Upload
melvin-gordon
View
132
Download
11
Tags:
Embed Size (px)
DESCRIPTION
3.6 Chain rule. When gear A makes x turns, gear B makes u turns and gear C makes y turns. ,. u turns 3 times as fast as x. So y turns 3/2 as fast as x. y turns ½ as fast as u. Rates are multiplied. The Chain Rule for composite functions. - PowerPoint PPT Presentation
Citation preview
When gear A makes x turns, gear B makes u turns and gear C makes y turns.,
3.6 Chain rule
1
2
dy
du
3du
dx
y turns ½ as fast as u
u turns 3 times as fast as xSo y turns 3/2 as fast as x
dy dy du
dx du dx
Rates are multiplied
The Chain Rule for composite functions
If y = f(u) and u = g(x) then y = f(g(x)) and
dy dy du
dx du dx multiply rates
( ( ) ( ( )) ( )dy d
f g x f g x g xdx dx
multiply rates
Find the derivative (solutions to follow)
2 71) ( ) (3 5 )f x x x
2 232) ( ) ( 1)f x x
2
73) ( )
(2 3)f t
t
4) ( ) sin(2 )f x x
25) ( ) tan( 1)f x x
Solutions
2 71) ( ) (3 5 )f x x x
2 232) ( ) ( 1)f x x
dy dy du
dx du dx
2
7 6
3 5 3 10
7
u x x x
y
du
dy
du
x
u u
d
6 3 10 )7 (dy
ud
xx
2 67(3 2 ) (3 10 )dy
x x xdx
2
2 1
3 3
1 2
2
3
u x xd
dy u u
y
du
u
dx
1
32
3(2 )
du
y
dx
x
12 32
( 1)3
(2 )dy
dxx x
12 3
4
3( 1)
dy x
dxx
Solutions dy dy du
dx du dx
2 3
2 3 2
7 14
u t
y u
d
d
d
u
d
u
x
uy
314 2udy
dx
33
2814(2 3) (2)
(2 3)
dyt
dx t
2 2
sin( ) cos( )
du
d
u x
yy
d
u udu
x
2co 2 (2 )s( )dy
ud
cos xx
22
73) ( ) 7(2 3)
(2 3)f t t
t
4) ( ) sin(2 )f x x
25) ( ) tan( 1)f x x 2
2
1 2
tan( ) sec ( )
u x x
y udy
du
du
dx
u
2 2 22 secsec ( )(2 )) ( 1dy
u x xx
xd
Outside/Inside method of chain rule
( ( ) ( ( )) ( )dy d
f g x f g x g xdx dx
insideoutside derivative of outside wrt inside
derivative of inside
think of g(x) = u
Outside/Inside method of chain rule example
1
2 33 1 ( ( )) ( )d
x x f g x g xdx
inside
outside
derivative of outside wrt inside
derivative of inside
1 2
2 23 313 1 3 1 6 1
3
dx x x x x
dx
2
2 3
6 1
3 3 1
x
x x
Outside/Inside method of chain rule
33sin sin ( ( )) ( )d d
f g x g xdx dx
inside
outside
derivative of outside wrt inside
derivative of inside
3 2sin 3 sin sin
d d
dx dx
23sin cos
Outside/Inside method of chain rule
2csc( 3) ( ( )) ( )d
f g x g xdx
insideoutsidederivative of outside wrt inside
derivative of inside
2 2 2csc( 3) csc( 3)cot( 3)2d
dx
2 22 csc( 3)cot( 3)
1
2 2 2( ) 1f x x x
More derivatives with the chain rule2 2( ) 1f x x x
1 1
2 2 2 22 2( ) 1 1d d
f x x x x xdx dx
1 1
2 2 22 21( ) 1 ( 2 ) 1 2
2f x x x x x x
132 2
12 2
( ) 1 2
1
xf x x x
x
1 12 22 23 3 2 3 3
1 1 1 12 2 2 22 2 2 2
1 2 1 (1 )2 2 2( )
1 1 1 1
x x xx x x x x x xf x
x x x x
3
12 2
3 2( )
1
x xf x
x
product
Simplify terms
Combine with common denominator
More derivatives with the chain rule3
2
3 1( )
3
xf x
x
2
2 2
3 1 3 1( ) 3
3 3
x d xf x
dxx x
2 2
2 22
3 1 ( 3)3 (3 1)(2 )3
3 3
x x x x
x x
2 2 2
2 22
3 1 (3 9) (6 2 )3
3 3
x x x x
x x
2 2 2
2 22
2 2
42
3(3 1) ( 3 2 9)3 1 3 9 6 23
3 3 3
x x x x
x x
x x x
x
Quotient rule
The formulas for derivatives assume x is in radian measure.sin (x°) oscillates only /180 times as often as sin (x) oscillates. Its maximum slope is /180.
