WM giải tích 1 lê xuân đại derivatives sinhvienzone com

C o e on ELECTRONIC VERSION OF LECTURE nZ Dr Lê Xuân Đại hV ie HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics Email: ytkadai@hcmut.edu.vn in m DERIVATIVES Dr Lê Xuân Đại (HCMUT-OISP) HCMC — 2016 https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o e DERIVATIVES HIGHER DERIVATIVES LINEAR APPROXIMATIONS AND DIFFERENTIALS RATES OF CHANGE AND RELATED RATES MATL AB hV ie nZ on in m OUTLINE Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o Tangents nZ on e DEFINITION 1.1 The tangent line to the curve y = f (x) at the point P(a, f (a)) is the line through P with slope f (x) − f (a) x→a x−a (1) ie m = lim hV provided that this limit exists in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 ie nZ on e C o Tangents hV in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o Tangents e EXAMPLE 1.1 Find an equation of the tangent line to the parabola y = x2 at the point P(1, 1) on SOLUTION The slope of tangent line to the parabola y = x2 is hV ie nZ f (x) − f (1) x2 − m = lim = lim = x→1 x→1 x − x−1 (x − 1)(x + 1) = lim(x + 1) = + = = lim x→1 x→1 x−1 The equation of the tangent line at (1, 1) is y − = 2(x − 1) ⇔ y = 2x − in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o Velocities nZ on e Suppose an object moves along a straight line according to an equation of motion s = f (t), where s is the directed distance of the object from the origin at the time t In the time interval from t = a to t = a + h the change in position is f (a + h) − f (a) The average velocity over this time interval is f (a + h) − f (a) h hV ie average velocity = in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o Velocities nZ on e Now suppose we compute the average velocities over shorter and shorter time intervals [a, a + h] We let h approach The instantaneous velocity v(a) at time t = a is defined by f (a + h) − f (a) h→0 h (2) hV ie v(a) = lim in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o Velocities hV ie nZ on e EXAMPLE 1.2 Suppose that a ball is dropped from the upper observation deck of CN Tower, 450 m above the ground What is the velocity of the ball after seconds? How fast is the ball travelling when it hits the ground? in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 / 55 C o Velocities SOLUTION e f (a + h) − f (a) 4.9(a + h)2 − 4.9a2 v(a) = lim = lim h→0 h→0 h h 2 4.9(a + 2ah + h − a2 ) 4.9(2ah + h2 ) = lim = lim = h→0 h→0 h h = lim 4.9(2a + h) = 9.8a on The equation of motion s = f (t) = 12 g.t = 4.9t , where g− acceleration of gravity nZ h→0 hV The velocity after 5s is v(5) = 9.8 × = 49m/s Since the observation deck is 450 m above the ground, the ball will hit the ground at the time t1 when s(t1 ) = 450 ⇒ 4.9.t12 = 450 ⇒ t1 = 450 ≈ 9.6s ⇒ 4.9 v(t1 ) = 9.8t1 ≈ 94m/s (the velocity of the ball as it https://fb.com/sinhvienzonevn hits the ground) ie in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) DERIVATIVES HCMC — 2016 / 55 C o Definition on e DEFINITION 1.2 The derivative of a function f at a number a, denoted by f (a), (read: f prime of a) is f (a + h) − f (a) h→0 h (3) ie nZ f (a) = lim hV if this limit exists Other Notations: f (a) = y (a) = in m Derivatives Dr Lê Xuân Đại (HCMUT-OISP) dy dx = x=a d f (a) dx https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 10 / 55 e C o The n−th order differentials nZ on DEFINITION 3.4 The n−th order differential of y = f (x) at a is defined in terms of dx by equation dn f (a) = f (n) (a)dxn ie (13) hV in m Linear approximations and Differentials Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 41 / 55 C o Rates of change in the natural and social sciences hV ie nZ on e RATES OF CHANGE in m Rates of change and Related rates Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 42 / 55 C o Rates of change in the natural and social sciences RATES OF CHANGE on e If x changes from x1 to x2 , then the change in x is ∆x = x2 − x1 and the corresponding change in y is ∆y = f (x2 ) − f (x1 ) The difference quotient hV ie nZ ∆y f (x2 ) − f (x1 ) = ∆x x2 − x1 is the average rate of change of y with respect to x over the interval [x1 , x2 ] The instantaneous rate of change of y with respect to x or the slope of the tangent line at P(x1 , f (x1 )) is in m Rates of change and Related rates Dr Lê Xuân Đại (HCMUT-OISP) dy ∆y = lim dx ∆x→0 ∆x https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 43 / 55 C o Rates of change in the natural and social sciences e PHYSICS ie ∆s represents the average velocity over a time ∆t period ∆t ds ∆s = lim represents the instantaneous dt ∆t→0 ∆t nZ on If s = f (t) is the position function of a particle that is moving in a straight line, then hV velocity the instantaneous rate of change of velocity with respect to time is acceleration: a(t) = v (t) = s (t) in m Rates of change and Related rates Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 44 / 55 C o Rates of change in the natural and social sciences e EXAMPLE 4.1 The position of a particle is given by the equation on s = f (t) = t − 6t + 9t, hV ie nZ where t is measured in seconds and s in meters Find the velocity at time t What is the velocity after 2s? When is the particle at rest? When is the particle moving forward (that is, in the positive direction) and backward? Find the total distance travelled by the particle during the first five seconds Find the acceleration at time t and after 4s When https://fb.com/sinhvienzonevn is the particle speeding up, slowing down? in m Rates of change and Related rates Dr Lê Xuân Đại (HCMUT-OISP) DERIVATIVES HCMC — 2016 45 / 55 C o Rates of change in the natural and social sciences The velocity function e on v(t) = s (t) = 3t − 12t + ⇒ v(2) = −3m/s The particle is at rest when nZ v(t) = ⇔ 3t − 12t + = ⇔ t = 1s The particle t = 3s hV ie moves forward when v(t) > ⇔ t >3 t ⇔ (3t − 12t + 9)(6t − 12) > Dr Lê Xuân Đại (HCMUT-OISP) https://fb.com/sinhvienzonevn DERIVATIVES HCMC — 2016 48 / 55 e C o Rates of change in the natural and social sciences t >3 1