Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =


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1 Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a f (x) = lim x a Find the vertical asymptotes of f (x) = x 3 x 2 9. Solution: lim f (x) = x a + f (x) = x a + Candidates to consider are points that give division by zero: x = ±3. x 3 We can use a word argument to deduce lim x 3 x 2 9 =. Thus, x = 3 is a vertical asymptote. On the other hand, x = 3 is NOT a vertical asymptote since the limit is: x 3 lim x 3 x 2 9 = lim 1 x 3 x + 3 = 1 6 and is not equal to ±. x = 3 is a hole in the function since it cancels. AMAT 217 (University of Calgary) Fall / 24
2 Horizontal asymptotes Definition of Horizontal Asymptote The line y = L is a horizontal asymptote of f (x) if L is finite and either lim f (x) = L or lim f (x) = L. x x Find the horizontal asymptotes of f (x) = x x. Solution: We must compute two limits: x and x. x. Notice that for x arbitrarily large that x > 0 so that x = x. In particular, for x in the interval (0, ) we have lim x x. x x = lim x x x = 1. Notice that for x arbitrarily large negative that x < 0 so that x = x. x For x in the interval (, 0) we have lim x x = lim x x x = 1. Therefore there are two horizontal asymptotes, namely, y = 1 and y = 1. AMAT 217 (University of Calgary) Fall / 24
3 Derivative Rules Know these! (c) = 0 (ax) = a (x n ) = nx n 1 (cf ) = cf (f ± g) = f ± g (fg) = f g + fg ( ) f = f g fg g g 2 [f (g(x))] = f (g(x)) g (x) (e x ) = e x (ln x) = 1 x (sin x) = cos x (cos x) = sin x (tan x) = sec 2 x (function) (function) Use logarithmic differentiation Inverse Trig Derivatives ( sin 1 x ) = 1 1 x 2 ( cos 1 x ) = 1 1 x 2 ( tan 1 x ) = x 2 Hyperbolic Derivatives (sinh x) = cosh x (cosh x) = sinh x (tanh x) = sech 2 x AMAT 217 (University of Calgary) Fall / 24
4 Find constants a, b so that f (x) = { ax 2 + 2x if x < 2 4b if x 2 is differentiable at x = 2. For differentiability at x = 2 we need f (x) to: (i) be continuous at x = 2, [Recall: differentiable continuous but continuous differentiable] (ii) have matching right and left derivatives at x = 2. The first condition requires that left and right hand limits be equal: lim f (x) = x 2 lim 2 x 2 (ax + 2x) = 4a + 4 and lim f (x) = x 2 + Thus, we require 4a + 4 = 4b, that is, a + 1 = b. lim = 4b x 2 +(4b) The second condition requires that left and right derivatives are equal: f (2) = d ( ax 2 + 2x) = (2ax + 2) dx = 4a + 2 and f +(2) = d x=2 dx x=2 (4b) = 0 x=2 Thus, we require 4a + 2 = 0, that is, 2a + 1 = 0. We have two equations and solving gives a = 1/2 and b = 1/2. Therefore, for f (x) to be differentiable at x = 2 we require a = 1/2 and b = 1/2. AMAT 217 (University of Calgary) Fall / 24
5 (From past exam/midterm) { 1 4x if x < 1 Suppose f (x) = x if x 1 Is f (x) differentiable at x = 1? If so, find f ( 1). Solution: First, we need to verify f (x) is continuous at x = 1. Thus, we need left and right hand limits be equal: lim f (x) = x 1 lim 4x) = 3 and lim x 1 ( 1 f (x) = x 1 + Thus, f (x) is continuous at x = 1 since f ( 1) = lim f (x) = 3. x 1 Now we need left and right hand derivatives be equal: From the left of x = 1, the derivative is: f ( 1) = d dx ( 1 4x) x= 1 = 4 lim + 2) = 3 x 1 +(x4 From the right of x = 1, the derivative is: f + ( 1) = d ( x ) dx x= 1 = 4x 3 x= 1 = 4( 1) 3 = 4 Since continuous and left/right derivatives match, we have f ( 1) = 4. AMAT 217 (University of Calgary) Fall / 24
6 Implicit Differentiation (with respect to x) 1 Differentiate both sides as usual, 2 Whenever you do the derivative of terms with y, multiply by dy dx. Find the equation of the tangent line to tan(x + y) = x + 1 at P = (0, π/4). Recall: The slope of the tangent line is the derivative dy dx P=(0,π/4). Differentiate both sides with respect to x thinking of y as a function of x: ( sec 2 (x + y) 1 + dy ) = dx } {{ } derivative of x + y The left side used the chain rule combined with (tan x) = sec 2 x. We have x = 0, y = π/4 and m = dy dx P=(0,π/4), so: sec 2 (0 + π/4)(1 + m) = 1 1 Since sec(π/4) = cos(π/4) = 1 1/ 2 = 2, we have: 2(1 + m) = m = 1 m = Thus, the equation of the tangent line is then (y π/4) = ( 1/2)(x 0). AMAT 217 (University of Calgary) Fall / 24
