Date post: | 27-Dec-2015 |
Category: |
Documents |
Upload: | sibyl-eaton |
View: | 240 times |
Download: | 0 times |
Find the derivative of the function f(x) = x2 – 2x
0
( ) ( )lim'( )h
f x h f x
hf x
2 2
0
( ) 2( ) ( 2 )limh
x h x h x x
h
2 2 2
0
2 2 2 2limh
x xh h x h x x
h
2
0
2 2limh
xh h h
h
0lim 2 2h
x h
2 2x
State Standard – 4.4 Students derive derivative formulas and use them to find the derivatives of algebraic, trig, exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
RULE 1 Derivative of a Constant Function
If f has the constant value f(x) = c, then
( ) 0df d
cdx dx
If the derivative of a function is its slope, then for a constant function, the derivative must be zero.
Example 1a
Find the derivative of f(x) = 8
df
dx(8)
d
dx 0
Example 1b
Find the derivative of
df
dx 2
d
dx
0
( )2
f x
RULE 2 Power Rule for Positive Integers
If n is a positive integer, then
1n ndx nx
dx
In the Warm-Up we saw that if , .
2 2y x x This is part of a pattern.
1n ndx nx
dx
examples:
4f x x
34f x x
8y x78y x
power rule
2 2y x
y x
1
21
2y x
1
2y x
1
2
1
2
y
x
1
2y
x
RULE 3 Constant Mulitple Rule
If u is a differentiable function of x, and c is a constant, then
d ducu c
dx dx
d ducu c
dx dx
examples:
1n ndcx cnx
dx
constant multiple rule:
5 4 47 7 5 35d
x x xdx
RULE 4 The Sum Rule
If f and g are both differentiable, then
( )d du dv
u vdx dx dx
Example
Find the derivative of y = x4 + 12x
4( ) (12 )dy d d
x xdx dx dx
34 12x
RULE 5 The Difference Rule
If f and g are both differntiable, then
( )d du dv
u vdx dx dx
Example
Find the derivative of y = x3 – 3x
3( ) (3 )dy d d
x xdx dx dx
23 3x
RULE 6 The Derivative of the Natural Exponential Function
( )x xde e
dx
Pg. 191
3 – 31 odd
Find the derivative of the function f(x) = 3x2 – 5x + 1
6 5x
State Standard – 4.4 Students derive derivative formulas and use them to find the derivatives of algebraic, trig, exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
Derivative of Sine and Cosine Functions:
sin cosd
x xdx
cos sind
x xdx
THE PRODUCT RULE:
( ) ( ) ( ) ( ) ( ) ( )d
f x g x f x g x g x f xdx
Example 1
Find the derivative of f(x) = (2x+5)(3+4x)
( ) ( ) ( ) ( ) ( )f x f x g x g x f x
( ) (2 5)(4) (3 4 )(2)f x x x
( ) 16 26f x x
f(x) g(x)
( ) 8 20 6 8f x x x
Example 2
Find the derivative of f(x) = (4x3)(sin x)
( ) ( ) ( ) ( ) ( )f x f x g x g x f x
3 2( ) (4 )(cos ) (sin )(12 )f x x x x x
3 2( ) 4 cos 12 sinf x x x x x
f(x) g(x)
Example 3
Find the derivative of y = (5x2)(cos x) + (3x)(sinx)
( ) ( ) ( ) ( )y f x g x g x f x
2(5 )( sin ) (cos )(10 ) (3 )(cos ) (sin )(3)y x x x x x x x
25 sin 13 cos 3siny x x x x x
25 sin 10 cos 3 cos 3siny x x x x x x x
3-25 odd
Find the derivative of the function f(x) = (x – 4)(x + 3)
( ) ( ) ( ) ( ) ( )f x f x g x g x f x
( ) ( 4)(1) ( 3)(1)f x x x
( ) 2 1f x x
( ) 4 3f x x x
State Standard – 4.4 Students derive derivative formulas and use them to find the derivatives of algebraic, trig, exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
THE QUOTIENT RULE:
2
( ) ( ) ( ) ( ) ( )
( ) ( )
d f x g x f x f x g x
dx g x g x
Example 1
Find the derivative of
2
( ) ( ) ( ) ( )( )
( )
g x f x f x g xf x
g x
2
( )(2) (2 1)(1)( )
x xf x
x
2
1( )f x
x
f(x)
g(x)
2
2 2 1( )
x xf x
x
2 1( )
xf x
x
Example 2
Find the derivative of
2
( ) ( ) ( ) ( )( )
( )
g x f x f x g xf x
g x
2
(5 2)(3) (3 4)(5)( )
(5 2)
x xf x
x
2
26( )
(5 2)f x
x
f(x)
g(x)
2
15 6 15 20( )
