Todays Agenda Upcoming Homework Section 2.1: Derivatives and Rates - PowerPoint PPT Presentation

Today’s Agenda • Upcoming Homework • Section 2.1: Derivatives and Rates of Change Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 1 / 9

Upcoming Homework • Written HW B: Section 1.5 #31; Section 1.6 #34 and #38. Due 9/11/2015. • WeBWorK #6: Sections 2.1 and 2.2. Due 9/14/2015. • Written HW C: Section 2.1 #14ab, 32, 36; Section 2.2 #26, 40, 47. • WeBWorK #7: Section 2.3. Due 9/18/2015. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 2 / 9

Section 2.1 Recall the definitions of tangent lines and secant lines to circles from Euclidean plane geometry: Definition 2.1.1 Let C be a circle in the plane with center x , and let p be a point on C . We say that a line ℓ is tangent to C at p if ℓ passes through p and if ℓ is perpendicular to the line connecting x and p . Definition 2.1.2 Let C be a circle in the plane. A secant line ℓ is a line in the plane that intersects C at exactly two (distinct) points p and q on C . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 3 / 9

Section 2.1 Take a moment with a partner to discuss how you could use the concept of secant lines to answer the following question: Suppose we have a function f : R → R . (a) How do we know if a tangent line to the plane curve y = f ( x ) at the point x 0 exists? (b) If we have already determined that a tangent line as described in part a does indeed exist, how can we calculate it? (The equation for the tangent line should be of the form g ( x ) = mx + b for constants m and b , and it must be true that g ( x 0 ) = f ( x 0 ).) Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 4 / 9

Section 2.1 Let’s calculate the equation for the tangent line to the graph of y = f ( x ), where f ( x ) = x 2 , at the point x 0 = 1. First we calculate a secant line that passes through ( x 0 , f ( x 0 )) = (1 , 1) and a point ( x 0 + t , f ( x 0 + t )) = (1 + t , (1 + t ) 2 ) for an arbitrary small number t > 0, using point-slope form: Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 5 / 9

Section 2.1 Let’s calculate the equation for the tangent line to the graph of y = f ( x ), where f ( x ) = x 2 , at the point x 0 = 1. First we calculate a secant line that passes through ( x 0 , f ( x 0 )) = (1 , 1) and a point ( x 0 + t , f ( x 0 + t )) = (1 + t , (1 + t ) 2 ) for an arbitrary small number t > 0, using point-slope form: g t ( x ) − 1 = (1 + t ) 2 − 1 ( x − 1) . t Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 5 / 9

Section 2.1 Let’s calculate the equation for the tangent line to the graph of y = f ( x ), where f ( x ) = x 2 , at the point x 0 = 1. First we calculate a secant line that passes through ( x 0 , f ( x 0 )) = (1 , 1) and a point ( x 0 + t , f ( x 0 + t )) = (1 + t , (1 + t ) 2 ) for an arbitrary small number t > 0, using point-slope form: g t ( x ) − 1 = (1 + t ) 2 − 1 ( x − 1) . t Similarly, we can calculate a secant line that passes through ( x 0 , f ( x 0 )) = (1 , 1) and a point ( x 0 − t , f ( x 0 − t )) = (1 − t , (1 − t ) 2 ): h t ( x ) − 1 = 1 − (1 − t ) 2 ( x − 1) . t Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 5 / 9

Section 2.1 Now we calculate t → 0 g t ( x ) = 2 x − 1 lim and t → 0 h t ( x ) = 2 x − 1 . lim Therefore, the tangent line to the curve y = f ( x ) = x 2 at the point x 0 = 1 is given by y = 2 x − 1 . This idea of approximating quantities and then taking limits is one of the central ideas of calculus, and so it is very important! Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 6 / 9

Section 2.1 Definition 2.1.3 The derivative of a function f at a number a , denoted by f ′ ( a ), is f ( a + h ) − f ( a ) f ′ ( a ) = lim , h h → 0 provided this limit exists. Notice that the derivative only gives us the slope of the tangent line. The derivative does NOT give us the equation for the tangent line! Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 7 / 9

Section 2.1 Definition 2.1.4 Consider the function y = f ( x ). The average rate of change of y with respect to x over the interval [ x 1 , x 2 ] is given by ∆ y ∆ x = f ( x 2 ) − f ( x 1 ) . x 2 − x 1 The instantaneous rate of change of y with respect to x at the point x 1 is given by ∆ y f ( x 2 ) − f ( x 1 ) lim ∆ x = lim . x 2 − x 1 x 2 → x 1 ∆ x → 0 The derivative f ′ ( a ) is the instantaneous rate of change of y = f ( x ) with respect to x when x = a (in the definition on the previous page, let a + h = x 2 and a = x 1 to get the same definition as on this page). Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 8 / 9

Section 2.1 Solve the following problem with a partner. Suppose an arrow is shot upward on the moon with an initial velocity of 58 m/s. Its height (in meters) after t seconds is given by H = 58 t − 0 . 83 t 2 . 1 Find the average velocity of the arrow over the following time intervals: [0 , 1 / 2], [1 / 2 , 1], [0 , 1]. 2 Find the instantaneous velocity of the arrow at time t = 1. 3 Find the instantaneous velocity of the arrow at time t = a . 4 At what time will the arrow return to the ground? 5 With what instantaneous velocity will the arrow hit the ground? Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 9 September 2015 9 / 9