1.2 the Slope of the Tangent a Lines the Slope of a Line

Calculus and Vectors – How to get an A+

1.2 The Slope of the Tangent

A Lines Ex 1. The equation of the line L1 is: 2x − 3y + 6 = 0 .

a) Find the slope of the line L1 . 2 2 y = x + 2 ⇒ ∴ m = 3 1 3

b) Find the equation of the line L2 , that is parallel to the

line L1 and passes through the point P(2,−1) . 2 2 4 y − (−1) = (x − 2) ⇒ y = x −1− ⇒ 3 3 3

2 7 rise ∆y y − y ∴L2 : y = x − The slope of a line: m = = = 2 1 3 3 run ∆x x − x 2 1 c) Find the equation of the line L3 , that is perpendicular Slope y-intercept equation of a line: y = mx + b to the line L1 and passes through the point Q(4,2) . Slope-point equation of a line: y − y1 = m(x − x1 ) 1 3 3 m = − = − ⇒ y − 2 = − (x − 4) ⇒ For parallel lines, slopes are equal: m = m 3 m 2 2 1 2 1 For perpendicular lines, slopes are negative 3 reciprocal: ∴L3 : y = − x + 8 2 1 m2 = − or m1m2 = −1 m1

B Function Notation Ex 2. Let f (x) = x 2 −1 . Find: y = f (x) a) f (1) where: x is the argument, input or independent variable f (1) = 12 −1 = 0 ⇒ ∴ f (1) = 0 y is the value, output or dependant variable b) f (a) f is the name of the function. ∴ f (a) = a 2 −1 c) f (a + 1)

f (a + 1) = (a + 1)2 −1 = a 2 + 2a + 1 −1 = a 2 + 2a

∴ f (a + 1) = a 2 + 2a

C Secant Line 2 Ex 3. Consider f (x) = . Find the equation of the Let y = f (x) be a function and P(x1, y1 ) and Q(x2 , y2 ) x + 1 two points on its graph. secant line that passes through the points A(0,2) and The slope of the secant line that passes through the B(−3,−1) . points P and Q is given by: y − y (−1) − (2) m = 2 1 = =1 rise ∆y y2 − y1 f (x2 ) − f (x1 ) m = = = = x2 − x1 (−3) − (0) run ∆x x − x x − x 2 1 2 1 y − 2 =1(x − 0)

∴ y = x + 2

If P(a, f (a)) and Q(a + h, f (a + h)) then the slope of the secant line is given by: f (a + h) − f (a) m = h D Tangent Line As the point Q approaches the point P , the secant line approaches the tangent line at P . See the diagram on the right side. The slope of the tangent line at P(a, f (a)) is:

f (a + h) − f (a) m = lim (1) h→0 h

E Graphical Computation Ex 4. Find the slope of the tangent line to the curve 1. Draw the tangent line using a ruler y = 2 x −1 at the point P(5,4) . 2. Choose two points on the tangent line A(x , y ) and 1 1 B(x2 , y2 ) 3. Use the formula: y − y m ≅ 2 1 x2 − x1 to estimate the slope of the tangent line.

Note. The slope of the tangent line can be calculated using: m = tanα where α is the angle between the tangent line and the positive direction of the x-axis.

A(1,2), B(9,6)

x1 =1, y1 = 2, x2 = 9, y2 = 6 y − y 6 − 2 1 1 m = 2 1 = = ∴m = x2 − x1 9 −1 2 2

F Numerical Computation Ex 5. Consider y = f (x) = x 3 . Estimate numerically the f (a + h) − f (a) 1. Use the formula m ≅ (2) slope of the tangent line at P(1,1) using h h = 0.1, h = 0.0001, h =10−20 . 2. Choose a sequence h1, h2 , h3 , ... → 0 . 1 2 3 a =1, f (1) =1 3. Compute m1, m2 , m3 , .. . 4. Observe the pattern and conclude. (1+ 0.1)3 −1 m ≅ ≅ 3.31 5. Be careful at “difference catastrophe”. 1 0.1 (1+ 0.0001)3 −1 Note. The difference catastrophe is related to the m2 ≅ ≅ 3.0003 limited capacity of memorizing numbers by any 0.0001 technologic device (scientific calculator, computer, (1+10−20 )3 −1 m3 ≅ ≅ 0 (difference catastrophe) etc). If two numbers are very close, then they have the 10−20 same internal representation in the memory. ∴m ≅ 3

