Calculus (1) Math 105 Past Paper PDF

Calculus (1) (Math 105) 3.1 Rates of Change and Tangents to Curves Page 1 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Chapter 3: The Derivative 3.1 Rates of Change and Tangents to Curves 3.2 The Derivative at a Point 3.3 The Derivative as a Function 3.4 Di...

Calculus (1) (Math 105) 3.1 Rates of Change and Tangents to Curves Page 1 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Chapter 3: The Derivative 3.1 Rates of Change and Tangents to Curves 3.2 The Derivative at a Point 3.3 The Derivative as a Function 3.4 Differentiation Rules 3.5 The Derivative as a Rate of Change 3.6 Derivatives of Trigonometric Functions 3.7 The Chain Rule Page 2 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Introduction Many of the ideas of calculus originated with the following two geometric problems: The Tangent line Problem Given a function 𝑓 and a point 𝑃(𝑥0 , 𝑦0 ) on its graph, find an equation of the line that is tangent to the graph at 𝑃 (See the Figure below). The Area Problem Given a function 𝑓, find the area between the graph of 𝑓 and an interval [𝑎, 𝑏] on the 𝑥-axis (See the Figure below). Page 3 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi  Traditionally, that portion of calculus arising from the tangent line problem is called differential calculus, and that arising from the area problem is called integral calculus. However, we will see later that the tangent line and area problems are so closely related that the distinction between differential and integral calculus is somewhat artificial. In this section we will discuss three ideas: tangent lines to curves, the velocity of an object moving along a line, and the rate at which one variable changes relative to another. Our goal is to show how these seemingly unrelated ideas are, in actuality, closely linked. Page 4 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi 3.1 Tangent Lines and Rates of Change ▪ Slope of Secant Lines A secant line is a straight line that connects two points on the curve of a function 𝑦 = 𝑓(𝑥). Definition 1: Using the formula for the slope of a line, we can write the slope (𝑚sec )of the secant line passing through 𝑃(𝑥0 , 𝑓(𝑥0 )) and 𝑄(𝑥1 , 𝑓(𝑥1 )) on the curve of 𝑦 = 𝑓(𝑥) as change in 𝑦 Δ𝑦 𝑓(𝑥1 ) − 𝑓(𝑥0 ) 𝑚sec = = = − − − (1), change in 𝑥 Δ𝑥 𝑥1 − 𝑥0 where Δ𝑥 = 𝑥1 − 𝑥0 , Δ𝑦 = 𝑓(𝑥1 ) − 𝑓(𝑥0 ). Δ𝑥 and Δ𝑦 are called increments of 𝑥 and 𝑦, respectively. Page 5 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Example 1 Find the slope of the secant line passing through 𝑃(0,2) and 𝑄(2,6) on the curve of 𝑓(𝑥) = 𝑥 2 + 2. Solution 𝑓(𝑥1 ) − 𝑓(𝑥0 ) 𝑓(2) − 𝑓(0) 6 − 2 𝑚sec = = = = 2. 