Tangent and Secant Lines

Monday, January 9, 2017 Tangent and Secant Lines

• Suppose you are given some curve, in this L case a circle, and a point P.

• The line L passing through the point P is L 1 called a tangent line. Q1 P • Given another point Q1 on the curve, let L1 be the line passing through the points P

and Q1.

• L1 is called a secant line and is at times written PQ1. Monday, January 9, 2017 Tangent and Secant Lines

L 2 • Pick a point Q closer to P. L 2

• Notice that the slope of the secant line L2 is

Q2 closer to the slope of L compared to the L1 slope of L1. Q1 P • In fact, as we pick points closer and closer to P, the slope of the associated secant lines will get closer and closer to the slope of L.

Q: So, how can we differentiate between secant lines and tangent lines? Is it enough to say that a tangent line goes through one point, while a secant line goes through two points?

A: No! Tangent and Secant Lines

• L is the tangent line to P, even though it intersects the curve more than once. P • However, we can remedy L this by saying that there is a neighborhood of the point P for which L doesn’t intersect the curve at any other point within this neighborhood.

Q: Is this enough? Do we have enough information now to distinguish between tangent and secant lines?

A: Not quite. Tangent and Secant Lines

• L1 is a secant line for the curve through the point P. L1 • We can find a neighborhood of P such

P that L1 doesn’t intersect any other point of the curve within the neighborhood.

What then is the real distinction between tangent lines and secant lines? Tangent and Secant Lines • If we zoom in on the point P, the curve at P starts to look more and P L more like the tangent line at P.

• We can think of the tangent line as a linear approximation of the curve at P.

• The notion of linear approximation is a really powerful tool in math. We will study it a bit later in the course.

Warning: The tangent line to a curve at a given point Ex: Cusps may NOT exist. This is equivalent to saying that there P L is no good linear approximation of the curve at the No matter how close we given point. zoom in, L will never give a linear approximation of the curve at the point P. Tangent and Secant Lines • If we zoom in on the point P, the curve at P starts to look more and P L more like the tangent line at P.

• We can think of the tangent line as a linear approximation of the curve at P.

• The notion of linear approximation is a really powerful tool in math. We will study it a bit later in the course.

L1 P • A secant line through the point P, on the other hand, doesn’t give a linear approximation of the curve at P.

• Nonetheless, secant lines have their use! Quick Review of Lines

L • Recall that the equation of a line is of the form y = mx + b, where m is the slope and b is y the y-intercept. 2 (x2,y2) y – y • Let L be a line, (x1,y1) and 2 1 (x2,y2) two points on L such that x > x . y1 2 1 (x1,y1)

• The slope m = (y2 - y1)/(x2 – x1).

• To find the y-intercept, plug in x2 – x1 either (x1,y1) or (x2,y2) into the equation y = mx + b and solve x x2 for b. 1 Quick Review of Lines

Ex: Find the equation of the line passing through the points (5,7) and (1,8).

• Since 5 > 1, the slope is given by: m = (7-8)/(5-1) = -1/4

• Thus, the equation of the line is going to be of the form: y = (-1/4)x + b

• Since the point (1,8) is on the line, it must be that 8 = (-1/4)(1) + b which means that b = 33/4

• Hence, the equation of the line is given by y = (-1/4)x + 33/4 Quick Review of Lines: Class Problems.

Question 1: What is the slope of the line passing through (1,1) and (2,4)?

