Presentation on theme: "Derivative as a Rate of Change Chapter 3 Section 4."— Presentation transcript:
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 the equation 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 Thus, when D = 10 meters the area is changing with respect to the diameter at the rate of
Motion Along a Line Displacement of object over time Δs = f(t + Δt) – f(t) Average velocity of object over time interval
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 s distance fallen g is acceleration due to Earth’s gravity (appx: 32 ft/sec 2 or 9.8 m/sec 2 ) – Same constant acceleration – No jerk
Example 2: Free Fall Example How many meters does the ball fall in the first 2 seconds? Free Fall equation s = 4.9t 2 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 – So at time t = 2, the velocity is
Example 2: Free Fall Example What is its velocity, speed and acceleration when t = 2? – Velocity = derivative of position at any time t – So at time t = 2, the speed is
Example 2: Free Fall Example What is its velocity, speed and acceleration when t = 2? – Velocity = derivative of position at any time t – The acceleration at any time t – So at t = 2, acceleration is (no air resistance)
Derivatives of Trigonometric Functions Chapter 3 Section 5
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: s = 5cos(t) – Velocity: 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 = x 2 + 3 – This is explicit differentiation Other types of functions x 2 + y 2 = 25 or y 2 – 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
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 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
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
Factor out dy/dx Divide out 3
Evaluate at (2,4) and (4,2) Slope at the point (2,4) Slope at the point (4,2)
Folium of Descartes Can you find the slope of the folium of Descartes 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