(5 2)
x xf x
x
3 4( )
5 2
xf x
x
Example 3
Find the derivative of
2
( ) ( ) ( ) ( )( )
( )
g x f x f x g xf x
g x
3 2
3 2
( )( sin ) (cos )(3 )( )
( )
x x x xf x
x
4
sin 3cos( )
x x xf x
x
f(x)
g(x)
2
6
( sin 3cos )( )
x x x xf x
x
3
cos( )
xf x
x
3 – 25 odd
State Standard – 4.4 Students derive derivative formulas and use them to find the derivatives of algebraic, trig, exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
Derivatives of the remaining trig functions:
sin cosd
x xdx
cos sind
x xdx
2tan secd
x xdx
2cot cscd
x xdx
sec sec tand
x x xdx
csc csc cotd
x x xdx
Example 1
Find the derivative of y = (sec x)(tan x)
( ) ( ) ( ) ( )y f x g x g x f x
2(sec )(sec ) (tan )(sec tan )y x x x x x
32sec secy x x
3 2sec sec tany x x x 3 2sec sec (sec 1)y x x x
3 3sec sec secy x x x
Higher Order Derivatives:
dyy
dx is the first derivative of y with respect to x.
2
2
dy d dy d yy
dx dx dx dx
is the second derivative.
(y double prime)
dyy
dx
is the third derivative.
4 dy y
dx is the fourth derivative.
We will learn later what these higher order derivatives are used for.
4 210 33 15y x x
' 340 66y x x '' 2120 66y x
''' 240y x
4 240y
WS
1) sin cosd
x xdx
2) cos sind
x xdx
23) tan secd
x xdx
24) cot cscd
x xdx
5) sec sec tand
x x xdx
6) csc csc cotd
x x xdx
Find the Derivative:
State Standard – 4.4 Students derive derivative formulas and use them to find the derivatives of algebraic, trig, exponential, and logarithmic functions.
Objective – To be able to find the derivative of a function.
So far we have been memorizing the derivatives of the trig functions. And today we will be investigating this further. This will help us as we go into the next section of using the Chain Rule.
2) Find the derivative of f(x) = (x)(sin x)
( ) ( ) ( ) ( ) ( )f x f x g x g x f x
( ) ( )(cos ) (sin )(1)f x x x x
( ) cos sinf x x x x
ON WHITE BOARD
a) Find the derivative of f(x) = (x)(cos x)
b) Find the derivative of f(x) = (x)(tan x)
( ) ( ) ( ) ( ) ( )f x f x g x g x f x ( ) ( )( sin ) (cos )(1)f x x x x
( ) sin cosf x x x x
( ) ( ) ( ) ( ) ( )f x f x g x g x f x 2( ) ( )(sec ) (tan )(1)f x x x x
2( ) sec tanf x x x x
4) Find the derivative of y = 2 csc x + 5 cos x
2( csc cot ) 5( sin )y x x x
2csc cot 5siny x x x
ON WHITE BOARD
a) Find the derivative of y = 4 sec x + 3 sin x
b) Find the derivative of y = 7 cot x + 2 tan x
4(sec tan ) 3(cos )y x x x
4sec tan 3cosy x x x
2 27( csc ) 2(sec )y x x
2 27csc 2secy x x
10) Find the derivative of
2
( ) ( ) ( ) ( )( )
( )
g x f x f x g xf x
g x
2
( cos )(cos ) (1 sin )(1 sin )
( cos )
x x x x xy
x x
2
cos
( cos )
x xy
x x
2 2
2
cos cos (1 sin )
( cos )
x x x xy
x x
1 sin
cos
xy
x x
2 2
2
cos cos 1 sin
( cos )
x x x xy
x x
1
ON WHITE BOARD
12) 2
( ) ( ) ( ) ( )( )
( )
g x f x f x g xf x
g x
2
2
(sec )(sec ) (tan 1)(sec tan )
(sec )
x x x x xy
x
2 2sec tan tan
sec
x x xy
x
3 2
2
sec (sec tan sec tan )
sec
x x x x xy
x
tan 1
sec
xy
x
3 2
2
sec sec tan sec tan
sec
x x x x xy
x
1 tan
sec
xy
x
2 21 tan secx x
Pg. 216
1 – 15 odd
cscy xcsc coty x x
2(csc )( csc ) (cot )( csc cot )y x x x x x
2csc(2csc 1)y x
32csc cscy x x
Find y´´
3 2csc csc cot )y x x x 3 2csc csc coty x x x
2 2csc (csc cot )y x x x 2 2csc (csc csc 1)y x x x
State Standard – 5.0 Students know the Chain Rule and its proof and applications to the calculation of the derivative of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve applications.
