Equation of the Tangent Line | Tangent Line Approximation |

Equation of the Normal Line | Rates of Change and Velocity/Acceleration |

Horizontal and Vertical Tangent Lines | More Practice |

Note that we visited **Equation of a Tangent Line **here in the **Definition of the Derivative** section. Also, there are some Tangent Line Equation problems using the Chain Rule here in **The Chain Rule** section.

We will talk about the **Equation of a Tangent Line** with** Implicit Differentiation** here in the **Implicit Differentiation and Related Rates** section.

## Equation of the Tangent Line

Let’s revisit the equation of a** tangent line**, which is a line that touches a curve at a point but doesn’t go through it near that point. The **slope** of the tangent line at this **point of tangency**, say “$ a$”, is the** instantaneous rate of change** at $ x=a$ (which we can get by taking the **Derivative** **of the curve**, as shown in the **Basic Differentiation Rules** section, and plugging in “$ a$” for “$ x$”). Since now we have the **slope** of this line, and also the coordinates of a **point** on the line, we can get the whole equation of this tangent line. We sometimes see this written as $ \displaystyle \frac{{dy}}{{dx}}\left| {_{{x=a}}} \right.$.

When we get the derivative of a function, we’ll use the $ x$-value of the point given to get the actual slope at that point. Then we’ll use the $ y$-value of the point to get the complete line, using either the **Point-Slope** $ (y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right))$ or **Slope-Intercept** $ (y=mx+b)$ method. I like to use point-slope. It’s really not too bad! (Weird fact: the equation of a tangent line for a linear function is just that function!)

Here are some examples. Note that in the last problem, we are given a **line parallel to the tangent line**, so we need to work backwards to find the point of tangency, and then find the equation of the tangent line.

Here are other types of problems you might see:

Here is one more; in this problem, we’ll find the **equation of a parabola** that goes through a certain point and is tangent to a line at another point.

## Equation of the Normal Line

The **normal line** to a curve at a point is the line through that point that is **perpendicular to the tangent**. Remember that a line is perpendicular to another line if their slopes are **opposite reciprocals** of each other; for example, if one slope is $ 4$, the other slope would be $ \displaystyle -\frac{1}{4}$.

We do this problem the same way, but use the opposite reciprocal of the slope when we go to get the line; here is an example:

## Horizontal and Vertical Tangent Lines

Sometimes we want to know at what point(s) a function has either a **horizontal **or **vertical** tangent line (if they exist). For a **horizontal tangent** line ($ 0$ **slope**), we want to get the **derivative**, and set it to $ 0$, which means setting the **numerator to **$ \boldsymbol {0}$. Then we can get the $ x$-value, and use the original function to get the $ y$-value; thus, we have the point(s) at the **horizontal tangent line**. We also have to make sure that that the denominator isn’t $ 0$ at these points.

For a **vertical tangent** line (**undefined slope**), we want to get the **derivative**, and set the bottom or **denominator to** $ \boldsymbol {0}$. Then we can get the $ x$-value, and use the original function to get the $ y$-value; thus, we have the point(s) at the **vertical tangent line**. We also to have to make sure the numerator isn’t $ 0$ at these points. **Note that where functions have vertical tangent lines, they are not differentiable at that point. **

## Tangent Line Approximation

Also known as **local linearization** and **linear approximation**, it turns out that we can **use a tangent line to approximate the value or graph of a function**. The reason they want you to learn this stuff is so you can “appreciate the math” and how they used to use Calculus many years ago before fancy calculators and computers. And, actually, it’s pretty neat! (Note that we will show **linear approximation** using **differentials** here in the **Differentials, Linear Approximation and Error Propagation** section.)

Let’s say we have a function, such as $ f\left( x \right)=\sqrt{x}$, and we want to know what the value of this function is ($ f\left( x \right)$), when $ x=4.023$. We know it’s close to $ 2$, but we can get even closer without a calculator using Calculus. We can do this by using the point-slope formula $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, where $ \left( {{{x}_{1}},\,\,{{y}_{1}}} \right)$ is a point that we can get easily, like $ \left( {4,2} \right)$, and $ m$* *is the derivative of the function at that point. We can then get our new $ y$-value, by plugging the original $ x$-value we have ($ 4.03$).

Here is this problem, along with some others; use **linear approximation** (local linearization, or tangent line approximation) to find an approximation for the indicated values:

## Rates of Change and Velocity/Acceleration

Note that there are more **position**, **velocity**, and **acceleration** problems here in the **Antiderivatives and Indefinite Integration** section. and in the **Integration as Accumulated Change** section.