𝑥1 − 𝑥0 2−0 2 Checkpoint 1 Find the slope of the secant line passing through two points on the curve of 𝑓(𝑥) = 1 at 𝑥 = 1 and 𝑥 = 2. 𝑥 Solution ▪ Average Rate of Change of a Function Definition 2: The average rate of change (𝑟avg ) (of 𝑦 = 𝑓(𝑥) with respect to 𝑥 over the interval [𝑥0 , 𝑥1 ] is Δ𝑦 𝑓(𝑥1 ) − 𝑓(𝑥0 ) 𝑟avg = = − − − (2) Δ𝑥 𝑥1 − 𝑥0 Example 2 Compute the average of change of the function 𝑓(𝑥) = 𝑥 2 − 1 over [−1,2]. Solution Page 6 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi The average of change of the function 𝑓(𝑥) given by 𝑓(𝑥1 ) − 𝑓(𝑥0 ) 𝑟ave =. 𝑥1 − 𝑥0 We have 𝑥0 = −1 and 𝑥1 = 2. Now compute 𝑓(𝑥0 ) and 𝑓(𝑥1 ). 𝑓(𝑥0 ) = 𝑓(−1) = (−1)2 − 1 = 0, 𝑓(𝑥1 ) = 𝑓(2) = (2)2 − 1 = 3. Then 𝑓(𝑥1 ) − 𝑓(𝑥0 ) 3−0 3 𝑟ave = = = =1 𝑥1 − 𝑥0 2 − (−1) 3 Checkpoint 2 𝑥 Compute the average of change of function 𝑓(𝑥) = + 2 over [2,5]. 𝑥 2 −1 Solution Page 7 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi 3.1.1 Tangent line To Curves Now we can give a mathematical definition of the tangent line to a curve 𝑦 = 𝑓(𝑥) at a point 𝑃(𝑥0 , 𝑓(𝑥0 )) on the curve. As illustrated in the Figure below, the slope 𝑚𝑃𝑄 of the secant line through 𝑃 and a second point 𝑄(𝑥, 𝑓(𝑥)) on the graph of 𝑓 is 𝑓(𝑥) − 𝑓(𝑥0 ) 𝑚𝑃𝑄 =. 𝑥 − 𝑥0 If we let 𝑥 approach 𝑥0 , then the point 𝑄 will move along the curve and approach the point 𝑃. Suppose the slope 𝑚𝑃𝑄 of the secant line through 𝑃 and 𝑄 approaches a limit as 𝑥 → 𝑥0. In that case we can take the value of the limit to be the slope 𝑚tan of the tangent line at 𝑃. Thus, we make the following definition. Page 8 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Definition 3: Suppose that 𝑥0 is in the domain of the function 𝑓. Then the tangent line to the curve 𝑦 = 𝑓(𝑥) at the point 𝑃(𝑥0 , 𝑓(𝑥0 )) on the graph of 𝑓 is the line passing through 𝑃 and having slope 𝑓(𝑥) − 𝑓(𝑥0 ) 𝑚tan = lim 𝑚sec = lim , (4) 𝑥→𝑥0 𝑥→𝑥0 𝑥 − 𝑥0 provided the limit exists. For simplicity, we will also call this the tangent line to 𝑦 = 𝑓(𝑥) at 𝑥0. The equation for the tangent line to the curve of 𝑦 = 𝑓(𝑥) at 𝑃(𝑥0 , 𝑓(𝑥0 )) is given by 𝑦 − 𝑓(𝑥0 ) = 𝑚tan (𝑥 − 𝑥0 ) or 𝑦 = 𝑓(𝑥0 ) + 𝑚tan (𝑥 − 𝑥0 ) See the appendix for more details on Equations of Lines. Example 3 Use Definition above to find an equation for the tangent line to the parabola 𝑦 = 𝑥 2 at the point 𝑃(1,1). Solution Applying Formula (4) with 𝑓(𝑥) = 𝑥 2 and 𝑥0 = 1, we have 𝑓(𝑥) − 𝑓(1) 𝑚tan = lim 𝑥→1 𝑥−1 2 𝑥 −1 = lim 𝑥→1 𝑥 − 1 (𝑥 − 1)(𝑥 + 1) = lim = lim (𝑥 + 1) = 2 𝑥→1 𝑥−1 𝑥→1 Thus, the tangent line to 𝑦 = 𝑥 2 at (1,1) has equation 𝑦 − 1 = 2(𝑥 − 1) or equivalently 𝑦 = 2𝑥 − 1. Page 9 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi There is an alternative way of expressing Formula (4) that is commonly used. If we let ℎ denote the difference ℎ = 𝑥 − 𝑥0 then the statement that 𝑥 → 𝑥0 