Tangent & Normal Lines

Let's focus on how to find the equation for the tangent line to a curve at a particular point, or the normal (meaning perpendicular) line to a curve at a particular point. These problems are straightforward once you get them down, and you are quite likely to see one on an exam, so practice here so that'll be totally routine for you. Each problem below has a complete solution one click away so you can immediately check your work.

PROBLEM SOLVING STRATEGY: Tangent & Normal Lines

These problems will always specify that you find the tangent or normal (= perpendicular) line at a particular point. We'll call that point $(𝑥 0, 𝑦 0)$ .

To answer these questions, you will almost always use the Point-Slope form of a line. Recall that if a line has slope m and contains the point $(𝑥 0, 𝑦 0)$ , then you can write its equation as:

$𝐏 𝐨 𝐢 𝐧 𝐭 - 𝐒 𝐥 𝐨 𝐩 𝐞 𝐟 𝐨 𝐫 𝐦 𝐨 𝐟 𝐚 𝐥 𝐢 𝐧 𝐞 : 𝑦 - 𝑦 0 = 𝑚 (𝑥 - 𝑥 0)$

The problem statement typically specifies the point $(𝑥 0, 𝑦 0)$ , and so really these problems come down to determining the slope m of the line — which we'll address below.

You will use that equation again and again; memorize it if you don't know it already.

(It's just a variant on the definition of slope: $𝑚 = 𝑦 - 𝑦 0 𝑥 - 𝑥 0$ ).

I. Tangent Line to a Curve

Very frequently in beginning Calculus you will be asked to find an equation for the line tangent to a curve at a particular point. We're calling that point $(𝑥 0, 𝑦 0)$ .

To find the line's equation, you just need to remember that the tangent line to the curve has slope equal to the derivative of the function evaluated at the point of interest:

$𝑚 t a n g e n t l i n e = 𝑓' (𝑥 0)$
That is, find the derivative of the function $𝑓' (𝑥)$ , and then evaluate it at $𝑥 = 𝑥 0$ . That value, $𝑓' (𝑥 0)$ , is the slope of the tangent line.

Hence we can write the equation for the tangent line at $(𝑥 0, 𝑦 0)$ as

$𝑦 - 𝑦 0 = 𝑚 t a n g e n t l i n e (𝑥 - 𝑥 0) 𝑦 - 𝑦 0 = 𝑓' (𝑥 0) (𝑥 - 𝑥 0)$

If those equations look abstract to you, don't worry. As soon as you work a few problems, the process will make sense — we promise.

II. Normal Line to a Curve

Sometimes instead a question will ask you instead to find the line normal to a curve. That's the same thing as asking for the line that is perpendicular to the curve.

You will again use the Point-Slope form of a line. But now to compute the slope of the line, recall that the slopes of perpendicular lines are the negative reciprocals of each other ( $𝑚 2 = - 1 𝑚 1$ ). We want the slope of the line that is perpendicular to the curve at a point, and hence that is perpendicular to the tangent line to the curve at that point:

$𝑚 n o r m a l l i n e = - 1 𝑚 t a n g e n t l i n e = - 1 𝑓' (𝑥 0)$

Hence we can write the equation for the normal line at $(𝑥 0, 𝑦 0)$ as

$𝑦 - 𝑦 0 = 𝑚 n o r m a l l i n e (𝑥 - 𝑥 0) 𝑦 - 𝑦 0 = - 1 𝑓' (𝑥 0) (𝑥 - 𝑥 0)$

We recommend not trying to memorize all of the formulas above. Instead, remember the Point-Slope form of a line, and then use what you know about the derivative telling you the slope of the tangent line at a given point. The problems below illustrate.

Question 2 illustrates the process of putting together different pieces of information to find the equation of a tangent line.

Question 3, and the rest of the questions, require that you find the pieces of information before you can put them together.

Practice Problem 1 (Warm-up): Graph of

𝑓

and its tangent at a point

Use the information in the graph to replace the question marks with correct values: Graph of f with point (3,10) highlighted; also its tangent line at that point. The line also passes through (5,16)

Graph of f with point (3,10) highlighted; also its tangent line at that point. The line also passes through (5,16)

(a)

𝑓 (? ――) = ? ――

(b)

𝑓' (? ――) = ? ――

Practice Problem 2: Given

𝑔 (5)

and

𝑔' (5)

, write the equation of the tangent line

For a particular function

𝑔

, we know

𝑔' (5) = 2

and

𝑔 (5) = - 3

. Write an equation for the line tangent to

𝑔

𝑥 = 5

Practice Problem 3: Given

𝑓 (𝑥)

, find tangent and normal lines at a point

Consider the curve given by

𝑦 = 𝑓 (𝑥) = 𝑥 3 - 𝑥 + 5

(a)

Find the equation to the line tangent to the curve at the point (1, 5).

