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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. - PowerPoint PPT Presentation
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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
2
4DA
2
2444
22 DDDdDdAD
dDdA
mmD /71.152
)10(2
2
Motion Along a Line
• Displacement of object over time Δs = f(t + Δt) – f(t)
• Average velocity of object over time interval
ttfttf
tsvaverage
)()(
timetravelnt displaceme
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 Equations distance falleng is acceleration due to Earth’s gravity
(appx: 32 ft/sec2 or 9.8 m/sec2)– Same constant acceleration– No jerk
2
21 gts
0121 2
dtdgt
dtd
dtdgt
dtds
dtd
dtdj
Example 2: Free Fall Example
• How many meters does the ball fall in the first 2 seconds?
• Free Fall equation s = 4.9t2 in meters
s(2) – s(0) = 4.9(2)2 - 4.9(0)2 = 19.6 m
22 9.4)8.9(21 tts
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
ttdtdtstv 8.99.4)(')( 2
sec/6.19)2(8.98.99.4)(')( 2 mttdtdtstv
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
ttdtdtstv 8.99.4)(')( 2
sec/6.19sec/6.19 mmvelocityspeed
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)
ttdtdtstv 8.99.4)(')( 2
2sec/8.98.9)('')(')( mtdtdtstvta
2sec/8.9)( mta
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: 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 = x2 + 3– This is explicit differentiation
• Other types of functionsx2 + 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
• Show that the points (2,4) and (4,2) lie on the curve and find their slopes and tangent line to curve
0933 xyyx
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
0933 xyyx
0)4)(2(942 33 072648
07272 00
0)2)(4(924 33
072864 07272
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
0933 xyyx
0933
dxdxy
dxdy
dxdx
dxd
0)(933 22
x
dxdy
dxdyx
dxdyyx
PRODUCT RULE
09933 22 ydxdyx
dxdyyx
09933 22 ydxdyx
dxdyyx
Factor out dy/dx
09393 22 yxxydxdy
22 3993 xyxydxdy
xyxy
dxdy
9339
2
2
xyxy
dxdy
33
2
2
Divide out 3
xyxy
dxdy
33
2
2
Evaluate at (2,4) and (4,2)
54
108
616412
)2(342)4(3
33
2
2
)4,2(2
2
)4,2(
xyxy
dxdy
Slope at the point (2,4)
45
810
124166
)4(324)2(3
33
2
2
)2,4(2
2
)2,4(
xyxy
dxdy
Slope at the point (4,2)
0933 xyyx
Find Tangents54
)4,2( dxdy
11 xxmyy
2544 xy
58
544 xy
458
54
xy
520
58
54
xy
512
54
xy
45
)2,4( dxdy
4452 xy
5452 xy
345
xy
512
54
xy
345
xy
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?
0933 xyyx
Derivatives of Inverse Functions and Logarithms
Chapter 3 Section 8
Examples 1 & 2