We now have a pretty good list of “shortcuts” to find derivatives of simple functions.
Of course, many of the functions that we will encounter are not so simple. What is needed is a way to combine derivative rules to evaluate more complicated functions.
Consider a simple composite function:
6 10y x
2 3 5y x
If 3 5u x
then 2y u
6 10y x 2y u 3 5u x
6dy
dx 2
dy
du 3
du
dx
dy dy du
dx du dx
6 2 3
one more:29 6 1y x x
23 1y x
If 3 1u x
3 1u x
18 6dy
xdx
2dy
udu
3du
dx
dy dy du
dx du dx
2y u
2then y u
29 6 1y x x
2 3 1dy
xdu
6 2dy
xdu
18 6 6 2 3x x This pattern is called the chain rule.
The Chain Rule can be written either in Leibniz notation:
Or in Prime Notation:
( ) ( ( )) ( )F x f g x g x
dy dy du
dx du dx
Example 1
Find of
4 3u x 9y u
dy
dx
827(4 3 )dy
xdx
9(4 3 )y x
dy du
du dx
89dy
udu
3du
dx
89u ( 3) 827u
On White Board
Find of
3 45u x x 7y u
dy
dx
3 4 6 2 37(5 ) (15 4 )dy
x x x xdx
3 4 7(5 )y x x
dy du
du dx
67dy
udu
2 315 4du
x xdx
67u 2 3(15 4 )x x
On White Board
Find of
43 7 5u x x 3y u
dy
dx
4 2 33(3 7 5) (12 7)dy
x x xdx
4 3(3 7 5)y x x
dy du
du dx
23dy
udu
312 7du
xdx
23u 3(12 7)x
9 – 42 mult of 3
secy xsec tany x x
2sec (sec ) tan (sec tan )y x x x x x
32sec secy x x
Find y´´
2 2sec (sec tan )y x x x 2 2sec (sec sec 1)y x x x
2sec (2sec 1)y x x
State Standard – 5.0 Students know the Chain Rule and its proof and applications to the calculation of the derivative of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve applications.
The Chain Rule can be written either in Leibniz notation:
Or in Prime Notation:
( ) ( ( )) ( )F x f g x g x
dy dy du
dx du dx
Example 1
Find of
sin 3u x 9y u
dy
dx
89cos (sin 3)dy
x xdx
9(sin 3)y x
dy du
du dx
89dy
udu
cosdu
xdx
89u (cos )x89cos ( )x u
On White Board
Find of
3
2
xu
sin cosy u u
dy
dx
3 3 3cos sin
2 2 2
dy x x
dx
3 3sin cos
2 2
x xy
dy du
du dx
cos sindy
u udu
3
2
du
dx
(cos sin )u u 3
2
On White Board
Find of
2 1u x cosy u
dy
dx
22 sin( 1)dy
x xdx
2cos( 1)y x
dy du
du dx
sindy
udu
2du
xdx
sin u (2 )x
9 – 42 mult of 3
Find the Derivative21) sin( 2 )x x
22) tan 4x3 23) sin(3 )x
24) cos10x25) cos x
26) cos )x37) sin(3 )x
38) cos( 4 )x x 29) sin 5x
3 210) cos( )x211) cos 4x
212) tan x213) sin )x
514) cos( )x
15) 3(tan 4 )x
16) sin 5x317) cos(2 )x
secy xsec tany x x
2sec (sec ) tan (sec tan )y x x x x x
32sec secy x x
Find y´´
2 2sec (sec tan )y x x x 2 2sec (sec sec 1)y x x x
2sec (2sec 1)y x x
State Standard – 5.0 Students know the Chain Rule and its proof and applications to the calculation of the derivative of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve applications.
