2
Limits and Derivatives
2.7
Tangent Lines, Velocity, and Derivatives
A tangent line to a circle is a line that intersects the circle at exactly one point.
We would like to take this idea of tangent line and apply it to other curves.
P (a, f (a)) y = f (x)
The word tangent is derived from the Latin word tangens, which means “touching.” Thus a tangent to a curve is a line that touches the curve. In other words, a tangent line should have the same direction as the curve at the point of contact.
Our goal will be to find an equation of the tangent line to a curve given by y = f (x) at the point P (a, f (a)). Note that when the x-coordinate is a, the y-coordinate is necessarily given by y = f (a).
Let’s begin by recalling the formula for the equation of a line.
of a line is
y − y1 = m(x − x1)
where (x1, y1) is a point on the line, and m is the slope of the line.
This is a good start. For our tangent line, we have a point, P (a, f (a)). So we can let x1 = a and y1 = f (a). Next we need to find the slope of the
tangent line, mtan.
Slope of a Line The formula for the slope of a line is
m = y2− y1 x2− x1
,
where (x1, y1) and (x2, y2) are two points on the line.
In order to use this formula for slope, we need to know the coordi-nates of two points on a line. So far, we have only one set of coordicoordi-nates, x1 = a, y1 = f (a).
We are going to get around this problem by introducing a secant line. This will be a line through the point P and another point Q(x, f (x)) on the curve. The idea is that the slope of the secant line will be almost the same as the slope of the tangent line provided that the point Q is close to P .
P (a, f (a)) Q(x, f (x))
y = f (x)
Tangent Line Secant Line
The secant line contains the point Q(x, f (x)) and the point P (a, f (a)). Therefore the slope of the secant line is given by
msec =
rise run =
f (x) − f (a) x − a
Again, we are going to use the slope of the secant line as an approxima-tion for the slope of the tangent line. To get a really good approximaapproxima-tion, we want to make Q as close to P as possible. This is achieved by finding the limit of the slope of the secant line, msec, as P approaches Q.
The slope of the tangent line, mtan, is then defined as the limit of the
slope of the secant line, msec, as Q approaches P . That is,
mtan = lim Q→Pmsec
To make Q approach P , we can let x approach a. Moreover, a formula for msec was given above. We now give the following definition for the slope
Definition The slope of the tangent line to a curve y = f (x) at (a, f (a) is given by the formula
mtan = limx→a
f (x) − f (a) x − a
This was sort of the whole point of studying limits in the previous chap-ter: we wanted to be able to make this limit calculation.
We can now do an example where we find the equation of the tangent line to a specific function.
Example 1 Find the equation of the tangent line to the curve f (x) = x2 at
the point (1, 1).
Solution The point-slope formula for a line is y − y1 = m(x − x1).
For this example, x1 = 1 and y1 = 1. Note that f (x) = x2 and a = 1.
We can now use the formula to find the slope of the tangent line, mtan.
mtan = x→alim
f (x) − f (a) x − a = lim x→1 f (x) − f (1) x − 1 = lim x→1 x2− 12 x − 1 = lim x→1 (x − 1)(x + 1) (x − 1) = lim x→1(x + 1) = 1 + 1 = 2
We have x1 = 1, y1 = 1 and mtan = 2. Therefore, when substituting into
the point-slope equation for the line we get
y − 1 = 2(x − 1) y = 2x − 1
Example 2 Find the slope of the tangent line at the given value of x = a using the formula given above for mtan.
1. f (x) = x2− 3x + 1, a = 2 2. f (x) =√x, a = 4
3. f (x) = x3, a = 3
Solution
1. f (x) = x2− 3x + 1, a = 2
mtan = x→alim
f (x) − f (a) x − a = lim x→2 f (x) − f (2) x − 2 = lim x→2 (x2− 3x + 1) − ((2)2− 3(2) + 1) x − 2 = lim x→2 (x2− 3x + 1) − (−1) (x − 2) = lim x→2 x2− 3x + 2 (x − 2) = lim x→2 (x − 1)(x − 2) (x − 2) = lim x→2(x − 1) = 2 − 1 = 1 2. f (x) =√x, a = 4 mtan = lim x→a f (x) − f (a) x − a = lim x→4 f (x) − f (4) x − 4 = lim x→4 √ x −√4 x − 4
= lim x→4 √ x − 2 (x − 4) = lim x→4 √ x − 2 (x − 4) · √ x + 2 √ x + 2 ! = lim x→4 (√x)2− (2)2 (x − 4)(√x + 2) = lim x→4 x − 4 (x − 4)(√x + 2) = lim x→4 1 √ x + 2 = 1 √ 4 + 2 = 1 4 3. f (x) = x3, a = 3 mtan = lim x→a f (x) − f (a) x − a = lim x→3 f (x) − f (2) x − 3 = lim x→3 x3− (3)3 x − 3 = lim x→3 (x − 3)(x2+ 3x + (3)3) (x − 3) = lim x→3(x 2+ 3x + 9) = (3)2+ 3(3) + 9 = 27
Example 3 Find the tangent line to the function f (x) = 1/x at (2,12).