7 Logarithmic Differentiation 1 Take ln of both sides of y = f (x) to get ln y = ln f (x). 2 Simplify using log properties. 3 Differentiate with respect to x using (ln y) = y y on left side. 4 Multiply by y and replace y with original function y = f (x). Find h (x) given h(x) = (x 2 + 1) 2x. We take the natural log of both sides: ln h(x) = ln((x 2 + 1) 2x ) Using log properties we get: ln h(x) = 2x ln(x 2 + 1) Taking the derivative gives: h (x) h(x) = 2 2x ln(x2 + 1) + 2x x Solving for h (x) and replacing h(x) = (x 2 + ( 1) 2x gives: ) h (x) = (x 2 + 1) 2x 2 ln(x 2 + 1) + 4x2 x AMAT 217 (University of Calgary) Fall / 24
8 By comparing powers we see that 3 = 300/T so that T = 100. AMAT 217 (University of Calgary) Fall / 24 Application: Halflife The halflife formula is ) t/t y(t) = y 0 ( 1 2 y(t) = amount at time t, y 0 = original amount, t = time (in years), T = halflife (in years). Find the halflife of a radioactive substance if after 300 years only 12.5% of the original amount remains. We are given that y(300) is 12.5% of the original amount y 0. That is, y(300) = y 0. Thus, substituting t = 300 into the formula gives: ( ) 1 300/T y 0 = y 0 2 Cancelling y 0 s gives: ( ) 1 300/T = 2 ( ) 1 = 8 ( ) 1 300/T 2 ( ) 1 3 = 2 ( ) 1 300/T 2
9 Application: Newton s Law of Cooling The Newton s Law of Cooling formula is T (t) = T e + (T 0 T e)e kt T (t) = temperature of object at time t, T 0 = original temperature of object, T e = temperature of environment (medium/room temp), k = a constant dependent on the material properties of the object. An object of temperature 60 C is placed in a room with temperature 20 C. At the end of 5 minutes the object has cooled to a temperature of 40 C. What is the temperature of the object at the end of 15 minutes? We are given that T 0 = 60 and T e = 20: T (t) = e kt We are also given that T (5) = 40 which will allow us to find k: We need to find T (15): 40 = e k(5) 20 = 40e 5k k = ln(1/2) 5 T (15) = e ln(1/2) (15) 5 = e ln(1/2)3 = Therefore, the temperature at the end of 15 minutes is 25 C. ( ) 1 3 = 25 2 AMAT 217 (University of Calgary) Fall / 24
10 Application: Related Rates (Related Rates #12 on Worksheet) When air expands isothermally (i.e., expands at constant temperature), the pressure P and volume V are related by Boyle s Law: PV = C, where C is a constant. At a certain instant, the pressure is 40 pounds per square inch, the volume is 8 cubic feet and is increasing at the rate of 0.5 cubic feet per second. At what rate is the pressure changing? We are given: P = 40 psi V = 8 ft 3 dv dt = 0.5 ft3 /s We need to compute dp dt. Since we are already given the relationship between P and V we differentiate PV = C with respect to time (using the product rule on PV ): Plugging in the given information gives: dp dt V + P dv dt = 0 dp dt (8) + (40)(0.5) = 0 dp = 2.5 psi/s dt Thus, the pressure is decreasing by 2.5 pounds per square inch per second. AMAT 217 (University of Calgary) Fall / 24
11 Application: Newton s Method Newton s Method Goal: To approximate a root of f (x) = 0 with an initial guess of x = x 0. If x n is the current approximation, then the next approximation is: x n+1 = x n f (xn) f (x n) Use Newton s method with an initial approximation of x 0 = 1 to find the approximation x 2 of the equation x 3 4 = 0 by taking f (x) = x 3 4. If x is the current approximation, the next approximation is given by: g(x) = x f (xn) f (x = x x3 4 n) 3x 2 = 2x x 2 Given that x 0 = 1, we compute x 1 by plugging x 0 = 1 into g(x): x 1 = g(x 0 ) = g(1) = = 2 3 As x 1 = 2, we compute x 2 by plugging x 1 = 2 into g(x): x 2 = g(x 1 ) = g(2) = = Thus, the approximation x 2 is equal to 5/3. AMAT 217 (University of Calgary) Fall / 24
