This Week. Basic Problem. Instantaneous Rate of Change. Compute the tangent line to the curve y = f (x) at x = a.

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1 This Week Basic Problem Compute the tangent line to the curve y = f (x) at x = a. Read Sections 2.7,2.8 and 3.1 in Stewart Homework #2 due 11:30 Thursday evening worksheet #3 in section on Tuesday slope of line through P and Q = f (b) f (a). b a f (x) f (a) f (a + h) f (a) m = slope of tangent line = lim = lim. x a x a h 0 h Instantaneous Rate of Change We say that f (x) f (a) lim x a x a is the instantaneous rate of change of y = f (x) f (x) f (a) f (a + h) f (a) m = slope of tangent line = lim = lim. x a x a h 0 h equation of tangent line : y f (a) = m(x a). For example if f (t) is the distance traveled at time t. Then f (t) f (a) lim t a t a is the instantaneous velocity at t = a.

2 Find the equation of the tangent line to y = 1/ 5 2x at x = a. f (x) f (a) m = lim x a x a 1 = lim 5 2x 1 5 2a x a x a 5 2a 5 2x = lim x a ( 5 2a 5 2x)(x a) ( 5 2a 5 2x)( 5 2a + 5 2x) = lim x a ( 5 2a 5 2x)(x a)( 5 2a + 5 2x) = lim x a (5 2a) (5 2x) ( 5 2a 5 2x)(x a)( 5 2a + 5 2x) m = lim x a 2(x a) ( 5 2a 5 2x)(x a)( 5 2a + 5 2x) 2 = lim x a ( 5 2a 5 2x)( 5 2a + 5 2x) 2 = ( 5 2a 5 2a)(2 5 2a) 1 = (5 2a) 3/2 Thus the equation of the tangent line is y 1/ 1 5 2a = (x a). (5 2a) 3/2 If the distance (in feet) traveled by a car at time t is given by 1 s(t) = 5 2t then the instantaneous velocity of the car at time t = a is 1 ft/sec. (5 2a) 3/2 A cylindrical tank holds 100,000 gallons of water which can be drained from the bottom of the tank in an hour. By Toricelli s law the volume V (t) remaining at time t (in minutes) is V (t) = 100, 000(1 t/60) 2 gallons, 0 t 60. What is the rate (as a function of time) at which water is flowing out of the tank? What is the rate at which water is flowing out half way through the hour?

3 We are looking for the instantaneous rate of change which is given by V (t) V (a) lim t a t a = lim t a 100, 000(1 t/60) 2 100, 000(1 a/60) 2 t a = 100, 000 lim t a (1 t/60) 2 (1 a/60) 2 t a = 100, 000 lim t a 1 2t/60 + (t/60) a/60 (a/60) 2 t a ( 1/30)(t a) + (1/3600)(t 2 a 2 ) = 100, 000 lim t a t a ( 1/30)(t a) + (1/3600)(t a)(t + a) = 100, 000 lim t a t a = 100, 000 lim(( 1/30) + (1/3600)(t + a)) t a = 100, 000(( 1/30) + (1/3600)(2a)) Thus the instantaneous rate of change is 100, 000(( 1/30) + (1/3600)(2a)). Halfway through the hour we are at t = 30 and 100, 000( 1/30 + (1/3600)(2(30))) 1, 667 gallons per minute. Derivatives At t = 0 a car is at the 15 mile marker on the freeway. For the next hour it travels at a constant speed of 55 miles per hour. The car slows gradually over a 2 minute period and comes to a stop. After 20 minutes it restarts gradually speeding up to 65 miles per hour over a two minute period. Then the car goes at 65 miles per hour for two hours and then it slows to a stop over a three minute period. Sketch a graph of the position of a car as a function of time. Given a function y = f (x) and a value x = a the derivative of f at x = a is f f (x) f (a) f (a + h) f (a) (a) = lim = lim. x a x a h 0 h As we saw last time this is the slope of the tangent line to y = f (x) at x = a. This is also the instantaneous rate of change of f (x) at x = a.

4 Notation We have a lot of different notation for the derivative of y = f (x). y d dx f (x) dy dx df dx f (x) ḟ(x) Df (x) Dxf (x) A ball is thrown up in the air with a speed of 40ft/sec. It s height (in feet) after t seconds is given by g(t) = 40t 16t 2. Calculate the instantaneous velocity as a function of t. When does it reach its maximum height? g (t) = lim h 0 g(t + h) g(t) h = lim h 0 40(t + h) 16(t + h) 2 (40t 16t 2 ) h = lim h 0 40t + 40h 16t 2 32th 16h 2 40t + 16t 2 h 40h 32th 16h 2 = lim h 0 h = lim 40 32t 16h h 0 = 40 32t Looking at the graph we can see that g(t) is increasing until t = 5/4 and decreasing afterwards. Notice that this is when the instantaneous velocity changes from positive to negative. Thus the ball reaches its maximum height of 25 feet at time t = 5/4.

