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You CAN Ace Calculus

17calculus > derivatives > tangent lines

### Calculus Main Topics

Derivatives

Derivative Applications

Optimization

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### Related Topics and Links

WikiBooks - Definition of Slope

Slope, Tangent Lines, Normal Lines and Linearization

Before studying this page, make sure you understand the difference between the slope and tangent. If you need to refresh your memory on equations of lines, you can find a quick review here.

### Difference Between Slope and Tangent

One thing I review in my class is knowing the difference between the slope and tangent. When you have a question that asks for one of these, read the question again carefully.

Slope
If the question is asking for the slope (or the slope of the tangent line), you just need to find m and your answer will be the derivative, or if they ask for the slope at a specific point, your answer will be number.

Tangent
If the question is asking for a tangent, they are asking you to find the equation of the tangent line. To do this, you need to find the slope first (by taking the derivative and evaluating it at the point), then plug the slope and the point into one of these line equations:

$$\begin{array}{rclcl} (y - y_0) & = & m(x - x_0) & --- & \text{Point-Slope Form} \\ y & = & mx + b & --- & \text{Slope-Intercept Form} \end{array}$$

Your final answer in this case is a linear equation. If you are in my class, I require your answer to be in slope-intercept form. Check the question carefully to see what form your teacher asks for.

Another twist to this is that not all teachers are consistent. Sometimes they will say 'tangent' and mean 'slope'. You will need to get to know how your teacher uses the term or ask them to clarify what they mean. However, if they say 'tangent line' (without saying 'the slope of'), this always means the equation of the tangent line.

Remember to read the question carefully and give the answer asked for, in the proper form. In my class, if you give the tangent line when I ask for the slope, you lose points and vice-versa. Hopefully, your teacher will help you learn the difference by doing this also.

### Search 17Calculus

Tangent Line

To find the equation of a tangent line, there are two main steps.
Step 1: Calculate the slope.
Step 2: Use the slope from step one and a point (usually given) and find the equation of the tangent line.
Note: This procedure works only if the given point is on the graph of the function. [ See here if the point is not on the graph. ]

Let's go through an example to demonstrate this process.
Example
Find the equation of the tangent line to the curve $$f(x)=x^3/3 + x^2 + 5$$ at the point $$(3,23)$$.

First, let's look at a graph of this function and get an idea of what the problem is asking for. This is the graph of $$f(x)$$ (blue line) and the point $$(3,23)$$ is shown with an arrow. The idea is that we need to take the derivative of the function $$f(x)$$ plug in the point $$(3,23)$$ into the derivative. Since the derivative gives the equation of the slope of $$f(x)$$, plugging in the point $$(3,23)$$ gives us the slope at that specific point. So, let's get started.
First, we need to make sure the point is on the graph. It looks like it is from the graph to the right but that is not good enough. It could be slightly off and not show up on this graph. To make sure, we evaluate $$f(3)$$. If it is equal to $$23$$, then we are good.
$$f(3) = 3^3/3 + 3^2 + 5 =$$ $$9 + 9 + 5 = 23$$
So, this tells us that the point $$(3,23)$$ is on the curve $$f(x)$$ and we can use the procedure outlined above.

Step 1: Calculate the slope.
Now, find the derivative of $$f(x)$$.
$$f'(x) = (3x^2)/3 + 2x + 0$$
In this step, we used the constant multiple rule, the power rule and constant rule.
Simplifying, we get $$f'(x) = x^2 + 2x$$.
Just to clarify, this equation, $$f'(x)$$ is the equation of the slope of $$f(x)$$ at every point. In this problem, we want the slope $$(3,23)$$, so we let $$x = 3$$.
So we get the slope $$m = f'(3) = 3^2 +2(3) = 15$$.

Step 2: Find the equation of the tangent line.
Now we can use either the point-slope form of the equation of a line or the slope-intercept form. I will use the point-slope form, $$y-y_1=m(x-x_1)$$.

$$\begin{array}{rcl} y-23 & = & 15(x-3) \\ y & = & 15x - 45 + 23 \\ y & = & 15x - 22 \end{array}$$

Final Answer: The equation of the tangent line to the curve $$f(x)=x^3/3 + x^2 + 5$$ at the point $$(3,23)$$ is $$y=15x-22$$

To verify that our answer is correct, we can graph the function and the tangent line on the same set of axes, to make sure the line looks tangent to the curve and goes through the point. Here is what the result looks like.

