Report

Derivative as a Rate of Change Chapter 3 Section 4 Usually omit instantaneous Interpretation: The rate of change at which f is changing at the point x Interpretation: Instantaneous rate are limits of average rates. Example • The area A of a circle is related to its diameter by A D the equation 4 • How fast does the area change with respect to the diameter when the diameter is 10 meters? • The rate of change of the area with respect to the diameter dA D 2 dA D 2 2 D D 2 dD 4 4 dD 4 2 • Thus, when D = 10 meters the area is changing with respect to the diameter at the rate of D 2 (10) 2 15.71 m 2 / m Motion Along a Line • Displacement of object over time Δs = f(t + Δt) – f(t) • Average velocity of object over time interval vaverage displacement s f (t t ) f (t ) travel time t t Velocity • Find the body’s velocity at the exact instant t – How fast an object is moving along a horizontal line – Direction of motion (increasing >0 decreasing <0) Speed • Rate of progress regardless of direction Graph of velocity f ’(t) Acceleration • The rate at which a body’s velocity changes – How quickly the body picks up or loses speed – A sudden change in acceleration is called jerk • Abrupt changes in acceleration Example 1: Galileo Free Fall • Galileo’s Free Fall Equation 1 2 s gt 2 s distance fallen g is acceleration due to Earth’s gravity (appx: 32 ft/sec2 or 9.8 m/sec2) – Same constant acceleration – No jerk d d ds 1 2 d d d j gt gt 1 0 dt dt dt 2 dt dt dt Example 2: Free Fall Example • How many meters does the ball fall in the first 2 seconds? 1 s (9.8)t 2 4.9t 2 2 • Free Fall equation s = 4.9t2 in meters s(2) – s(0) = 4.9(2)2 - 4.9(0)2 = 19.6 m Example 2: Free Fall Example • What is its velocity, speed and acceleration when t = 2? – Velocity = derivative of position at any time t v(t ) s ' (t ) d 4.9t 2 9.8t dt – So at time t = 2, the velocity is v(t ) s ' (t ) d 4.9t 2 9.8t 9.8(2) 19.6 m / sec dt Example 2: Free Fall Example • What is its velocity, speed and acceleration when t = 2? – Velocity = derivative of position at any time t v(t ) s ' (t ) d 4.9t 2 9.8t dt – So at time t = 2, the speed is speed velocity 19.6 m / sec 19.6 m / sec Example 2: Free Fall Example • What is its velocity, speed and acceleration when t = 2? – Velocity = derivative of position at any time t v(t ) s ' (t ) d 4.9t 2 9.8t dt – The acceleration at any time t a (t ) v' (t ) s ' ' (t ) d 9.8t 9.8 m / sec 2 dt – So at t = 2, acceleration is (no air resistance) a (t ) 9.8 m / sec 2 Derivatives of Trigonometric Functions Chapter 3 Section 5 Derivatives Application: Simple Harmonic Motion • Motion of an object/weight bobbing freely up and down with no resistance on an end of a spring • Periodic, repeats motion • A weight hanging from a spring is stretched down 5 units beyond its rest position and released at time t = 0 to bob up and down. Its position at any later time t is s = 5 cos(t) • What are its velocity and acceleration at time t? Application: Simple Harmonic Motion • Its position at any later time t is s = 5 cos(t) – Amp = 5 – Period = 2 • What are its velocity and acceleration at time t? – Position: – Velocity: s = 5cos(t) s’ = -5sin(t) • Speed of weight is 0, when t = 0 – Acceleration: s’’ = -5 cos(t) • Opposite of position value, gravity pulls down, spring pulls up Chain Rule Chapter 6 Section 6 Implicit Differentiation Chapter 3 Section 7 Implicit Differentiation • So far our functions have been y = f(x) in one variable such as y = x2 + 3 – This is explicit differentiation • Other types of functions x2 + y2 = 25 or y2 – x = 0 • Implicit relation between the variables x and y • Implicit Differentiation – Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x (always put dy/dx after derive y term) – Collect the terms with dy/dx on one side of the equation and solve for dy/dx Circle Example Folium of Descartes • The curve was first proposed by Descartes in 1638. Its claim to fame lies in an incident in the development of calculus. • Descartes challenged Fermat to find the tangent line to the curve at an arbitrary point since Fermat had recently discovered a method for finding tangent lines. • Fermat solved the problem easily, something Descartes was unable to do. • Since the invention of calculus, the slope of the tangent line can be found easily using implicit differentiation. Folium of Descartes • Find the slope of the folium of Descartes x 3 y 3 9 xy 0 • Show that the points (2,4) and (4,2) lie on the curve and find their slopes and tangent line to curve Folium of Descartes • Show that the points (2,4) and (4,2) lie on the curve and find their slopes and tangent line to curve 3 3 x y 9xy 0 23 43 9(2)(4) 0 8 64 72 0 72 72 0 00 43 23 9(4)(2) 0 64 8 72 0 72 72 0 00 Folium of Descartes • Show that the points (2,4) and (4,2) lie on the curve and find their slopes and tangent line to curve – Find slope of curve by implicit differentiation by finding dy/dx x y 9xy 0 3 3 d 3 d 3 d d x y 9 xy 0 dx dx dx dx dy dy d 3x 3 y 9 x y ( x) 0 dx dx dx 2 2 PRODUCT RULE dy dy 3x 3 y 9x 9 y 0 dx dx 2 2 dy dy 3x 3 y 9x 9 y 0 dx dx 2 2 Factor out dy/dx dy 3 y 2 9 x 3x 2 9 y 0 dx dy 3 y 2 9 x 9 y 3x 2 dx dy 9 y 3 x 2 dx 3 y 9 x 2 dy 3 y x 2 dx y 3 x 2 Divide out 3 dy 3 y x 2 2 dx y 3x Evaluate at (2,4) and (4,2) dy dx ( 2, 4 ) 3y x2 2 y 3x ( 2, 4 ) 3(4) 2 2 12 4 8 4 2 4 3(2) 16 6 10 5 Slope at the point (2,4) dy dx ( 4, 2 ) 3y x2 2 y 3x ( 4, 2 ) 3(2) 4 2 6 16 10 5 2 2 3(4) 4 12 8 4 Slope at the point (4,2) x3 y 3 9xy 0 dy dx ( 2, 4 ) 4 5 4 y 4 x 2 5 4 8 y4 x 5 5 4 8 y x 4 5 5 4 8 20 y x 5 5 5 4 12 y x 5 5 Find Tangents y y1 m x x1 dy dx y2 ( 4, 2 ) 5 4 5 x 4 4 5 y 2 x 5 4 5 y x 3 4 4 12 y x 5 5 5 y x3 4 Folium of Descartes • Can you find the slope of the folium of Descartes x 3 y 3 9 xy 0 • At what point other than the origin does the folium have a horizontal tangent? – Can you find this? Derivatives of Inverse Functions and Logarithms Chapter 3 Section 8 Examples 1 & 2