5 Speed and Velocity Find the formula for g (t) and graph g (t) as a function of time. Find the formula for g (t) and graph g (t) as a function of time. If f (t) is the position of a car at time t. Then f (a) is the velocity of a car at time t = a. Then f (a) is the speed of a car at time t = a. Law of Decreasing Marginal Returns Petroleum beneath the earth s surface The Law of diminishing marginal returns says that each additional person (dollar invested in machinery, etc.) will increase your production by less than the previous. Restate the law of diminishing marginal returns in terms of derivatives. Petroleum has accumulated under the earth s surface for millions of years. Let P(t) be the number of barrels of petroleum that are below the earth s surface t years after We are currently pumping 63 million barrels per day. The earth s proven reserves are at least 1.2 trillion barrels. What does this tell you about P(t)? about P (t)? We are currently pumping at a rate of 63 million barrels per day is about 23 billion per year. This is the instantaneous rate of change so P (10) 23 billion barrels per year. The earth s proven reserves are at least 1.2 trillion barrels so P(10) 1.2 trillion barrels.

6 ipad sales Apple started selling the ipad on April 3. Let I(t) be the number of ipads sold t days after the launch. What do the following statements mean in terms of I(t) and its derivatives. Apple will sell 12 million ipads in 2010 and 28 million in By April Apple will be selling 1.8 million ipad per month. Derivatives and Graphs Below is the graph of a function y = f (x) and y = f (x). What does this tell you about I(t)? about I (t)? I(273) = 12 million and I(638) = 40million I(393) I(363) = 1.8 million or I (363) I(393) I(363) 1, 800, 000 = = 60, Ways a function can fail to be differentiable if f (x) > 0 for all x on [a, b] then f is increasing on [a, b] if f (x) < 0 for all x on [a, b] then f is decreasing on [a, b] if f (x) = 0 for all x on [a, b] then f is constant on [a, b] We say that f is differentiable at x = a if f (a) exists. There are a number of ways that f (x) can fail to be differentiable at x = a. 1 f is not continuous 2 f has a sharp corner or 3 f has a vertical tangent line

7 Ways a function can fail to be differentiable Is f (x) = x differentiable at x = 0? To see if f f (h) f (0) h 0 (0) = lim = lim h 0 h h 0 h exists we take both one sided limits. h 0 h lim = lim h 0 + h h 0 + h = lim 1 = 1 h 0 + h 0 h lim = lim h 0 h h 0 h = lim 1 = 1 h 0 This function is not differentiable at x = 1, x = 0 and x = 1. As h 0 h 0 1 = lim lim = 1 h 0 + h h 0 h Thus limh 0 h 0 h does not exist and f (x) = x is not differentiable at x = 0. Graphically we can see that y = x doesn t have a well defined tangent line at x = 0. Below is the graph of a function y = f (x). Where is f differentiable? Sketch the graph of the function f (x).

8 Below is the graph of a function y = f (x). Below is the graph of a function y = f (x). Sketch the graph of the function f (x). Sketch the graph of the function f (x) if f (0) = 0. Estimating Derivatives from graphs Draw the graph of a function with the following properties. 1 f (x) > 0 on (, 3) and (2, ) 2 f (x) < 0 on ( 3, 2) 3 f (x) < 0 on ( 5, 1) 4 f (x) > 0 on (, 5) and (1, ) 5 f (x) < 0 on (, 3.5) and (.5, 1.5) 6 f (x) > 0 on ( 3.5,.5) and (1.5, ) Below is the graph of the function y = f (x). Estimate the values of f (1) and f (2) and f (3).

9 Estimating Derivatives from graphs To estimate the value of f (1) we draw the tangent line at x = 1. Estimating Derivatives from graphs To estimate the value of f (2) we draw the tangent line at x = 2. The tangent line at (1, f (1)) appears to go through (0,1.3) and (2,2) so its slope is about.35 and f (1).35. The tangent line at (2, f (2)) appears to go through (2,1.5) and (4,.1) so its slope is about.7 and f (2).7. Estimating Derivatives from graphs Four functions To estimate the value of f (3) we draw the tangent line at x = 3. Below is the graph of a function f (x) and its derivatives f (x), f (x) and f (x). Determine which of the functions is which. The tangent line at (3, f (3)) appears to go through (2,2) and (4,-1) so its slope is about 1.5 and f (3) 1.5.

