, this is a limit question. One of the main reasons that you study limits in calculus is so that you can determine the slope of a curve at a point or the slope of the tangent line.

For a graph of lines, it is easy to estimate the slopes of the tangent lines since the slope of the tangent is the same as the slope of the line. Estimate the slope of the following function at -3, -2, -1, 0, 1, 2, 3. Organize the slopes in a table.

By mentally drawing a tangent line at the following \begin{align*}x\end{align*}

\begin{align*}x\end{align*} |
slope |

-3 | 0 |

-2 | 0 |

-1 | -1 |

0 | -1 |

1 | 2 |

2 | 0 |

3 | 0 |

If you graph these points you will produce a graph of what’s known as the derivative of the original function. A **derivative** is a function of the slopes of the original function.

### Examples

#### Example 1

Earlier, you were asked about the instantaneous speed of a car. If you write the ratio of distance to time and use limit notation to allow time to go to zero you do seem to get a zero in the denominator.

\begin{align*}\lim \limits_{time \to 0}\left(\frac{distance}{time}\right)\end{align*}

The great thing about limits is that you have learned techniques for finding a limit even when the denominator goes to zero. Instantaneous speed for a car essentially means the number that the speedometer reads at that precise moment in time. You are no longer restricted to finding slope from two separate points.

#### Example 2

Estimate the slope of the function\begin{align*}f(x)=\sqrt{x}\end{align*} at the point \begin{align*}(1,1)\end{align*} by calculating 4 successively close secant lines. If you had to guess what the slope was at the point \begin{align*}(1,1)\end{align*} what would you guess the slope to be?

Calculate the slope between \begin{align*}(1,1)\end{align*} and 4 other points on the curve:

- The slope of the line between \begin{align*}\left(5,\sqrt{5}\right)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{\sqrt{5}-1}{5-1} \approx 0.309\end{align*}
- The slope of the line between \begin{align*}(4,\ 2)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_2=\frac{2-1}{4-1}\approx 0.333\end{align*}
- The slope of the line between \begin{align*}\left(3,\sqrt{3}\right)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_3=\frac{\sqrt{3}-1}{3-1} \approx 0.366\end{align*}
- The slope of the line between \begin{align*}\left(2,\sqrt{2}\right)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_4=\frac{\sqrt{2}-1}{2-1} \approx 0.414\end{align*}

Notice that the pattern in the previous example is leading up to \begin{align*}\frac{\sqrt{1}-1}{1-1}\end{align*} . Unfortunately, this cannot be computed directly because there is a zero in the denominator. Luckily, you know how to evaluate using limits.

To determine what the slope would be at the point \begin{align*}(1,1)\end{align*}, you would evaluate the limit \begin{align*}\lim \limits_{x \to 1} \left(\frac{\sqrt{x}-1}{x-1}\right)\end{align*}

\begin{align*}m&=\lim \limits_{x \to 1} \left(\frac{\left(\sqrt{x}-1\right)}{\left(x-1\right)} \cdot \frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\right)\\&=\lim \limits_{x \to 1}\left(\frac{\left(x-1\right)}{\left(x-1\right) \left(\sqrt{x}+1\right)}\right)\\&=\lim \limits_{x \to 1} \left(\frac{1}{\left(\sqrt{x}+1\right)}\right)\\&=\frac{1}{\sqrt{1}+1}\\&=\frac{1}{2}\\&=0.5\end{align*}

The slope of the function \begin{align*}f(x)=\sqrt{x}\end{align*} at the point \begin{align*}(1,1)\end{align*} is exactly \begin{align*}m=\frac{1}{2}\end{align*}.

#### Example 3

Sketch a complete cycle of a sine graph. Estimate the slopes at \begin{align*}0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\end{align*}.

\begin{align*}x\end{align*} | Slope |

0 | 1 |

\begin{align*}\frac{\pi}{2}\end{align*} | 0 |

\begin{align*}\pi\end{align*} | -1 |

\begin{align*}\frac{3\pi}{2}\end{align*} | 0 |

\begin{align*}2\pi\end{align*} | 1 |

You should notice that these are the exact values of cosine evaluated at those points.

#### Example 4

Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the distance vs. time graph and the rate vs. time graph.

Distance vs. Time:

Rate vs. Time: (this is the graph of the derivative of the original function shown above)

#### Example 5

Approximate the slope of \begin{align*}y=x^3\end{align*} at \begin{align*}(1,1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?

- The slope of the line between \begin{align*}(0.7,0.7^3)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{0.7^3-1}{0.7-1} \approx 2.19\end{align*}
- The slope of the line between \begin{align*}(0.8,0.8^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_2=\frac{0.8^3-1}{0.8-1} \approx 2.44\end{align*}
- The slope of the line between \begin{align*}(0.9,0.9^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_3=\frac{0.9^3-1}{0.9-1} \approx 2.71\end{align*}
- The slope of the line between \begin{align*}(0.95,0.95^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{0.95^3-1}{0.95-1} \approx 2.8525\end{align*}
- The slope of the line between \begin{align*} (0.975,0.975^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{0.975^3-1}{0.975-1} \approx 2.925625\end{align*}

The slope at \begin{align*}(1,1)\end{align*} will be slightly greater than the estimates because of the way the slope curves. The slope at \begin{align*}(1,1)\end{align*} appears to be about 3.

### Review

1. Approximate the slope of \begin{align*}y=x^2\end{align*} at \begin{align*}(1,1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?

2. Evaluate the following limit and explain how it confirms your answer to #1.

\begin{align*}\lim \limits_{x \to 1} \left(\frac{x^2-1}{x-1}\right)\end{align*}

3. Approximate the slope of \begin{align*}y=3x^2+1\end{align*} at \begin{align*} (1,4)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?

4. Evaluate the following limit and explain how it confirms your answer to #3.

\begin{align*}\lim \limits_{x \to 1} \left(\frac{3x^2+1-4}{x-1}\right)\end{align*}

5. Approximate the slope of \begin{align*}y=x^3-2\end{align*} at \begin{align*}(1,-1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?

6. Evaluate the following limit and explain how it confirms your answer to #5.

\begin{align*}\lim \limits_{x \to 1}\left(\frac{x^3-2-(-1)}{x-1}\right)\end{align*}

7. Approximate the slope of \begin{align*}y=2x^3-1\end{align*} at \begin{align*}(1,1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?

8. What limit could you evaluate to confirm your answer to #7?

9. Sketch a complete cycle of a cosine graph. Estimate the slopes at \begin{align*}0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\end{align*}.

10. How do the slopes found in the previous question relate to the sine function? What function do you think is the derivative of the cosine function?

11. Sketch the line \begin{align*}y=2x+1\end{align*}. What is the slope at each point on this line? What is the derivative of this function?

12. Logan travels by bike at 30 mph for 2 hours. Then she gets in a car and drives 65 mph for 3 hours. Sketch both the distance vs. time graph and the rate vs. time graph.

13. Explain what a tangent line is and how it relates to derivatives.

14. Why is finding the slope of a tangent line for a point on a function the same as the instantaneous rate of change at that point?

15. What do limits have to do with finding the slopes of tangent lines?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 14.8.