Solution First we find the slope mtan where f (x) = 1/x and a = 2.
mtan = x→alim
f (x) − f (a) x − a = lim x→2 f (x) − f (2) x − 2 = lim x→2 1 x − 1 2 x − 2
Multiply the numerator and denominator by 2x to clear the fraction.
= lim
x→2
(1x− 1 2)2x
= lim x→2 (2 − x) (x − 2)2x = lim x→2 −(x − 2) (x − 2)2x = lim x→2− 1 2x = 1 2 · 2 = − 1 4
We can use the point-slope formula with x1 = 2, y1 = f (2) = 1/2, and
mtan= −1/4. This gives
y − y1 = mtan(x − x1)
y − 1/2 = −1
4(x − 2)
y = −x 4 + 1
Velocity and Rate of Change
Suppose we drop a rock from a cliff in the High Sierras. There is an equation that we can use to determine the position of a falling object after t seconds. If s(t) is the position in meters of a falling object after t seconds then we have the formula
s(t) = 4.9t2 m.
For example, after 1 second, the rock has fallen s(1) = 4.9(1)2 = 4.9 m. After 2 seconds, the rock has fallen s(2) = 4.9(2)2 = 4.9(4) = 19.6 m.
In fact, this formula can be used to estimate the height of a cliff. We can make the approximation that s(t) = 5t2 m. Then we can drop a rock off of a cliff and record how long it takes for the rock to fall. Say the rock takes 5 seconds to fall. In that case the distance that the rock fell is approximately s(10) = 5 ∗ (5)2 = 125 m. So the distance from the top of the cliff to the base is about 125 meters.
Now let’s talk about velocity. Using this formula for position, or the dis-tance that a falling object has traveled, we can calculate the average velocity
of an object.
Suppose we are interested in knowing how fast a falling object is going after it has fallen a seconds. We have the following formula for average velocity.
vavg= s(t2) − s(t1)
t2− t1
So the units for average velocity will be meters per second in our example. To find how fast the falling object is going after it has traveled a seconds, we can let t1 = a and t2 = t. This gives
vavg = s(t) − s(a) t − a
Common sense tells us that to get the best approximation possible, we should make the interval over which we find the average velocity as small as possible. We would like to take the limit of the average velocity as t approaches a. This is how we define instantaneous velocity. The instan-taneous velocity of of a particle at time t = a with position function s(t) is given by
v(a) = lim
t→a
s(t) − s(a) t − a .
Example 4 Suppose that a ball is dropped from the upper deck of the CN Tower, 450 meters above the ground. The position function of the ball is given by s(t) = 4.9t2 meters after 2 seconds. What is the velocity of the ball after 5 seconds?
Solution We can find a formula for velocity at time 2 seconds using the formula. v(2) = lim t→2 s(t) − s(2) t − 2 = lim t→2 4.9t2− 4.9(2)2 t − 2 = lim t→2 4.9(t2− (2)2) t − 2
= lim t→2 4.9(t − 2)(t + 2) t − 2 = lim t→24.9(t + 2) = 4.9(2 + 2) = 4.9(4) = 19.6 m/s
Example 5 The position function of a particle is given by the equation s = t3 where t is measured in seconds and s in meters. Find the velocity at
time t = 2 seconds.
