Find the derivative of the function f(x) = x 2 – 2x

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



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



THE QUOTIENT RULE:

2

( ) ( ) ( ) ( ) ( )

( ) ( )

d f x g x f x f x g x

dx g x g x

Example 1


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


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


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



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:



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



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



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


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


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

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