G Algebraic Computation Ex 6. Find the slope of the tangent line to the graph of f (a + h) − f (a) 2 1. Use the formula m = lim . y = f (x) = x − 3x at the point P(1,−2) . h→0 h a =1, f (1) = −2 2. Do not substitute h by 0 because you will get the 2 0 f (a + h) = f (1 + h) = (1 + h) − 3(1 + h) = indeterminate case . 2 2 0 =1 + 2h + h − 3 − 3h = −2 − h + h 3. Compute algebraic the difference quotient f (a + h) − f (a) f (1 + h) − f (1) f (a + h) − f (a) DQ = = = DQ = until you succeed to cancel out h h h 2 2 (−2 − h + h ) − (−2) h − h the factor h . = = = h −1 h h 4. Substitute in the remaining expression h by 0 . m = lim(h −1) = 0 −1 = −1 ⇒ ∴m = −1 h→0

Ex 7. Find the equation of the tangent line to the graph Ex 8. Consider y = f (x) = x 2 − 2x . 1 of y = f (x) = at the point P(2,1) . x −1 a) Find the slope of the tangent line at the generic point a = 2, f (a) = f (2) =1 P(a, f (a)) . 1 1 2 f (a + h) = f (2 + h) = = f (a) = a − 2a 2 + h −1 1+ h f (a + h) = (a + h)2 − 2(a + h) = a2 + 2ah + h2 − 2a − 2h 1 1 1 −1 −1 +1 f (a + h) − f (a) (a2 + 2ah + h2 − 2a − 2h) − (a2 − 2a) f (a + h) − f (a) 1+ h 1+ h 1+ h DQ = = DQ = = = h h h h h 1 +1 2ah + h2 − 2h 1+ h = = 2a + h − 2 1 h −1 − h −1 m = lim(2a + h − 2) = 2a − 2 ⇒ ∴m = 2a − 2 = 1+ h = = x→0 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ h⎜ +1⎟ h(1+ h)⎜ +1⎟ (1+ h)⎜ +1⎟ b) Find the point where the tangent line is horizontal. ⎝ 1+ h ⎠ ⎝ 1+ h ⎠ ⎝ 1+ h ⎠ ⎧m = 0 −1 −1 1 ⎨ ⇒ 0 = 2a − 2 ⇒ a =1, f (1) = −1⇒ ∴P(1,−1) m = lim = = − m = 2a − 2 x→0 ⎛ 1 ⎞ ⎛ 1 ⎞ 2 ⎩ (1+ h)⎜ +1⎟ (1+ 0)⎜ +1⎟ ⎝ 1+ h ⎠ ⎝ 1+ 0 ⎠ c) Find the point P such that mP = 2 . ⎧ 1 ⎪m = − 1 1 2a − 2 = 2 ⇒ a = 2, f (2) = 0 ⇒ ∴P(2,0) ⇒ y −1 = − (x − 2) ⇒∴ y = − x + 2 ⎨ 2 ⎪ 2 2 ⎩P(2,1) d) Find the point P such that the tangent line at P is

perpendicular to the line L2 : x − 3y = 3 .

1 1 1 L2 : y = x −1⇒ m2 = ⇒ m = − = −3 3 3 m2 ⎧m = −3 1 ⎛ 1 ⎞ 1 5 ⎨ ⇒ −3 = 2a − 2 ⇒ a = − , f ⎜− ⎟ = +1 = ⎩m = 2a − 2 2 ⎝ 2 ⎠ 4 4 ⎛ 1 5 ⎞ ∴P⎜− , ⎟ ⎝ 2 4 ⎠

Reading: Nelson Textbook, Pages 10-18 Homework: Nelson Textbook: Page 18, #1a, 2a, 3ac, 4a, 5a, 6a, 8a, 9a, 10a, 11a, 15, 21, 25