Tangent Lines and Rates of Cange 9-2-2005 Given a function y = f(x), ow do you find te slope of te tangent line to te grap at te point P(a, f(a))? (I m tinking of te tangent line as a line tat just skims te grap at (a, f(a)), witout going troug te grap at tat point Tis is a vague description, but it will do for now) Here s te idea Pick a point (a +, f(a + )) nearby, and draw te line connecting (a, f(a)) to (a +, f(a + )) (A line connecting two points on a grap is called a secant line) (a+,f(a+)) (a,f(a)) P f(a+) f(a) a a+ Tus, represents ow muc you moved over in te x-direction Te line as slope f(a + ) f(a) (a + ) a = f(a + ) f(a) If you slide te second point (a +, f(a + )) along te grap toward P, te secant line gets closer and closer to te tangent line Algebraically, tis amounts to taking te it as 0 Tus, te slope of te tangent at x = a is f(a + ) f(a) m tan Example Let f(x) = x 2 (a) Find te slope of te secant line joining (0, f(0)) to (, f()) f() f(0) 0 = 9 0 0 = 9 = (b) Find te slope of te tangent line to f(x) at x = In tis case, I let a = in te equation for m tan and compute te it: f( + ) f() ( + ) 2 2 6 + 2 m tan (6 + ) = 6 0
Anoter form of te tangent line formula is f(x) f(a) m tan x a x a You can get tis formula from te previous one by letting = x a Ten x = a +, so 0 gives x a Example Find te slope of te tangent line to y = x at (a, f(a)) f(x) f(a) m tan x a x a x a x a x a x a ax x a x a a x x a ax(x a) x a ax = a 2 0 20 0 - -2-2 -0-20 -0 Te grap of y = is a rectangular yperbola Notice tat by not plugging in a specific number for a, x I ve obtained ( a formula wic I can use for any a For example, te slope of te tangent at a = (ie at te point, ) ) is m tan = 2 = 9 Tere is anoter interpretation of te slopes of te secant line and te tangent line Te slope of te secant line joining (a, f(a)) to (b, f(b)) is f(b) f(a) b a Tis is te cange in f divided by te cange in x, so it represents te average rate of cange of f as x goes from a to b (ie on te interval a x b) Wat is te slope of te tangent line at a? It represents te instantaneous rate of cange at x = a (Sometimes people get lazy and just say rate of cange to mean instantaneous rate of cange ) Example Let f(x) = x (a) Find te average rate of cange of f(x) on te interval x 4 f(4) f() 4 = 4 4 = 2 = (b) Find te instantaneous rate of cange of f(x) at x = 4 2
Te instantaneous rate of cange of f(x) at x = 4 is m tan at a = 4 I ll use te second formula for m tan : I set a = 4 and compute te it: f(x) f(4) m tan x 4 x 4 x 4 f(x) f(a) m tan x a x a x 4 x 4 x 4 x 4 (x 4)( x + 2) x 4 x 2 x 2 x + 2 x 4 x 4 x 4 x 4 = x + 2 x + 2 = 4 Tus, te instantaneous rate of cange of f(x) at x = 4 is Tis means tat if f continued to cange 4 at te same rate, ten for every 4 units tat x increased, te function would increase by unit Of course, te function does not continue to cange at te same rate In fact, te rate of cange of te function canges! te rate of cange of te function is a function itself Suppose tat te function under investigation gives te position of an object moving in one dimension (Tink of someting moving left or rigt along te x-axis, or an object tat is trown straigt upward, and wic eventually falls back to eart) For instance, suppose tat s(t) is te position of te object at time t Te average velocity of te object from t = a to t = b is te cange in position divided by te time elapsed: s(b) s(a) v avg = b a Notice tat tis is te same as te slope of te secant line to te curve, or te average rate of cange Te instantaneous velocity at t = a is s(a + ) s(a) v(a) Tis is te slope of te tangent line to te curve, or te instantaneous rate of cange You can also use te second formula s(t) s(a) v(a) t a t a Rougly speaking, te instantaneous velocity measures ow fast te object is travelling at a particular instant Example Te position of an object at time t is s(t) = t 2 5t + 6 (a) Find te average velocity from t = 4 to t = 5 v avg = s(5) s(4) 5 4 = 6 2 = 2 (b) Find te average velocity from t = to t = 4 v avg = s(4) s() 4 = 2 2 = 0