Radians Versus Degrees 1180
radians
d/dx[sin (x)] = cos (x)
d/dx [sin (x°) ] = /180 cos (x°)
3.7 Implicit Differentiation
Although we can not solve explicitly for y, we can assume that y is some function of x and use implicit differentiation to find the slope of the curve at a given point
y=f (x)
If y is a function of x then its derivative is dy
dx
y2 is a function of y, which in turn is a function of x.
2 2d dy
y ydx dx
sind
ydx
2 3dx y
dx
1
2dy
dx
1
21
2y
dy
dx
cos ydy
dx
2 2 33 2dy
dxx y y x
using the chain rule:
Find the following derivatives wrt x
Use product rule
Implicit Differentiation
4. Solve for dy/dx
2 2 sin( )y x xy
2 2 cos( ) (1)dy dy
y x xy x ydx dx
2 cos( )
2 cos( )
dy x y xy
dx y x xy
2 2 cos( )( ) cos( )dy dy
y x xy x xy ydx dx
2 cos( )( ) 2 cos( )dy dy
y xy x x xy ydx dx
(2 cos( )) 2 cos( )dy
y x xy x y xydx
1. Differentiate both sides of the equation with respect to x, treating y as a function of x. This requires the chain rule.
2. Collect terms with dy/dx on one side of the equation.
3. Factor dy/dx
Find equations for the tangent and normal to the curve at (2, 4).
Use Implicit Differentiation
find the slope of the tangent at (2,4)find the slope of the normal at (2,4)
1. Differentiate both sides of the equation with respect to x, treating y as a function of x. This requires the chain rule.
2. Collect terms with dy/dx on one side of the equation.
3. Factor dy/dx
4. Solve for dy/dx
3 3 9 0x y xy Solution
2 23 3 (9 9) 0dy dy
x y x ydx dx
2 23 9 9 3dy dy
y x y xdx dx
2 23 3 9 9 ) 0dy dy
x y x ydx dx
2
2
9 3
3 9 )
dy y x
dx y x
2 2(3 9 ) 9 3dy
y x y xdx
2
tan 2
9(4) 3(2) 24 4
30 53(4) 9(2)m
5
4normalm
Find dy/dx2 2 2( )x y x y
1. Write the equation of the tangent line at (0,1)
2. Write the equation of the normal line at (0,1)
1. Differentiate both sides of the equation with respect to x, treating y as a function of x. This requires the chain rule.
2. Collect terms with dy/dx on one side of the equation.
3. Factor dy/dx
4. Solve for dy/dx
Solution 2 2 2( )x y x y
2 21 2( )(2 2 )dy dy
x y x ydx dx
3 2 2 31 4 4 4 4dy dy dy
x x y xy ydx dx dx
2 3 3 24 4 1 4 4dy dy dy
x y y x xydx dx dx
2 3 3 2(1 4 4 ) 1 4 4dy
x y y x xydx
3 2
2 3
1 4 4
1 4 4
dy x xy
dx x y y
Find dy/dx2 2 2( )x y x y
1. Write the equation of the tangent line at (0,1)
2. Write the equation of the normal line at (0,1)
3 2
2 3
1 4 4
1 4 4
dy x xy
dx x y y
1 11 ( 0) 1
3 3y x or y x
1 3( 0) 3 1y x or y x
3.8 Higher Derivatives
The derivative of a function f(x) is a function itself f ´(x). It has a derivative, called the second derivative f ´´(x)
If the function f(t) is a position function, the first derivative f ´(t) is a velocity function and the second derivative f ´´(t) is acceleration.
2
2( )
d yf x
dx
The second derivative has a derivative (the third derivative) and the third derivative has a derivative etc.