(b)

Find the equation of the line normal (perpendicular) to the curve at the point (1,5).

Practice Problem 4: Find the normal line to a function's curve

Find an equation of the normal (perpendicular) line to the curve

𝑦 = \sqrt 25 - 𝑥 2

at the point

(3, 4)

Practice Problem 4: Find where the curve is perpendicular to a line

Find the equation of the tangent line to the curve

𝑦 = 𝑥 2 + 4 𝑥 - 2

at a point where the tangent is perpendicular to the line

- 𝑥 - 2 𝑦 + 6 = 0

Answer:

𝑦 = 2 𝑥 - 3

This question may seem a little hard to parse. We're essentially going to work our way backwards through what was given, following roughly these four steps:

Step 1. Determine the slope of the line the question tells us about in the last sentence,

- 𝑥 - 2 𝑦 + 6 = 0

. We'll call that line Line 1, and we'll then have its slope

𝑚 1

.

Step 2. We're interested in the point on the curve that's perpendicular to Line 1, so we'll find the slope

𝑚 2 = - 1 𝑚 1

.

Step 3. We'll find the point on the curve that has slope equal to

𝑚 2

.

Step 4. We'll write the equation of the line tangent to the curve at that point.

Let's implement our strategy:

Step 1.
We're given the equation of a line we're calling Line 1:

- 𝑥 - 2 𝑦 + 6 = 0

. We can easily determine its slope by putting the equation into slope-intercept form

𝑦 = 𝑚 𝑥 + 𝑏

- 𝑥 - 2 𝑦 + 6 = 0 - 2 𝑦 = 𝑥 - 6 𝑦 = - 12 𝑥 + 3

So Line 1 has slope

𝑚 1 = - 12

.

Step 2.
The line we're after is perpendicular to this line, and recall that line 2 is perpendicular to line 1 if their slopes are negative reciprocals:

𝑚 2 = - 1 𝑚 1

. Our line therefore has slope

𝑚 2 = - 1 𝑚 1 = 2

Step 3.
Next, we need to find the point on the curve

𝑦 = 𝑥 2 + 4 𝑥 - 2

that has slope equal to

𝑚 2 = 2

. The curve's slope at any point is given by its derivative there:

𝑦 = 𝑥 2 + 4 𝑥 - 2 𝑑 𝑦 𝑑 𝑥 = 2 𝑥 + 4

We're looking for the point where the curve's slope

𝑑 𝑦 𝑑 𝑥 = 2

𝑑 𝑦 𝑑 𝑥 = 2 = 2 𝑥 + 4 - 2 𝑥 = 2 𝑥 = - 1

So we're interested in the tangent line to the curve at

𝑥 = - 1

.

Step 4.
To write the equation of the tangent line through this point, we also need its

𝑦

-value:

𝑦 = 𝑥 2 + 4 𝑥 - 2 𝑦 = (- 1) 2 + 4 (- 1) - 2 = 1 - 4 - 2 = - 5

So our tangent line intersects (grazes) the curve at the point (-1,-5). And we determined earlier that our line has slope

𝑚 2

= 2. Hence, using the the point-slope form of a line that contains the point

(𝑥 𝑜, 𝑦 𝑜

, we can write the tangent line as

𝑦 - 𝑦 0 = 𝑚 (𝑥 - 𝑥 0) 𝑦 - (- 5) = (2) [𝑥 - (- 1)] 𝑦 + 5 = 2 𝑥 + 2 𝑦 = 2 𝑥 - 3 ✓

Practice Problem 6: Tangents to some trig graphs

Find the tangent lines as requested.

(a)

Find an equation of the tangent line to the curve

𝑦 = c o s 2 𝑥

𝑥 = 𝜋 4

(b)

Find an equation of the tangent line to the curve

𝑦 = t a n 2 𝑥

at the point

𝑥 = 𝜋 6

(a) $𝑦 = - 𝑥 + 𝜋 4 + 12$
(b) $𝑦 = 8 3 \sqrt 3 𝑥 - 4 𝜋 9 \sqrt 3 + 13$

Practice Problem 7: Given the tangent line, determine. . . (Based on an actual exam problem)

The tangent line to the graph of

𝑓

at the point

𝑥 = 1

𝑦 = 3 𝑥 + 9

(a)

(i) What is

𝑓 (1)

? (ii) What is

𝑓' (1)

(b)

Given that

𝑓 (𝑥) = 𝑎 𝑥 3 + 𝑏

, find the constants

𝑎

and

𝑏

Practice Problem 8: An actual unversity exam problem

Let

𝑓

be the real-valued function defined by

𝑓 (𝑥) = \sqrt 1 + 6 𝑥

(a)

Give the domain and range of

𝑓

(b)

Determine the slope of the line tangent to the graph of

𝑓

𝑥 = 4

(c)

Determine the

𝑦

-intercept of the line tangent to the graph of

𝑓

𝑥 = 4

(d)

Give the coordinates of the point on the graph of

𝑓

where the tangent line is parallel to

𝑦 = 𝑥 + 12

Practice Problem 9: Linet intersects cubic and parabola (actual exam problem)

The line

𝑥 = 𝑐

, where

𝑐 > 0,

intersects the cubic

𝑦 = 2 𝑥 3 + 3 𝑥 2 - 5

at point P, and the parabola

𝑦 = 4 𝑥 2 + 4 𝑥 + 3

at point Q.