Now, there are some things you need to watch for when you are asked to find a tangent line.
First, If you are given only the x-value, you can plug that value into the original function, $$f(x)$$ to get $$y$$.
A second twist is, given a point $$(x_1,y_1)$$, depending on how the question is worded, you are not guaranteed that the point is on the function. If it isn't, you need to do a whole different set of procedures (outlined in the panel below).

tangent line

For example, you may get a question worded like this:
Find the equation of the tangent line to the graph $$g(x)$$ through the point $$(x_1,y_1)$$.
In this case, the point $$(x_1,y_1)$$ may or may not be on the graph. To check it, we evaluate $$g(x_1)$$. If $$g(x_1) = y_1$$, we can use the procedure above. To be safe, it is always good to graph the function and evaluate the function at the x-value to make sure you get the y-value to determine if the given point is on the graph.

### Special Case When The Point is Not On The Graph

Finding the equation of the tangent line through a point that is NOT on the graph is an added twist to this problem. However, unlike most twists, this one takes quite a bit of thought to work through. So it is good to think about it before it comes up on an exam. First, let's discuss what the problem looks like. [ Hints: Do not read through this quickly. Stop and think about what this says at each step or you will not understand it. It may also help to draw a graph as you go through it. And you may need to read this several times to get your head around it. ]

The idea is that we have a function and we want to find a tangent line through a point that is not on the function. We can't just plug the x-value into the derivative to find the slope because the tangent line does not go through that x-value. We need to find the point on the function.

To demonstrate this procedure, we will use the same function as the previous example except we want the tangent line to go through the point $$(4,30)$$. To check that the point is not on the function, we plug $$x=4$$ into the function to get
$$f(4) = ((4^3)/3+4^2+5 = 64/3+16+5 = 127/3 \neq 30$$
This is verified by the graph on the right (never rely on the graph to give this information).
Here is what we can use from the previous example above.
The function is $$f(x)=x^3/3 + x^2 + 5$$.
The derivative is $$f'(x) = x^2 + 2x$$.

We are not given information about where the tangent line touches the function. So we will call this unknown point $$(a,b)$$. We know the slope at this point $$m = a^2+2a$$ and the function value at this point is $$b = (a^3)/3+a^2+5$$ (since the point $$(a,b)$$ is on the graph of the function). We also know that the point given in the problem statement $$(4,30)$$ as well as the unknown point $$(a,b)$$ are on the tangent line. Using the point-slope form of the equation of the line $$y-y_1 = (x-x_1)m$$ we can put all the information into one equation.
$$\begin{array}{rcl} y-y_1 & = & (x-x_1)m \\ b-30 & = & (a-4)(a^2+2a) \\ (a^3)/3+a^2+5 - 30 & = & a^3 + 2a^2 - 4a^2 - 8a \\ (a^3)/3+a^2-25 - a^3 +2a^2+8a = 0 \\ (-2a^3)/3 + 3a^2 + 8a-25 & = & 0 \end{array}$$
This last equation is not factorable, so we use the root finding capability on our calculator or graphing utility to solve for $$a$$. We know from algebra that there is either one real root or three real roots. We determine that there are three real roots whose values are
$$\begin{array}{rcr} a_1 & = & -3.13678 \\ a_2 & = & 2.19815 \\ a_3 & = & 5.43863 \end{array}$$
So which one do we choose? We were not given information in the problem statement about which one, so we can't eliminate any of them. So what do we do with all three? Since there are three different points where there is a tangent line through the point $$(4,30)$$, there are actually three tangent lines as well. So we need to find all three equations.

The details involve only algebra now, so we will present the answers and the graph.

 point tangent line $$(-3.13678, 4.55495 )$$ $$y = 3.56582x+15.73661$$ $$( 2.19815, 13.37225 )$$ $$y = 9.22816x-6.912637$$ $$(5.43863, 88.20122)$$ $$y = 40.455956x-131.82375$$

Note: We solved this analytically, i.e. not by guessing from the graph. Most instructors will want you to work it this way. Check with your instructor to see what they require.