Repeated Use of the Chain Rule
We sometimes have to use the Chain Rule two or more times to find a derivative.
Example 1
Find the derivative of g(t) = tan (5 – sin 2t)
( ) (tan(5 sin 2 ))d
g t tdt
5 sin 2u t
dy du
du dt
2secdy
udu
(5 sin 2 )du d
tdt dt
2sec u (5 sin 2 )d
tdt
tany u
2sec (5 sin 2 )t (5 sin 2 )d
tdt
2( ) sec (5 sin 2 ) ( 2cos 2 )g t t t 2u tcos
dyu
du
2du
dt
5 siny u
Example 2
Find the derivative of f(x) = cos ( sin 3x)
( ) (cos(sin 3 ))d
f x tdx
sin 3u x
dy du
du dx
sindy
udu
(sin 3 )du d
xdx dx
sin u (sin 3 )d
xdx
cosy u
sin(sin 3 )x (sin 3 )d
xdx
( ) sin(sin 3 ) (3cos3 )f x x x 3u xcos
dyu
du
3du
dx
siny u
Find the Derivative
1) sin(cos(2 5))y x
3
2) 1 tan12
xy
23) 1 cos( )y x
9 – 42 mult of 3
Find the Derivative
2
2 2
3
2
1) tan(sin(3 5))
2) sin(cos( ))
3) cos (3 2 )
4) 3 sec3
5) 4 cos 3
y x
y x
y x x
xy
y x x
2
2 3 2 2
4
3 2
6) 1 tan( )
7) 1 cos10
8) sin ( 4)cos ( )
9) tan (3 1)
10) cos (cos (cos ))
y x
xy
y x x
y x
y x
Find y´´
1) sin cosd
x xdx
2) cos sind
x xdx
23) tan secd
x xdx
24) cot cscd
x xdx
5) sec sec tand
x x xdx
6) csc csc cotd
x x xdx
Find the Derivative:
State Standard – 6.0 Students use implicit differentiation in a wide variety of problems.
Objective – To be able to use implicit differentiation
Implicit Differentiation
(Takes Four Steps)
1) Differentiate both sides of the equation with respect to ‘x’, treating ‘y’ as a differentiable function of ‘x’.
2) Collect the terms with the dy/dx on one side of the equation.
3) Factor out the dy/dx.
4) Solve for dy/dx.
2 2 1x y This is not a function, but it would still be nice to be able to find the slope.
2 2 1d d d
x ydx dx dx
Do the same thing to both sides.
2 2 0dy
x ydx
Note use of chain rule.
2 2dy
y xdx
2
2
dy x
dx y
dy x
dx y
22 siny x y
22 sind d d
y x ydx dx dx
This can’t be solved for y.
2 2 cosdy dy
x ydx dx
2 cos 2dy dy
y xdx dx
22 cosdy
xydx
2
2 cos
dy x
dx y
This technique is called implicit differentiation.
1 Differentiate both sides w.r.t. x.
2 Solve for .dy
dx
Example 1
Find of
4 6 ( ) 0x y y 4x
dy
dx
2
3
xy
y
2 22 3 4x y
6y
6 ( ) 4y y x
4x
6y
Example 2
Find of
2 2 ( ) 5 4( )x y y y 2x
dy
dx
5 2
2 4
xy
y
2 2 5 4x y x y
2 4y
2 ( ) 4( ) 5 2y y y x
2x
2 4y
4( )y 4( )y
(2 4) 5 2y y x
Example 3
Find of
2 23 3 ( ) 3 (1)( ) (1)( )x y y x y y
2x
dy
dx
2
2
x yy
y x
3 3 3x y xy
2y x
2 2 ( ) ( )x y y x y y 2x
2y x
( )x y ( )x y2 2( ) ( )y y x y x y
3 3
2 2( )y y x x y
Example 4
Find of
2 (sin )( sin )( ) (cos )(cos ) 0x y y y x
(cos )(cos )y x
dy
dx
cos cos
sin sin
y xy
x y
2sin cos 1x y
(sin )( sin )x y
(sin )( sin )( ) (cos )(cos ) 0x y y y x (cos )(cos )y x
(sin )( sin )x y
(sin )( sin )( ) (cos )(cos )x y y y x
2 2
Pg. 233
1a, 2a, 3a, 5 – 9, 11, 13, 14, and 20
From www.Dictionary.com
Explicit-
6.Mathematics. (of a function) having the dependent variable expressed directly in terms of the independent variables, as y = 3x + 4.