Wat does tis mean? Notice tat s() = 2 and s(4) = 2 In oter words, te object moved around from t = to t = 4, but it wound up back were it started Since te net cange in position was 0, te average velocity was 0 (c) Find te instantaneous velocity wen t = I set a = in I get s(t) s(a) v(a) t a t a s(t) s() t 2 5t + 6 0 t 2 5t + 6 (t )(t 2) v() (t 2) = t t t t t t t t t People wo ave seen calculus before know tat m tan is usually called te derivative of f(x) at a It is denoted by f (a) or y dy (a) or dx or df(x) or D x f(a) dx Tat is, te derivative of y = f(x) at x = a is given by f f(a + ) f(a) (a) f (a) gives te instantaneous rate of cange of f at a, or te slope of te tangent line to te grap of y = f(x) at (a, f(a)) Te derivative is a function in its own rigt Since x is usually used to denote te input variable for a function, it s common to write te definition of te derivative in tis form: f f(x + ) f(x) (x) f is differentiable at x if f (x) exists tat is, if te it above is defined Example Compute f (x) for f(x) = x f f(x + ) f(x) (x) 0 x + x 0 x x + x x + x x + x x + = x x + x + x + x x + x (x + ) x + x + x x + ( x + x + ) x x + ( x + x + ) = 0 x x + ( x + x + ) = 2x /2 Example Suppose Is f differentiable at x =? { 2 x if x f(x) = if x > x 4
f f( + ) f() () However, te definition of f(+) depends on weter is positive or negative I need to take te leftand rigt-and its at Te rigt-and it is f( + ) f() 0 + 0 + + 0 + ( + ) + 0 + + = Te left-and it is 0 + ( + ) 0 + + = f( + ) f() 2 ( + ) 0 0 0 ( ) = 0 Since te left- and rigt-and its agree, te two-sided it exists Tus, Tis sows tat f is differentiable at x = f f( + ) f() () = Example A differentiable function is continuous Geometrically, a differentiable function as a tangent line at eac point of its grap You d suspect tat tis would rule out gaps, jumps, or vertical asymptotes typical discontinuities In fact, te requirement tat a differentiable function ave a tangent line at eac point means tat its grap as no corners all of te curves and turns are smoot To see algebraically wy tis result is true, suppose f(x) is differentiable at a point c By definition, f f(x) f(c) (c) x c Ten ( ) ( f (c) (x c) ) f(x) f(c) ( ) x c (x c) On te one and, (x c) = 0, so te left side is 0 On te oter and, te product of te its is te it of te product, so I can rewrite tis as Tis says tat f is continuous at c f(x) f(c) 0 (x c) (f(x) f(c)) x c f(x) f(c) = f(c) 5
Example Te picture below sows tat grap of a function y = f(x) Sketc te grap of f (x) I ll do eac piece separately from left to rigt Te left and piece starts out wit a small positive slope Te slope increases till it is large and positive at te asymptote Te piece in te middle starts out wit a big positive slope at te left-and asymptote It decreases to 0 tere s a orizontal tangent at te top of te bump It continues to decrease, becoming big and negative at te rigt-and asymptote Finally, te rigt-and piece starts out wit a big negative slope near te asymptote As you go out to te rigt, te slope continues to be negative, but te curve flattens out tat is, te slope approaces 0 Putting tese observations togeter produces a picture like tis: Example An often-used rule of tumb is: Te derivative is undefined at a place were a grap as a corner Here s an example wic illustrates tis point Suppose f(x) = { 2x if x < 0 x x 2 if x 0 6
Here s te grap: -2-2 - -2 - -4 It looks as toug tere migt be a corner at x = 0, but it s ard to tell Compute te derivative at 0: f f(0 + ) f(0) f() (0), since f(0) = 0 Since f is defined in two pieces, I ave to compute te it on te left and rigt: f() 0 2 0 2 = 2, 0 f() 0+ 2 ( ) = 0+ 0+ Te left- and rigt-and its do not agree Terefore, te two-sided it f (0) is undefined f is not differentiable at x = 0 Te left- and rigt-and its I computed are sometimes called te left- and rigt-and derivatives of f at x = 0 Intuitively, tey give te slope of te tangent as you come in from te left and rigt, respectively Tus, te left-and derivative at 0 is 2 and te rigt-and derivative at 0 is c 2005 by Bruce Ikenaga 7