The derivative has many applications in “real life”; one of the most useful is to find the **rate of change** of one variable with respect to another. Think of a **rate of change**, or sometimes called an **instantaneous rate of change** as how fast something is changing at a certain point, like a point in time. For example, if a particle is moving along a horizontal line, it’s **position**, relative to say the origin, is a function, but the **derivative** of this function is its **instantaneous** **velocity** (how fast it’s moving) at a certain point, and the **derivative of its velocity** is its **acceleration** (how fast its velocity is changing). Note that the acceleration function is a **Higher Order Derivative, **since we need to take the derivative of the position function twice to get the acceleration function. You can also think of the instantaneous velocity as the average velocity as the difference in $ x$’s (for example, change in time) gets closer and closer to $ 0$; thus, we are taking a limit, or a derivative.

On the other hand, the **average rate of change** or **average velocity** is $ \displaystyle \frac{{\text{change in position}}}{{\text{change in time}}}=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$ from interval $ a$ to $ b$, and you can think of this as $ \displaystyle \text{rate}=\frac{{\text{distance}}}{{\text{time}}}$. These problems typically involve a **time interval**, as opposed to an instantaneous point. Note that we don’t need calculus to find this – just some algebra!

Again, typically when we are finding a **rate of change**, or **instantaneous rate of change (“IROC”)**, we will be taking a **derivative** at a certain point, and when we are finding an **average rate of change (“AROC”)**, we will be using the $ \displaystyle \frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$ **(slope) formula **over an interval.

Here is a graph to show this; note how the instantaneous change is the **slope of a tangent line** at a point:

For these problems, find the **average rate of change** over the interval. Compare this to the **instantaneous rate of change at the endpoints** of the interval:

When we’re talking about an object traveling with respect to a time ($ t$) with a** position **$ s\left( t \right)$, **velocity **$ v\left( t \right)$, and **acceleration **$ a\left( t \right)$, here are some concepts. If you’ve taken Physics, you’ve probably dealt with this stuff**: **

**Initially**is when $ t=0$,**at the origin**is when $ s(t)=0$, and**at rest**is when $ v(t)=0$.- If a particle is moving along a horizontal line, if its
**position**$ s(t)$ (relative to say the origin) is represented by a function, the**derivative**of this function is its**instantaneous****velocity**$ v(t)$ (how fast it’s moving), and the**derivative of its velocity**is its**acceleration**$ a(t)$ (how fast its velocity is changing). - The
**position**of an object is a**vector**, since it has both a magnitude (a scalar, such as distance) and a direction. A change in position is the**displacement vector**, which is how far out of place the object is, compared to where it started. The**distance**it has traveled is the total amount of ground an object has covered during its motion, a scalar, and is the**absolute value of the displacement vector**. - The
**velocity**function is the**derivative**of the position function, and can be negative, zero, or positive. If the**derivative**(velocity) is**positive**, the object is**moving to the right**(or up, if that’s how the coordinate system is defined); if**negative**, it’s**moving to the left**(or down); if the**velocity is 0**, the**object is at rest**. Thus, if the**velocity changes sign**, the**object changes direction**. - The
**velocity**of an object is actually a**vector**, whereas the**speed**is the**absolute value of the velocity vector**, and is a scalar. The**speed**of an object cannot be negative, whereas velocity can. **Total distance****traveled**is the sum of the absolute values of the differences in positions of all the resting points.**Average Velocity**is**total****displacement**during that interval over**total time elapsed**.**Acceleration**(the**derivative of velocity**, which is also a vector) can cause speed to increase, decrease, or stay the same.- More information about
**speed**(remember that**speed**is the**absolute value of velocity**):- If
**velocity**and**acceleration**have the**same sign**, speed is**increasing**. (Example: a velocity from**40**mph to**45**mph, both positive, or from**–40**mph to**–45**mph, both negative). - If
**velocity**and**acceleration**have**opposite signs**, speed is**decreasing**. (Example: a velocity from**45**mph to**40**mph, velocity positive, acceleration negative, or from**–45**mph to**–40**mph, velocity negative, acceleration positive).

- If
**Farthest left**(or down) means minimum and**farthest right**(or up) means maximum.- To find the
**maximum or minimum velocity or acceleration**, find where the derivative of these functions fails or is zero. - The
**position of free-falling objects**is $ \displaystyle s\left( t \right)=\frac{1}{2}g{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$, where $ {{v}_{0}}$ is the initial velocity, $ {{s}_{0}}$ is the initial height, and $ g$ is the acceleration from gravity. On earth, the acceleration due to gravity is about**–32**feet per second per second or**–9.8**meters per second per second. In algebra, we saw this equation as $ g\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$, since we typically used feet per second. When the object hits the ground, $ s\left( t \right)=0$ (remember to always count the position**from the ground up**).

Here are some problems using the **position**, **velocity**, and **acceleration** functions:

Here’s one more problem, where we have to distinguish between the **instantaneous velocity** and the **average velocity**:

**Learn these rules, and practice, practice, practice!**

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the **Mathway** site, where you can register for the **full version** (steps included) of the software. You can even get math worksheets.

You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to **The Chain Rule**** **– you are ready!