(a)

If a line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q, find the value of

𝑐

where

𝑐 > 0

(b)

Write the equations of the two tangent lines described in (a).

(a) $𝑐 = 1$
(b) Cubic tangent: $𝑦 = 12 𝑥 - 12$ ; Parabola tangent: $𝑦 = 12 𝑥 - 1$

Practice Problem 10: Tangent line intersects

𝑥

-axis (actual exam problem)

Where does the tangent line to the graph of

𝑦 = 𝑓 (𝑥)

at the point

(𝑥 0, 𝑦 0)

intersect the x-axis?

Practice Problem 11: Normal line passes through (0, 3/4)

A line normal (perpendicular) to the curve

𝑦 = 2 𝑥 2

at a point in the first quadrant also passes through the point

(0, 34)

. Find an equation for this line.

As usual, a quick figure helps a lot.
Curve y=2x^2, and line passing through (0, 3/4) and the point (x_1, y_1) on the curve

Here we've shown the curve

𝑦 = 𝑥 2,

and the line that passes through the point

(0, 34)

. This line intersects the curve at the point we're calling

(𝑥 1, 𝑦 1) .

Note that naming this point on the curve with coordinates like this is crucial to our solution.

There are several pieces of information we have to put together to solve this problem.
(1) The first is that the slope of a line that is normal (perpendicular) to this curve at the point

(𝑥 1, 𝑦 1)

is given by

𝑚 n o r m a l, a t 𝑥 = 𝑥 1 = - 1 𝑓' (𝑥 1)

Since

𝑓 (𝑥) = 2 𝑥 2,

we have

𝑓' (𝑥) = 4 𝑥 .

For the particular point of interest,

(𝑥 1, 𝑦 1),

where the line intersects the curve, we can write the slope of the normal line as

𝑚 n o r m a l, a t 𝑥 = 𝑥 1 = - 1 4 𝑥 1 ◃ (1)

Keep that in mind for a moment.

(2) We also know that the line contains the points

(0, 34)

and

(𝑥 1, 𝑦 1) .

Hence we can write its slope as

𝑚 l i n e = 𝑦 1 - 34 𝑥 1 - 0 ◃ (2)

Now, the magic: the line must actually meet both conditions (1) and (2), and so we must have

𝑚 n o r m a l, a t 𝑥 = 𝑥 1 = 𝑚 l i n e

This requirement will let us solve for

𝑦 1,

as you'll see:

𝑚 n o r m a l, a t 𝑥 = 𝑥 1 = 𝑚 l i n e - 1 4 𝑥 1 = 𝑦 1 - 34 𝑥 1 - 14 = 𝑦 1 - 34 - 𝑦 1 = - 34 + 14 - 𝑦 1 = - 12 𝑦 1 = 12 ◃

We now have the y-value where the line intersects the curve. That leaves us with a few steps to go, since the question asked for the equation of the line. So let's next determine

𝑥 1

: We know that the point

(𝑥 1, 𝑦 1)

lies on the curve

𝑦 = 2 𝑥 2,

and so since

𝑦 1 = 12

we must have

12 = 2 𝑥 21 𝑥 21 = 14 𝑥 1 = \pm \sqrt 14 = \pm 12

Ah, but the problem specifies "at a point in the first quadrant," and so we choose the positive solution:

𝑥 1 = 12 ◃

And then we immediately know the slope of the normal line, since we decided (1) above

𝑚 n o r m a l, a t 𝑥 = 𝑥 1 = - 1 4 𝑥 1

we must have

𝑚 n o r m a l, a t 𝑥 = 1 / 2 = - 1 4 (12) = - 12 ◃

Now we're actually almost done: we know that the slope of the normal line is

𝑚 n o r m a l = - 12,

and since we were told the line passes through the point

(0, 34)

we know it has y-intercept

𝑏 = 34 .

Hence using the slope-intercept form of a line:

𝑦 = 𝑚 n o r m a l 𝑥 + 𝑏 𝑦 = - 12 𝑥 + 34 ✓

Curve y=2x^2 and the line passing through the point (0, 3/4) and the point (1/2, 1/2) on the curve.

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