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 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?

#### Watch This

http://www.youtube.com/watch?v=7CvLzpzGhJI Brightstorm: Definition of a Derivative

#### Guidance

The slope at a point \begin{align*}P\end{align*} (also called the slope of the tangent line) can be approximated by the slope of secant lines as the “run” of each secant line approaches zero.

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 (the slope of a tangent line).

**Example A**

Estimate the slope of the following function at -3, -2, -1, 0, 1, 2, 3. Organize the slopes in a table.

**Solution:** By mentally drawing a tangent line at the following \begin{align*}x\end{align*} values you can estimate the following slopes.

\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.

**Example B**

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.

**Solution:** 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*}

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?

**Example C**

Evaluate the following limit and explain its connection with Example B.

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

**Solution:** Notice that the pattern in the previous problem 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.

\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*}.

**Concept Problem Revisited**

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.

#### Vocabulary

A ** tangent line** to a function at a given point is the straight line that just touches the curve at that point. The slope of the tangent line is the same as the slope of the function at that point.

A ** secant line** is a line that passes through two distinct points on a function.

A ** derivative** is a function of the slopes of the original function.

#### Guided Practice

1. 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*}.

2. 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.

3. 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?

**Answers:**

1.

\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.

2. Distance vs. Time:

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

3.

- 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.

#### Practice

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?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 14.8.