This section covers:

**Equation of the Tangent Line****Equation of the Normal Line****Horizontal and Vertical Tangent Lines****Tangent Line Approximation****Rates of Change and Velocity****More Practice**

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

Note that 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 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 \(\frac{{dy}}{{dx}}\left| {_{{x=a}}} \right.\).

So 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

**value of the point to get the complete line, using either the**

*y***point-slope**\((y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right))\) or

**slope-intercept**\((y=mx+b)\) method. I like to use point-slope.

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 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 \(-\frac{1}{4}\).

So we do this problem the same way, but use the opposite reciprocal of the slope when we go to get the line.

# 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**, set it to 0 (or set the **numerator to 0**), get the *x* value, and then use the original function to get the *y *value; we then have the point. (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**, set the bottom or **denominator to 0**, get the *x* value, and then use the original function to get the *y* value; we then have the point. (We also to have to make sure the numerator isn’t 0 at these points).

# 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.)

So 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* is 4.03. 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*, by plugging the original *x* we have (*x* = 4.03).

Let’s do 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

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

So 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 would be its **instantaneous** **velocity** (how fast it’s moving) at a certain point, and the **derivative of its velocity** would be 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.)

On the other hand, the **average rate of change** or **average velocity** in an example like this is \(\frac{{\text{change in position}}}{{\text{change in time}}}=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}\) from interval *a* to *b* (think \(\text{rate}=\frac{{\text{distance}}}{{\text{time}}}\)), and these problems typically involve a **time interval** (as opposed to an instantaneous point).

(You might 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).

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

Here is a graph to show this:

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 (*y*) with respect to a time (*x*) with a** position**, **velocity**, and **acceleration**, here are some concepts. If you’ve taken Physics, you’ve probably dealt with this stuff**: **

- If a particle is moving along a horizontal line, its
**position**(relative to say the origin) is a function, but the**derivative**of this function would be its (**instantaneous**)**velocity**(how fast it’s moving) at a certain point, and the**derivative of its velocity**would be its**acceleration**(how fast its velocity is changing). - The
**position**of an object is actually a**vector**, since it has both a magnitude (a scalar, such as distance) and a direction. A change in position is a**displacement**, 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 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**. This is also called the**direction**of the object. - 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. **Acceleration**(the**derivative of velocity**, which is also a vector) can cause speed to increase, decrease, or stay the same. Negative acceleration means slowing down (velocity decreasing) and positive acceleration means speeding up (velocity increasing).- (Probably too much information at this point, but here goes): If
**velocity****is****positive****and****decreasing**, then**speed is decreasing**(the size of the velocity is becoming smaller, say from 40 to 35). If**velocity****is****negative**and**acceleration is positive**, then**speed is decreasing**(velocity is becoming less negative, so the size of velocity (speed) is becoming less, say from –45 to –40). If**velocity is positive**and**speed is decreasing**, then**acceleration is negative**(the speed is decreasing so the we are getting less positive velocity, say from 40 to 35). If**speed is increasing**and**acceleration is negative**, then**velocity is negative and decreasing**(the size of velocity is increasing, but the only way a larger number becomes a smaller quantity is if the number is negative, like from –40 to –45). If**velocity is negative****and increasing**, then**speed is decreasing**(speed is increasing but is negative, like from –45 to –40, but the magnitude (size) of 40 is less than 45). (With these, it helps to remember that the**speed**is the**absolute value**of the**velocity**). - The
**position of free-falling objects**is \(s\left( t \right)=\frac{1}{2}g{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}\), 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!