is equivalent to the statement ℎ → 0, so we can rewrite (4) in terms of 𝑥0 and ℎ as 𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 ) 𝑚tan = lim (5) ℎ→0 ℎ The Figure below shows how Formula (5) expresses the slope of the tangent line as a limit of slopes of secant lines. Example 4 2 Find an equation for the tangent line to the curve 𝑦 = at the point (2,1) on this 𝑥 curve. Solution 2 First, we will find the slope of the tangent line by applying Formula (5) with 𝑓(𝑥) = 𝑥 and 𝑥0 = 2. These yields Page 10 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi  𝑓(2 + ℎ) − 𝑓(2) 𝑚tan = lim ℎ→0 ℎ 2 2 − (2 + ℎ) −1 ( ) 2 + ℎ 2+ℎ = lim = lim ℎ→0 ℎ ℎ→0 ℎ −ℎ −1 1 = lim = lim =− ℎ→0 ℎ(2 + ℎ) ℎ→0 2 + ℎ 2 Thus, an equation of the tangent line at (2,1) is 1 1 𝑦 − 1 = − (𝑥 − 2) or equivalently 𝑦 = − 𝑥 + 2 2 2 (see the Figure below). Example 5 Find equations of the straight lines that are tangent and normal to the curve 𝑦 = √𝑥 at the point (4,2). Solution The slope of the tangent at (4,2) (See the Figure below) is Page 11 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi  √4 + ℎ − 2 (√4 + ℎ − 2)(√4 + ℎ + 2) 𝑚𝑡𝑎𝑛 = lim = lim ℎ→0 ℎ ℎ→0 ℎ(√4 + ℎ + 2) 4+ℎ−4 = lim ℎ→0ℎ(√4 + ℎ + 2) 1 1 = lim =. ℎ→0 √4 + ℎ + 2 4 The tangent line has equation 1 1 𝑦 = (𝑥 − 4) + 2 or 𝑦 = 𝑥 + 1 4 4 and the normal has slope −4 and, therefore, equation 𝑦 = −4(𝑥 − 4) + 2 or 𝑦 = −4𝑥 + 18. Checkpoint 3 Find the slopes of the tangent lines to the graph of 𝑓(𝑥) = 𝑥 2 + 1 at the points (0,1) and (−1,2), as shown in the Figure below. Answer: 𝑚𝑡𝑎𝑛 = 0 at (0,1). 𝑚𝑡𝑎𝑛 = −2 at (−1,2). Page 12 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi  Appendix Page 13 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi  Average Instantaneous Concept Formula Terminology Formula 𝑚tan = lim 𝑚sec 𝑥→𝑥0 𝑓(𝑥) − 𝑓(𝑥0 ) = lim Slope of Secant Δ𝑦 𝑓(𝑥1 )−𝑓(𝑥0 ) Tangent line 𝑥→𝑥0 𝑥 − 𝑥0 𝑚sec = = or Lines Δ𝑥 𝑥1 −𝑥0 𝑚tan = lim 𝑚sec ℎ→0 𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 ) = lim ℎ→0 ℎ 𝑟inst = lim 𝑟ave 𝑥→𝑥0 𝑓(𝑥) − 𝑓(𝑥0 ) = lim 𝑥→𝑥0 𝑥 − 𝑥0 Average Rate of Instantaneous or Δ𝑦 𝑓(𝑥1 )−𝑓(𝑥0 ) Change 𝑟ave = = Rate of change Δ𝑥 𝑥1 −𝑥0 𝑟inst = lim 𝑟ave ℎ→0 𝑓(𝑥0 + ℎ) − 𝑓(𝑥0 ) = lim ℎ→0 ℎ 𝑓(𝑡) − 𝑓(𝑡0 ) 𝑣inst = lim 𝑡→𝑡0 𝑡 − 𝑡0 Average Velocity Δ𝑠 𝑓(𝑡1 )−𝑓(𝑡0 ) Instantaneous 𝑣ave = = (Average speed) Δ𝑡 𝑡1 −𝑡0 Velocity oror or 𝑓(𝑡0 + ℎ) − 𝑓(𝑡0 ) 𝑣inst = lim ℎ→0 ℎ Page 14 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Page 15 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Page 16 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Page 17 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Page 18 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi Page 19 of 19 Math 105 (Semester 1 2024/2025) Dr. Omar Alsuhaimi