10 Below is the graph of the function y = f (x). Put the terms below in increasing order. Normal line to a parabola The graph of the equation y = x 2 8x + 9 forms a parabola. For a point (a, a 2 8a + 9) on the graph find the 1 equation of the tangent line. 2 equation of the normal line. 3 other point on the graph where the normal line intersects the graph. f (3) (3) < 0 < f f (3) (5) < f f (5). (3) Applications Draw the graph of a function satisfying the following conditions 1 f (x) > 0 on [0. ] 2 f (x) < 0 on [0. ] 3 f (x) > 0 on [0. ]

11 Continuity and differentiability Derivative Rules f is continuous at x = a does not imply that f is differentiable at x = a but f is differentiable at x = a does imply that f is continuous at x = a. Limit laws give us the following. Suppose f and g are differentiable the 1 (f + g) (x) = f (x) + g (x) 2 (f g) (x) = f (x) g (x) 3 (cf ) (x) = cf (x) Power Rule Find the derivative of the function f (x) = 1/x. Suppose f (x) = x n for some positive integer n. Find f (x) f (x + h) n x n (x) = lim h 0 h = lim h 0 x n + nx n 1 h + + nxh n 1 + h n x n h nx n 1 h + + nxh n 1 + h n = lim h 0 h = lim nx n nxh n 2 + h n 1 h 0 = nx n 1. 1 f (x) = lim h 0 x+h x h = lim h 0 1 x+h 1 x x(x + h) h x(x + h) x (x + h) = lim h 0 x(x + h)h = lim h 0 = lim h 0 = 1 x 2 h x(x + h)h 1 x(x + h)

12 Find the derivative of the function f (x) = x. x + h h f (x) = lim h 0 h x + h x x + h + x = lim h 0 h x + h + x x + h x = lim h 0 h( x + h + x) h = lim h 0 h( x + h + x) 1 = lim h 0 ( x + h + x) 1 = 2 x We just showed that d dx (x 1 ) = ( 1)x 2 and d dx (x 1/2 ) = (1/2)x 1/2 This suggests that the power rule may be true for more than just positive integers. Power Rule Derivatives of Polynomials In fact the following is true Theorem If n is any real number d dx (x n ) = nx n 1 Using the rules 1 (f + g) (x) = f (x) + g (x) 2 (cf ) (x) = cf (x)) 3 (x n ) = nx n 1 we can calculate the derivative of any polynomial.

13 Examples Find the derivative of f (x) = f (x) = 7x 4 + 3x 3 + 2x 2 + 5x + 9 ( ) 7x 4 + 3x 3 + 2x 2 + 5x + 9 = (7x 4 ) + (3x 3 ) + (2x 2 ) + (5x) + (9) = 7(x 4 ) + 3(x 3 ) + 2(x 2 ) + 5(x) + 0 = 7(4x 3 ) + 3(3x 2 ) + 2(2x) + 5(1) = 28x 3 + 9x 2 + 4x + 5. Examples Find the derivative of f (x) = = f (x) = 3 x + 2 x + 2x x.1 ( 3 x + 2 ) + 2x x.1 x ( 3x 1/2 + 2x 1/2 + 2x x.1) = ( 3x 1/2 ) + (2x 1/2 ) + (2x 4.3 ) + (5x.1 ) = 3(x 1/2 ) + 2(x 1/2 ) + 2(x 4.3 ) + 5(x.1 ) = 3((1/2)x 1/2 ) + 2(( 1/2)x 3/2 ) + 2(( 4.3)x 5.3 ) + 5((.1)x.9 ) = 1.5x 1/2 x 3/2 8.6x x.9. Derivatives of Exponentials Find the derivative of g(x) = e cx. g e c(x+h) e cx (x) = lim h 0 h e cx e ch e cx = lim h 0 h e cx (e ch 1) = lim h 0 h = ce cx e ch 1 lim h 0 ch e ch 1 = cg(x) lim ch 0 ch e y 1 = cg(x) lim. y 0 y So if we can figure out the value of the last limit then we know the derivative at every point. Here is a graph of m(y) = ey 1 y. We can see that as y 0 m(y) 1 so d dx (ex ) = e x. This is the reason that we choose to use base e.

14 Derivatives of Exponentials Theorem If f (x) = e cx then f (x) = ce cx. As a = e ln(a) and a x = e ln(a)x another way to write this is Find the derivative of g(t) = 5 4 t + π t. ( g (t) = 5(4 t ) + π( ) t) = 5(4 t ) + π( t) = 5(ln(4))4 t + π(t 1/2 ) = 5(ln(4))4 t + π 2 t 1/2. Theorem If f (x) = a x then f (x) = ln(a)a x. Find the derivative of g(t) = 5 e 2t + a/t 2. g (t) = (5 e 2t + a/t 2) = 5(e 2t ) + a(t 2 ) = 5( 2)e 2t 2at 3. At what point on the curve y = f (x) = 2 x is the tangent line parallel to the line y = 3x? We want to find x such that f (x) = 3. f (x) = (ln(2))2 x. Setting f (x) = 3 we get 3 = ln(2)2 x or 2 x = 3/ ln(2). ( ) 3 ln(2)x = ln ln(2) or x = 1 ( ) 3 ln(2) ln. ln(2)

15 ( ( ) ) Thus the point is 1 ln 2 ln 3 ln(2), 3/ ln(2) and the line is ( y 3/ ln(2) = 3 x 1 ( )) 3 ln 2 ln. ln(2)