Question 2: What is the slope of the line passing through (1,1) and (a,a2), for a ≠ 1? Calculating Tangent and Secant Lines

• Suppose you are given the graph of a function f and a secant line through points (a,f(a)) and (b,f(b)). f f(b) • We can calculate the equation of the secant line using the technique we reviewed in the previous slide. f(a) • The slope is given by: m=(f(b) – f(a))/(b – a)

a b Calculating Tangent and Secant Lines

• Suppose you are given the graph of a function f and a secant line through points (a,f(a)) and (b,f(b)). f f(a + h) • We can calculate the equation of the secant line using the technique we reviewed in the previous slide. f(a) • The slope is given by: m=(f(b) – f(a))/(b – a)

a a + h • Since b = a + h, for some quantity h, we can replace the formula for the slope by m = (f(a + h) – f(a))/((a + h) – a) = (f(a + h) – f(a))/h

Slope of the secant line = Calculating Tangent and Secant Lines

Slope of the secant line =

• If we allow the quantity h to get smaller, the point (a+h,f(a+h)) is getting closer to f(a) (a,f(a)), which in turn means that the slope of the secant line through (a+h,f(a+h)) is getting closer to the slope of the tangent a line at (a,f(a)).

• Thus, the slope of the tangent line is given by taking the limit as h goes to 0.

Slope of the tangent line at (a,f(a)) = Calculating Tangent and Secant Lines

Ex: Calculate the equation of the tangent line of f(x)=x3 at the point (2,8).

• The slope of the tangent line is given by: ! !!! !!(!) lim h→0 ! • Given that the slope is 12, we can (2 + ℎ)!−2! plug in the point (2,8) to find the y- = lim h→0 ℎ intercept 8 = 12(2) + b, and so !!!"#! !"!! !! !! b = -16 = lim h→0 ! • Thus, the equation of the line is given !"#! !"!! !! by y = 12x -16. = lim h→0 !

2 = limh→0 (12 + 6h + h ) = 12 Calculating Tangent and Secant Lines: Class Problems

Consider the function f(x) = x2

Question 3: What is the slope of the line passing through (1,1) and (1 + h,(1 + ℎ)!)?

Question 4: What is the slope and equation of the tangent line at (1,1)?

Question 5: What is the slope and equation of the tangent line at (a,a2)? Limits of Functions

L • Given a function f, and a number a, the limit of f as x approaches a is the value, L, f(x) approaches as x goes to a. f • In this case, we write

limx→a f(x) = L

a Limits of Functions f(a)

Note that L doesn’t have to be the same as f(a). f

a Limits of Functions • We can take limits from the left hand side and the right hand side.

• Here, the limit from the left f hand side is L1. We denote this by

L2 limx→a- f(x) = L1 L1 • The limit from the right

hand side is L2. We denote this by

limx→a+ f(x) = L2 a Note: If, as in this picture, the left hand limit and the right hand limit as x approaches a are different, then the limit of f as x approaches a does not exist. Limits of Functions • It’s possible that the limit of f as x approaches a is not a number.

• In this picture

limx→a f(x) = ∞

• The line x = a is an example f of a vertical asymptote.

• In general, we say that f has a vertical asymptote at x = a if the limit as x approaches a a from the left, right, or both is ∞ or -∞. Limits of Functions In this picture, we have

limx→a- f(x) = -∞

limx→a+ f(x) = ∞

And so f

limx→a f(x) does not exist.

a Limits of Functions

Some Limit Laws

• limx→a [f(x) + g(x)] = limx→a f(x) + limx→a g(x)

• limx→a [f(x) - g(x)] = limx→a f(x) - limx→a g(x)

• limx→a [f(x) · g(x)] = limx→a f(x) · limx→a g(x)

!(!) limx→a f(x) • limx→a = , as long as limx→a g(x) ≠ 0 !(!) limx→a g(x)

! ! • limx→a [� � ] = [limx→a f(x)] for n ∈ ℤ Limits of Functions

Limit Theorems Direct Substitution property: If f is a polynomial or a rational function defined at a, then limx→a f(x) = f(a)

Theorem: If f(x) ≤ g(x) in some neighborhood of a

and limx→a f(x) and limx→a g(x) exist, then limx→a f(x) ≤ limx→a g(x). g

a Limits of Functions

Limit Theorems Squeeze Theorem: If f(x) ≤ h(x) ≤ g(x) in some neighborhood of a and limx→a f(x) = limx→a g(x) = L, then limx→a h(x) = L

a Limits of Functions: Class Problems

Question 6: Next Class

• Next time, we will give a precise definition of limits.