Solution We can find a formula for velocity at time 2 seconds using the formula. v(2) = lim t→2 s(t) − s(2) t − 2 = lim t→2 (t3) − ((2)3) t − 2 = lim t→2 (t − 2)(t2+ 2t + 4) t − 2 = lim t→2(t 2+ 2t + 4) = (2)2+ 2(2) + 4 = 12 m/s Derivatives
In the previous subsection we gave a formula for the slope of the tangent line, mtan. The slope of the tangent line is equal to what is called the derivative
of a function f at a number a.
The slope of the tangent line to a curve at a point (a, f (a)) equals the derivative. However, there are many other applications for the derivative of a function. The bulk of this chapter will be spent studying the derivative as a pure math entity.
Definition The derivative of a function f at a number a, denoted by f0(a), is given by the following limit, provided the limit exists.
f0(a) = lim
x→a
f (x) − f (a) x − a
Example 6 Find the derivative of f (x) = x2+ 2x + 5 at a = 2. Solution f0(2) = lim x→2 f (x) − f (2) x − 2 = lim x→2 ((x)2+ 2(x) + 5) − ((2)2 + 2(2) + 5) x − 2 = lim x→2 x2+ 2x − 8 x − 2 = lim x→2 (x + 4)(x − 2) (x − 2) = lim x→2(x + 4) = 2 + 4 = 6
The derivative can be different for different values of a. We would like to be able to find a general formula for the derivative f0(a) and then evaluate it for different values of a.
Example 7 Let f (x) = x2.
1. Compute a formula for f0(a) using the definition of derivative.
2. Calculate f0(−1), f0(0) and f0(1). Solution 1. f0(a) = lim x→a f (x) − f (a) x − a = lim x→a x2− a2 x − a = lim x→a (x − a)(x + a) x − a = lim x→a(x + a) = a + a = 2a. 2. f0(−1) = 2(−1) = −1, f0(0) = 2(0) = 0, and f0(1) = 2(1) = 2.
f (x) = x2 slope=−2 (−1, 1) slope=0 (0, 0) slope=2 (1, 1)
Derivative equals slope of the tangent line. The slope of the tangent line can take dif-ferent values for difdif-ferent points on the graph. We see that for f (x) = x2, we have
f0(−1) = −2, f0(0) = 0, and f0(1) = 2.
There is an equivalent formula for derivative. If we make the substitution h = x − a, then x = a + h, and as x → a, h → 0. So that
f0(a) = lim x→a f (x) − f (a) x − a = limh→0 f (a + h) − f (a) h
An Equivalent Definition for Derivative The derivative of a function f at a number x = a is given by
f0(a) = lim
h→0
f (a + h) − f (a) h
For certain functions, this is an easier formula to work with.
Example 8 Find the derivative using the equivalent definition of derivative of f (x) = x2 at a = 2.
f0(a) = lim h→0 f (2 + h) − f (2) h = limh→0 (2 + h)2− (2)2 h = lim h→0 4 + 4h + h2− 4 h = lim h→0 4h + h2 h = lim h→0 h(4 + h) h = lim h→0(4 + h) = 4 + 0 = 4
Example 9 Find a formula for f0(a) for f (x) = x2 using the equivalent
definition of derivative. Solution f0(a) = lim h→0 f (a + h) − f (a) h = limh→0 (a + h)2− a2 h = lim h→0 a2+ 2ah + h2− a2 h = limh→0 2ah + h2 h = limh→0 h(2a + h) h = limh→0(2a + h) = 2a + 0 = 2a
We see that the derivative of a function depends on the number a and is therefore itself a function. Rather that use the letter a, we can use the letter x.
Definition The derivative of a function f denoted f0 is given by
f0(x) = lim
h→0
f (x + h) − f (x)
h ,
provided that the limit exists.
Example 10 Find a formula for f0(x) using the equivalent definition of derivative, where f (x) =√x. Solution f0 (x) = lim h→0 f (x + h) − f (x) h = lim h→0 1 h √ x + h −√x
= lim h→0 1 h √ x + h −√x √ x + h +√x √ x + h +√x = lim h→0 1 h (x + h) − x √ x + h +√x ! = lim h→0 1 h h √ x + h +√x ! = lim h→0 1 √ x + h +√x = √ 1 x + 0 +√x = 1 2√x
Finding the derivative directly from the definition is long and difficult. We will introduce some rules for finding the derivative that will make the job much easier.