Implicit-4.Mathematics. (of a function) having the dependent variable not explicitly expressed in terms of the independent variables, as x2 + y2 = 1.
Find of
6 2 ( ) 2 5( )x y y y 6x
dy
dx
2 6
2 5
xy
y
2 23 2 5x y x y
2 5y
2 ( ) 5( ) 2 6y y y x
6x
2 5y
5( )y 5( )y
(2 5) 2 6y y x
State Standard – 4.4 Students derive derivative formulas and use them to find the derivatives of inverse trig functions.
Objective – To be able to take derivatives of Inverse Trig Functions.
1
2
1(sin )
1
du u
dx u
1sin arcsinx x
1
2
1(cos )
1
du u
dx u
12
1(tan )
1
du u
dx u
1
2
1(csc )
1
du u
dx u u
1
2
1(sec )
1
du u
dx u u
12
1(cot )
1
du u
dx u
1
2
1(sin )
1
du u
dx u
1sin x y
sinx yd d
dx dx
1 cosdy
ydx
2 2 1x b
1
cos
dy
dx y
1
y
x
b1
2 21b x 21b x
21
1
x
adj
hyp 2
1
1
dy
dx x
Example 1
Find the derivative of
2 2
1( ) 2
1 ( )f x x
x
4
2( )
1
xf x
x
1 2( ) sinf x x2u x
1
2
1(sin )
1
du u
dx u
Example 2
Find the derivative of
1
221
2
1 1( )
21
f x x
x
1( )
(2 2)f x
x x
1( ) tanf x x
1
2u x
12
1(tan )
1
du u
dx u
11 2( ) tanf x x
1 1( )
1 2f x
x x
1
( )(2)(1 )
f xx x
Find the Derivative.1 21) cos ( )y x
1 12) cosy
x
13) sin 2y x
14) sin 1y x
15) sec (2 1)y s
16) tan 3 1y x
17) csc2
xy
12
38) siny
x
In Sec. 3.2 we learned the basics of Higher Order Derivatives. So Find the first four derivatives of
23 6y x x
(4) 0y
3 23 2y x x
6 6y x
6y
First Derivative:
Second Derivative:
Third Derivative:
Fourth Derivative:
Velocity and Acceleration
State Standard – 7.0 Students compute derivatives of higher order.
Objective – To be able to solve problems involving multiple derivative steps.
Higher Order Derivatives:
dyy
dx is the first derivative of y with respect to x.
2
2
dy d dy d yy
dx dx dx dx
is the second derivative.
(y double prime)
dyy
dx
is the third derivative.
4 dy y
dx is the fourth derivative.
We will learn later what these higher order derivatives are used for.
Example 1
Find the second derivative of
2 sin 4cosy x x x
2 siny x x
(2 )(cos ) (sin )(2)y x x x
2 cos 2siny x x x
(2 )( sin ) (cos )(2) 2cosy x x x x
2 sin 2cos 2cosy x x x x
Example 2
Find the second derivative of
4
2 6xy
x
2
1xy
x
2
2 2
(1)( ) ( 1)(2 )
( )
x x xy
x
2 2
4
2 2x x xy
x
2
4
2x xy
x
3 2
3 2
( 1)( ) ( 2)(3 )
( )
x x xy
x
2
( ) ( ) ( ) ( )
[ ( )]
f x g x f x g xy
g x
3 3 2
6
( 3 6 )x x xy
x
3
2x
x
3 3 2
6
3 6x x xy
x
3 2
6
2 6x xy
x
Acceleration is the first derivative of Velocity.
Velocity is the first derivative of position.
Acceleration is the second derivative of position.
( ) ( )ds
v t s tdt
( ) ( ) ( )a t v t s t
( )s s t is the position function
Example 3
The position of a particle is given by the equation:
Where t is measured in seconds and s in meters.
a) Find the acceleration at time t. What is the acceleration after 4 seconds.
3 2( ) 6 9s f t t t t
2( ) 3 12 9ds
v t t tdt
( ) 6 12dv
a t tdt
(4) 6(4) 12a 2
(4) 12m
as
24 12
Pg. 240
5 – 15 odd, 29, 31, 43 and 44
The position of a particle is given by the equation:
Where t is measured in seconds and s in meters.
a) Find the acceleration at time t. What is the acceleration after 5 seconds.