• Make sure to visit Canvas and do the assignments for next class. Math 3 Group Work Solutions, 1/9

1. What is the slope of the line passing through (1, 1) and (2, 4)? [Answer: 3]

Solution:

y2 y1 4 1 3 The slope of the line is = = =3. x2 x1 2 1 1

2. What is the slope of the line passing through (1, 1) and (a, a2),fora =1? [Answer: a +1] 6

Solution:

2 y2 y1 a 1 (a+1)(a 1) The slope of the line is = = . Since a =1,thisisequaltoa +1. x2 x1 a 1 a 1 6

3. What is the slope of the line passing through (1, 1) and (1 + h, (1 + h)2), where h>0? [Answer: h +2]

Solution:

2 2 y2 y1 (1h) 1 h +2h The slope of the line is = = . Since h =1,wecanrewritethisash +2. x2 x1 1+h 1 h 0 6

4. What is the slope and equation of the tangent line at (1, 1)? [Answer: 2; f(x)= 2x 1] Solution:

We can look at the slope of the line through (1, 1) and (1 + h, (1 + h)2)ash gets closer h2+2h and closer to 0, In problem 3, we calculated this to be h , and this goes to 2 as h gets closer to 0. Thus, the slope is 2. Now, we know the slope of our desired line is 2, we know a point on it, namely, (1, 1), so we can write the equation in point-slope form y 1=2(x 1), or y =2x 1. Thus the equation of the tangent line is f(x)=2x 1.

5. What is the slope and equation of the tangent line at (a, a2)? [Answer: 2a; f(x)= 2ax a2] Solution:

Using the same method as for the point (1, 1), we look at the slope of the line through (a, a2)and(a + h, (a + h)2). In other words, we look at the slope of the secant lines through the points (a, a2)and(a + h, (a + h)2), a point very close to (a, a2). The slope of this line is (a+h)2 a2 2ah+h2 a+h a = h . This gets closer and closer to 2a as h gets closer to 0. Thus, the slope is 2a. 1 Again, using point slope form, we know that the point on the tangent line is (a, a2)and the slope is 2a, so we get y a2 =2a(x a), or y =2ax a2. Thus, the equation of the tangent line at (a, a)isf(x)=2 ax a2.

6. Determine whether the following limits exist and if so, calculate the value:

3 i. limx 2 2x +1[answer:17] ! ii. The limit as x 1: [answer: -1] !

iii. limx ⇡ sin(1/x)[answer:doesnotexist] !

iv. limx 0 1/x [answer: does not exist] ! x3 6x2+3x+10 v. limx 1 x 2 [answer: -8] !

Solution:

i. This is a polynomial that is deﬁned at x =2,sohere,wecanjustpluginx =2toget 2(23)+1=17.

ii. Again here, we can just look at the value of f(1), which is 1. This is because the points with x-value near 1 have f(x)near 1. iii. The limit does not exist: we can always ﬁnd values of x close to 0 with sin(1/x)=1 and values of x close to 0 with sin(1/x)=0,namely,foranyn, we can take x =1/⇡n,and sin(1/x)=sin(⇡n)=0.But,form odd, we can take x =2/⇡m, and sin(1/x)=1or 1. The values of m and n can be as big as we want, so x can be as close to 0 as we want and we will get very di↵erent values of f(x).

2 iv. The value of 1/x gets arbitrarily large as x gets closer and closer to 0, so it does not approach any limit point. This limit does not exist.

v. This one is easiest to do by plugging in values of x close to 1 to see what value f(x) takes. However, a systematic way to do this is to factor the polynomial in the numerator as (x 2)(x+1)(x 5) (x 2)(x +1)(x 5). Then, we calculate limx 1 =lim(x 1(x +1)(x 5) = ! x 2 ! (1 + 1)(1 5) = 2( 4) = 8. We can do this because we are only concerned about points near x =1.