12 Application: Linearizations The linearization of a function is the equation of its tangent line. Linearization of f(x) at x=a L(x) = f (a) + f (a)(x a). Find the linearization of f (x) = tan(sin(x 1)) 4x + 1 at the point on the curve where x = 1. Recall that the slope of the tangent line at x = 1 is equal to f (1). We have f (x) = sec 2 (sin(x 1)) cos(x 1) 1 4 Thus, when x = 1 we have: f (1) = sec 2 (sin(0)) cos(0) 4 Since sin 0 = 0, cos 0 = 1 and sec 0 = 1 cos 0 = 1 1 = 1: f (1) = sec 2 (0) 4 = 1 4 = 3 Now we use the pointslope formula for a straight line: y y 1 = m(x x 1 ). The slope is m = 3, and x 1 = 1. To get y 1 we sub x = 1 into f (x): f (1) = tan(sin(0)) 4(1) + 1 = 3 Thus, the equation of the tangent line at x = 1 is: y ( 3) = ( 3)(x 1) L(x) = 3x AMAT 217 (University of Calgary) Fall / 24
13 Geometry Some application problems require geometry, so it helps to go over the basics: Pythagorean Theorem know how to get equations using similar triangles circumference of a circle of radius r is C = 2πr area of a rectangle is length times width (A = lw) area of a circle of radius r is A = πr 2 area of a triangle with base length b and height h is A = bh 2 volume of a rectangular box is length times width times height (V = lwh) volume of a cylinder is V = πr 2 h volume of a right circular cone with base radius r and height h is V = 1 3 πr 2 h surface area of a box: you must add up the areas of six rectangles surface area of a cylinder: you must add up the areas of two circles and one rectangle perimeter: you must add up the length of a bunch of line segments trigonometric ratios and SOH CAH TOA cost associated with length: (cost of material) times (length) cost associated with area: (cost of material) times (area) : If building a rectangular fence with front fencing costing $4/m while sides and back fencing is $2.50/m (since you want the front to look nice and use more expensive material), the cost function is then: C = 4(x) + 2.5(y + x + y) where x is length of front and back, y is length of left and right sides. AMAT 217 (University of Calgary) Fall / 24
14 Absolute Extreme Points To find absolute extrema of a continuous function f (x) on a closed interval [a, b], plug in critical points, singular points and endpoints into f (x) (that are in the interval) to get the highest and lowest points of the function. Find the absolute extrema of f (x) = x on the interval [ 1, 2]. The derivative is: f (x) = 2x x = 0 is a critical point (there are no singular points) Plugging in endpoints and critical points belonging to [ 1, 2] gives: Critical Points: f (0) = = 1 Singular Points: None Endpoints: f ( 1) = 2 f (2) = 5 Comparing gives: The absolute maximum of f (x) is 5 which occurs at x = 2 The absolute minimum is 1 which occurs at x = 0 AMAT 217 (University of Calgary) Fall / 24
15 Local Extreme Points To find local extrema of f (x), find the critical points and singular points and classify them using the first derivative test. The Increasing and Decreasing Test 1 If f (x) > 0 on some interval I, then f (x) is increasing on I. 2 If f (x) < 0 on some interval I, then f (x) is decreasing on I. Determine local extrema of f (x) = x 3 + 6x 2 15x + 7. Solution:The derivative is: f (x) = 3(x 1)(x + 5), with critical points x = 5, 1 and intervals of increasing/decreasing as illustrated: Therefore, x = 1 is a local minimum and x = 5 is a local maximum. AMAT 217 (University of Calgary) Fall / 24
16 Concavity Test If f (x) > 0 on some interval I, then f (x) is concave up on I. If f (x) < 0 on some interval I, then f (x) is concave down on I. Possible inflection points are where f (x) = 0 or DNE (check the concavity on both sides). Identify the intervals of concavity and inflection points for f (x) = x 3 + 6x 2 15x + 7. The process for determining concavity is almost identical to that of determining increasing/decreasing (except we use the second derivative instead). The second derivative is: f (x) = 6(x + 2). We first determine possible inflection points (where f (x) = 0 or f (x) does not exist). In our case, f (x) = 0 when x = 2. Draw a number line and sub test points into f (x): f (x) is concave down on (, 2) and concave up on ( 2, ). Since concavity changes at x = 2, then x = 2 is an inflection point. AMAT 217 (University of Calgary) Fall / 24