3
3( )
d yf x
dx
4(4)
4( )
d yf x
dx ( ) ( )
nn
n
d yf x
dx
Find the second derivative for ( )1
xf x
x
2 2
( 1)(1) (1) 1( )
( 1) ( 1)
x xf x
x x
22
1( ) ( 1)
( 1)
d df x x
dx x dx
33
2( ) 2( 1) (1)
( 1)f x x
x
Find the third derivative for ( )1
xf x
x
In algebra we study relationships among variables
•The volume of a sphere is related to its radius •The sides of a right triangle are related by Pythagorean Theorem•The angles in a right triangle are related to the sides.
In calculus we study relationships between the rates of change of variables.
How is the rate of change of the radius of a sphere related to the rate of change of the volume of that sphere?
Examples of rates-assume all variables are implicit functions of t = time
Rate of change in radius of a sphere
Rate of change in volume of a sphere
Rate of change in length labeled x
Rate of change in area of a triangle
dr
dt
dV
dt
dx
dt
d
dt
Rate of change in angle,
dA
dt
3.9
Solving Related Rates equations
1. Read the problem at least three times.2. Identify all the given quantities and the quantities
to be found (these are usually rates.)3. Draw a sketch and label, using unknowns when
necessary.4. Write an equation (formula) that relates the
variables.5. ***Assume all variables are functions of time and
differentiate wrt time using the chain rule. The result is called the related rates equation.
6. Substitute the known values into the related rates equation and solve for the unknown rate.
Figure 2.43: The balloon in Example 3.Related RatesA hot-air balloon rising straight up from a level field is tracked by a range finder 500 ft from the liftoff point. The angle of elevation is increasing at the rate of 0.14 rad/min. How fast is the balloon rising when the angle of elevation is is /4?
Given: 0.14 / min
drad
dt
500x ft
500x
Find:
4
dywhen
dt
y
Figure 2.43: The balloon in Example 3.Related RatesA hot-air balloon rising straight up from a level field is tracked by a range finder 500 ft from the liftoff point. At the moment the range finder’s elevation angle is /4, the angle is increasing at the rate of 0.14 rad/min. How fast is the balloon rising at that moment?
x
tan500
y
y
tan500
d d y
dt dt
2 1sec
500
d dy
dt dt
2 1sec ( )(.14)
4 500
dy
dt
2
1 1(.14)
500cos ( )4
dy
dt 140 / sec
dyft
dx
Figure 2.44: Figure for Example 4.Related RatesA police cruiser, approaching a right angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. The cruiser is moving at 60 mph and the police determine with radar that the distance between them is increasing at 20 mph. When the cruiser is .6 mi. north of the intersection and the car is .8 mi to the east, what is the speed of the car?
Given: 20
60
dsmph
dtdy
mphdt
Find:.8, .6
dxwhen x y
dt
Figure 2.44: Figure for Example 4.
Given: 20
60
dsmph
dtdy
mphdt
Find:.8, .6
dxwhen x y
dt
2 2 2s x y
2 2 2ds dx dy
s x ydt dt dt
then s = 1
2(1)(20) 2(.8) 2(.6)( 60)dx
dt
70dx
mphdt
Figure 2.45: The conical tank in Example 5.Related RatesWater runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base of radius 5 ft. How fast is the water level rising when the water is 6 ft. deep?
Given: 39 / min
10 , 5
dVft
dtH ft R ft
Find:
6 .dy
when y ftdt
Figure 2.45: The conical tank in Example 5.Water runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base of radius 5 ft. How fast is the water level rising when the water is 6 ft. deep?
Given: 39 / min
10 , 5
dVft
dtH ft R ft
Find:
6 .dy
when y ftdt
21
3V x y
5
10
x
y 2y x
2 31 2(2 )
3 3V x x x
22dV dx
xdt dt
29 2 (3)dx
dt
x=3
1
2
dx
dt
1 12* 2* .32 / min
2
dy dxft
dt dt
3. .10 The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent.
Differentiability
Differentiability and Linearization
Approximating the change in the function f by the change in the tangent line of f.
Linearization
Write the equation of the straight line approximation co (0 1)s ,y x x at
1 sin 1 0 (0,1)1 ay x t
1 1( )y y m x x
( ) ( )( )y f a f a x a
Point-slope formula
y=f(x)
1 1( 0)
1
y x
y x