Example 11 The limit below represents the derivative of some function f at some number a. State such an f and a.
f0(a) = lim
h→0
sec(h) − 1 h
Solution The formula used for f0(a) appears to be the one involving h. We compare. f0(a) = lim h→0 sec(h) − 1 h and f 0 (a) = lim h→0 f (a + h) − f (a) h
Let’s only look at the numerator of the two fractions.
f (a + h) − f (a) and sec(h) − 1
The terms f (a+h) and sec(h) should agree. Note that sec(h) = sec(0+h). We should have f (a + h) = sec(0 + h). Therefore, f (x) = sec(x) and a = 0. It should follow that f (0) = 1. We see that f (a) = sec(0) = 1.
Answer: f (x) = sec x, a = 0
Example 12 The limit below represents the derivative of some function f at some number a. State such an f and a.
f0(a) = lim
x→π/3
cos(x) − 1/2 x − π/3
Solution We make a comparison. f0(a) = lim x→a f (x) − f (a) x − a and f 0 (a) = lim x→π/3 cos(x) − 1/2 x − π/3
We have x → a and x → π/3. We conclude that a = π/3.
We can compare just the numerators of the fractions: f (x) − f (a) and cos(x) − 1/2. We conclude that f (x) = cos x. Moreover, we should have f (a) = cos(a) = cos(π/3) = 1/2.
Math 180 Homework
§2.7 The Tangent Lines, Velocity, and Derivatives
Sketch a plausible tangent line at the given point. 1. At x = π
2. At x = 0
3. At x = 0
5. Estimate the slope of the tangent line to the curve at
(a) x = 0 (b) x = 0.5
(c) x = 1 (d) x = 1.5
6. Estimate the slope of the tangent line to the curve at
(a) x = 1 (b) x = 2 (c) x = 3
7. List the points A, B, C, and D in order of increasing slope of the tangent line.
Find the equation of the tangent line to the curve at the given point.
8. f (x) = x2− 2, (1, −1) 9. f (x) = x2− 2, (0, −2)
10. f (x) = 1/(x − 2), (3, 1)
11. f (x) =√x, (1, 1)
12. If a ball is thrown into the air with velocity of 40 ft/s, its height (in feet) after t seconds is given by y = 40 − 16t2. find the velocity when
t = 2.
13. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 1/t2, where t is measured in seconds. find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.
For these homework problems, use either definition of derivative given in this section. f0(a) = lim h→0 f (a + h) − f (a) h or f0(a) = lim x→a f (x) − f (a) x − a
• For the functions f given below, find a formula for f0(a). Use the formula
for f0(a) to find f0(0), f0(5) and f0(8).
14. f (x) = x3
15. f (x) = x2+ x − 2.
16. f (x) =√3x + 1.
• Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
17. f0(a) = lim h→0 (1 + h)10− 1 h 18. f0(a) = lim h→0 4 √ 16 + h − 2 h 19. f0(a) = lim x→5 2x− 32 x − 5 20. f0(a) = lim x→π/4 tan x − 1 x − π/4 21. f0(a) = lim h→0 cos(π + h) + 1 h 22. f0(a) = lim t→1 t4+ t − 2 t − 1
Math 180 Homework
The Tangent Lines, Velocity, and Derivatives
Solutions
1. .
2. .
3. There is no tangent line at x = 0
4. .
5. (a) 1 (b) 1/2
(c) −1/2 (d) −1
6. (a) −1 (b) 0 (c) 1 7. C, B, A, D 8. y = 2x − 3 9. y = −2 10. y = −x + 4 11. y = x2 + 12
12. The ball will be falling toward the ground at a velocity of 64 ft/s.
13. v(a) = −2/a3 m/s, v(1) = −2 m/s, v(2) = −1/4 m/s, v(3) = −2/27 m/s. 14. f0(a) = 3a2. f0(0) = 0, f0(5) = 75, f0(8) = 192. 15. f0(a) = 2a + 1. f0(0) = 1, f0(5) = 11, f0(8) = 17. 16. f (x) = 3 2√3x + 1. f 0(0) = 3/2, f0(5) = 3/8, f0(8) = 3/10. 17. f (x) = x10, a = 1 18. f (x) = √4 x, a = 16 19. f (x) = 2x, a = 5 20. f (x) = tan x, a = π/4 21. f (x) = cos x, a = π 22. f (t) = t4+ t, a = 1