17 Antiderivatives Integral Rules Power Rules: Trig Rules: Hyperbolic: Constant Multiple Rule: Sum/Difference Rule: Inverse Trig Rules: Constant Rule: x n dx = xn+1 n C, Exponent Rule: sin x dx = cos x + C. sinh x dx = cosh x + C. kf (x) dx = k k dx = kx + C. f (x) ± g(x) dx = (n 1), x 2 dx = tan 1 x + C f (x) dx, e x dx = e x + C. cos x dx = sin x + C. cosh x dx = sinh x + C. f (x) dx ± k is constant. g(x) dx. 1 dx = ln x + C, (x 0). x sec 2 x dx = tan x + C. sech 2 x dx = tanh x + C. 1 1 x 2 dx = sin 1 x + C. AMAT 217 (University of Calgary) Fall / 24
18 How to evaluate integrals? Integration Techniques 1 The basics (a) Use a formula (b) Expand products (c) Split up fractions (d) Complete the square 2 Substitution Rule Evaluate (z + 3 z)(4 z 2 ) dz Solution: Remember that we don t have a rule to deal with integrals of products. In this case we expand products: (z + 3 z)(4 z 2 ) dz = 4z z 3 + 4z 1/3 z 7/3 dz ( z 2 = 4 2 ) ( z z 4/3 4/3 ) = 2z 2 z z4/3 3z10/3 + C 10 z10/3 10/3 + C AMAT 217 (University of Calgary) Fall / 24
19 Substitution Rule: If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f (g(x))g (x) dx = f (u) du. A general strategy to follow is as follows. Substitution Rule Strategy 1 Choose a possible u = u(x), try one that appears inside a function. 2 Calculate du = u (x) dx. 3 Either replace u (x) dx by du, or replace dx by 4 Write the rest of the integrand in terms of u. 5 Integrate. 6 Rewrite the result back in terms of x. du u, and cancel. (x) AMAT 217 (University of Calgary) Fall / 24
20 Evaluate sec 4 x tan 6 x dx We try u = tan x, so that du = sec 2 x dx, i.e., dx = du sec 2 x : sec 4 x tan 6 x dx = = = = = sec 4 x (u 6 du ) sec 2 Using the substitution x sec 2 x(u 6 ) du Cancelling (1 + tan 2 x)(u 6 ) du Using sec 2 x = 1 + tan 2 x (1 + u 2 )(u 6 ) du Using the substitution u = tan x (u 6 + u 8 ) du Expanding. = u7 7 + u9 9 + C Since = tan7 x 7 + tan9 x 9 x n dx = xn+1 + C, n 1 n C Replacing u back in terms of x AMAT 217 (University of Calgary) Fall / 24
21 Definite Integrals  Areas 1 Compute x dx. 1 b f (x) dx = Net Area under f (x) from x = a to x = b a Recall that y = x forms a V shape when graphing it. The integral is equal to the net area under the V from x = 1 to x = 1. This gives two triangles, each with base of 1 and height of 1. 1 The net area is then 1, thus, x dx = Evaluate the following integral x dx and hence, computing the area under y = x from x = 0 0 to x = 1 (the area of a triangle with base and width 1). 1 0 x dx = x2 2 1 = AMAT 217 (University of Calgary) Fall / 24
22 The Fundamental Theorem of Calculus  Part I FTC I + Chain Rule: d dx v(x) f (t) dt = f (v(x))v (x) f (u(x))u (x) u(x) Differentiate the following integral with respect to x: Solution: x 2 t 3 sin(1 + t) dt. 10x We have f (t) = t 3 sin(1 + t), u(x) = 10x and v(x) = x 2. Then u (x) = 10 and v (x) = 2x. Thus, d x 2 t 3 sin(1 + t) dt dx 10x = [ (x 2 ) 3 sin(1 + (x 2 )) ] [2x] [ (10x) 3 sin(1 + (10x)) ] [10] = 2x 7 sin(1 + x 2 ) 10000x 3 sin(1 + 10x) AMAT 217 (University of Calgary) Fall / 24
23 Area Between Two Curves Problem: Given two curves y = f (x) and y = g(x), determine the area enclosed between them. Solution: The area A of the region bounded by the curves y = f (x) and y = g(x) and the lines x = a and x = b is: b A = f (x) g(x) dx. a Informally this can be thought of as follows: b Area = (top curve) (bottom curve) dx, a x b. a AMAT 217 (University of Calgary) Fall / 24
24 Determine the area enclosed by y = x 2, y = x, x = 0 and x = 2. Solution: From the last problem we have the following sketch: Since the top curve changes at x = 1, we need to use the formula twice. For A 1 we have a = 0, b = 1, the top curve is y = x and the bottom curve is y = x 2. For A 2 we have a = 1, b = 2, the top curve is y = x 2 and the bottom curve is y = x. Area = A1 + A2 = 1 ( 2 x x 2 ) dx + (x 2 x) dx = AMAT 217 (University of Calgary) Fall / 24
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