3 2( ) 2 5 4s f t t t t
2( ) 6 10 4ds
v t t tdt
( ) 12 10dv
a t tdt
(5) 12(5) 10a 2(5) 50
ma
s60 10
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of:
Where t is measured in seconds and s in feet.
a) How high does the rock go?
2160 16 sec.s t t ft after t
( ) 160 32 / secds
v t t ftdt
0 160 32t 32 160t
( ) 45 00s ft5sect
What is the velocity at the rock’s height?25 5( ) 160( ) 516( )s
( ) 80 05 0 4 0s
State Standard – 4.4 Students find the derivatives of logarithmic functions.
Objective – Students will be able to solve problems involving logarithmic functions.
Derivative of Log Functions
1ln
dx
dx x
) ln8a y x1
88
yx
1ln
du u
dx u
Example 1
1y
x
2) ln( 2)b y x
2
12
2y x
x
2
2
2
xy
x
Derivative of Log Functions
1ln
dx
dx x
) lna y x x
1( ) (ln )(1)y x x
x
1ln
du u
dx u
Example 2
1 lny x
3) (ln )b y x
2 13(ln )y x
x
23(ln )xy
x
Produ
ct Rule
Chain R
ule
More Derivatives
x xde e
dx
2xy e
2( ) 2xy e
u ude e u
dx
Example 3
22 xy e
Derivative of Log Functions (Base other than e)
(ln )( )x xda a a
dx
) 2xa y
(ln 2)(2 )xy
(ln )u uda a a u
dx
Example 4
2 (ln 2)xy
3) 2 xb y 3(ln 2)2 3xy
3(3ln 2)(2 )xy
Derivative of Log Functions (Base other than e)
1log
(ln )a
dx
dx a x
2) loga y x
1log
(ln )a
du u
dx a u
Example 5
1
(ln 2)y
x
43) logb y x
34
14
(ln 3)y x
x
4
(ln 3)y
x
Pg. 249
2 – 5, 7, 9, 21 – 23, and 30
Find the First Derivative:
1)
Find the second Derivative:
2)
6xy
cos 2y x( sin 2 )(2)y x
6 ln 6xy
2sin 2y x
( 2)(cos 2 )(2)y x 4cos 2y x
State Standard – 4.2 Students demonstrate an understanding of the interpretation of the derivative as an instantaneous rate of change.
Objective – Students will be able to solve problems involving rate of change.
If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. We will find an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time.
34
3V r
3
: 100dv cm
Givendt s
24(3 )
3
dv drr
dt dt
24dv dr
rdt dt
Example 1
3
2
1100
4 (25)
dr cm
dt s
2
1
4
dr dv
dt r dt
1
25
dr cm
dt s
Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/s. How fast is the radius of the balloon increasing when the diameter is 50 cm?
: 25dr
Unknown when r cmdt
24 r24 r
Steps for Related Rates Problems:
1. Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.
2V r h3000
min
dv L
dt
21000 ( )dv dh
rdt dt
23000 1000dh
rdt
Example 2
2
3dh
dt r
2
3000
1000
dh
dt r
How rapidly will the fluid level inside a vertical cylindrical tank drop if we pump the fluid out at the rate of 3000 L/min?
h
21000 r21000 r
?dh
dt
But since there are 1000L in a cubic meter.
21000V r h
2
3
min
m
r
The fluid level will drop at the rate of
21
3V r h
3
2min
dv m
dt
21
3 2
hV h
2(3 )12
dv dhh
dt dt
Example 3
2
4dh dv
dt h dt
A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of 2 m3/min, find the rate of rising water when the water is 3 m deep.
h?
dh
dt
2
4
r
h
8
9 min
dh m
dt
4r
2
2
hr
3
12V h
2
4
dv h dh
dt dt
2
4
h2
4
h
3
2
42
(3) min
dh m
dt
0.28min
m
21
3V r h
3
9min
dv ft
dt
21
3 2
hV h
2(3 )12
dv dhh
dt dt
Example 4
2
4dh dv
dt h dt
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 radius of 5 ft. How fast is the water level rising when the water is 6 ft. deep?
h?
dh
dt
5
10
r
h
1
min
dh ft
dt
10r
5
2
hr
3
12V h
2
4
dv h dh
dt dt
2
4
h2
4
h
3
2
49
(6) min
dh ft
dt
0.32min
ft
Pg. 260
1, 3 – 5, 7, 10, and 19