Linearization

Linearization is essentially using the equation of the tangent line to approximate the function near a point.

Normal Line

The normal line is perpendicular to the curve and, therefore, also perpendicular to the tangent line. To find the equation of the normal line at a point, follow the same procedure above, expect after finding the slope of the tangent line, take the negative reciprocal of the slope to get the slope of the normal line. Then use the point to the find the equation of normal line.

In the example above, the slope of the normal line is $$m=-1/15$$. Using this and the point $$(x_1, y_1) = (3,23)$$, the equation of the normal line is

$$\begin{array}{rcl} y-y_1 & = & m(x-x_1) \\ y-23 & = & (-1/15)(x-3) \\ y & = & -x/15 + 1/5 + 23 \\ y & = & -x/15 + 116/5 \end{array}$$

normal line

The graph of this normal line is shown in the plot on the right. It is shown as the green almost horizontal line. Notice that it doesn't look exactly perpendicular to the tangent line. This is because the scales on the axes are not equivalent. However, you know from the equations that they are perpendicular. The slope of the tangent line is $$m_t = 15$$ and the slope of the normal line is $$m_n = -1/15$$ and since $$m_n = -1/m_t$$, they are perpendicular.

Okay, time for some practice problems.

filters

slope (1)

tangent (14)

normal (5)

Practice Problems

Instructions - - Unless otherwise instructed, find the equation of the tangent line (tangent), normal line (normal) or slope of the following functions through the given points, giving your answers in slope-intercept form and in exact form. The given point may or may not be on the graph. Use the derivative to calculate slope.
Note: In these problems, you are not usually asked to graph. However, it will help you a lot to visualize what is going on and to check your answer if you take a few seconds to graph the equation and the tangent line on the same set of axes.

 Level A - Basic

Practice A01

$$g(x)=x^3-3x+1$$
tangent $$(1,-1)$$

solution

Practice A02

$$\displaystyle{f(x)=\frac{2}{3}x^3+x^2-12x+6}$$
tangent $$x=0$$

solution

Practice A03

$$\displaystyle{h(x)=2x+\frac{1}{\sqrt{x}}}$$
tangent $$(1,3)$$

solution

Practice A04

$$\displaystyle{g(x)=\frac{2}{3}x^3+\frac{5}{2}x^2+2x+1}$$
tangent $$x=3$$

solution

Practice A05

$$g(x)=(x^2+1)(2-x)$$
tangent $$(2,0)$$

solution

Practice A06

$$y=1-9x^2$$
tangent $$(2,-35)$$

solution

Practice A07

$$y=4-9x^2$$
slope $$(6,-320)$$

solution

Practice A08

$$\displaystyle{y=(2+x)e^{-x}}$$
tangent $$(0,2)$$

solution

Practice A09

$$y=x^2$$
normal $$(-2,4)$$

solution

Practice A10

$$y=2x^2+3x-5$$
normal $$(2,9)$$

solution

Practice A11

$$y=5-x-2x^2$$
normal $$(-1,4)$$

solution

Practice A12

$$y=x^4+2e^x$$
normal $$(0,2)$$

solution

Practice A13

$$y=(2+x)e^{-x}$$
normal $$(0,2)$$

solution

Practice A14

$$f(x)=(1+x)^2$$
tangent $$x=0$$

solution

Practice A15

$$\displaystyle{f(x)=(1-2x)^{3/2}}$$
tangent $$x=0$$

solution

Practice A16

$$f(x)=\sin(x)$$
tangent $$x=0$$

solution

Practice A17

$$f(x)=\cos(x)$$
tangent $$x=\pi/2$$

solution

Practice A18

$$\displaystyle{f(x)=\frac{x^2-x+1}{3x+2}}$$
tangent $$(0,1/2)$$

solution

Practice A19

$$f(x)=(2x-1)^5(x+1)$$
tangent $$x=1$$

Find the horizontal and vertical tangent lines to the function $$f(x)=x\sqrt{1-x^2}$$ on the interval $$-1 \leq x \leq 1$$.