Implicit Differentiation. Number of heart beats per minute, t seconds after the beginning of a race...

Implicit Implicit DifferentiationDifferentiation

Number of heart Number of heart beats per minute, t beats per minute, t seconds seconds

after the beginning of a race is after the beginning of a race is given bygiven by

What is your heart rate at the What is your heart rate at the beginning of the race?beginning of the race?

21400 3 204( )20

t tR t

t


after the beginning of a race is after the beginning of a race is given bygiven by

Find R(0).Find R(0).

R(0) = R(0) = beats/min. beats/min.

21400 3 204( )20

t tR t

t

20 20 89.44


How fast is your heart beating How fast is your heart beating after 100 seconds?after 100 seconds?

How fast is your heart rate How fast is your heart rate increasing after 100 sec? increasing after 100 sec? [Acceleration][Acceleration]

21400 3 204( )20

t tR t

t


How fast is your heart beating How fast is your heart beating after 100 seconds? R(100) after 100 seconds? R(100)

How fast is your heart rate How fast is your heart rate increasing after 100 sec? increasing after 100 sec? [Acceleration][Acceleration]

21400 3 204( )20

t tR t

t


How fast is your heart beating How fast is your heart beating after 100 seconds? R(100) after 100 seconds? R(100)

How fast is your heart rate How fast is your heart rate increasing after 100 sec? increasing after 100 sec? [Acceleration] R’(100)[Acceleration] R’(100)

21400 3 204( )20

t tR t

t


What is your heart rate at the What is your heart rate at the end of the race?end of the race?

21400 3 204( )20

t tR t

t


What is your heart rate at the What is your heart rate at the end of the race?end of the race? 2 21

400 3 20 /4lim ( )( 20) /t

t t tR t

t t

21400 3 204( )20

t tR t

t


What is your heart rate at the What is your heart rate at the end of the race?end of the race?

21400 3 204( )20

t tR t

t

2 21 1400 3 20 / 4004 2lim ( ) 200( 20) / 1t

t t tR t

t t

If h(x) = [g(x)]If h(x) = [g(x)]n n thenthen

h’(x) = n [g(x)] h’(x) = n [g(x)]n-1n-1g’(x)g’(x)

We review the power rule.We review the power rule.

If y = 4(2x + 2)If y = 4(2x + 2)55 then y’ then y’ = =

A.A. 20 (2x + 2)20 (2x + 2)44 2 = 40 (2x + 2) 2 = 40 (2x + 2)4 4

B.B. 20 (2x + 2)20 (2x + 2)55 2 = 40 (2x + 2) 2 = 40 (2x + 2)55

C.C. 20(2)20(2)44 = 20 (16) = 320 = 20 (16) = 320

D.D. 20 (2x + 2)20 (2x + 2)44 = 20 (2x + 2) = 20 (2x + 2)44

If y = 4(2x + 2)If y = 4(2x + 2)55 then y’ then y’ = =

A.A. 20 (2x + 2)20 (2x + 2)44 2 = 40 (2x + 2) 2 = 40 (2x + 2)4 4

B.B. 20 (2x + 2)20 (2x + 2)55 2 = 40 (2x + 2) 2 = 40 (2x + 2)55

C.C. 20(2)20(2)44 = 20 (16) = 320 = 20 (16) = 320

D.D. 20 (2x + 2)20 (2x + 2)44 = 20 (2x + 2) = 20 (2x + 2)44

If h(x) = [3x + 1/x]If h(x) = [3x + 1/x]77

h’(x) = h’(x) =

A.A. 21 x - 7/x21 x - 7/x22

B.B. 7 [3x + 1/x]7 [3x + 1/x]6 6 [3 + 1/x[3 + 1/x22]]

C.C. 7 [3x + 1/x]7 [3x + 1/x]6 6 [3 – 1/x[3 – 1/x22]]

D.D. 7 [3 – 1/x]7 [3 – 1/x]66

If h(x) = [3x + 1/x]If h(x) = [3x + 1/x]77

h’(x) = h’(x) =

A.A. 21 x - 7/x21 x - 7/x22

B.B. 7 [3x + 1/x]7 [3x + 1/x]6 6 [3 + 1/x[3 + 1/x22]]

C.C. 7 [3x + 1/x]7 [3x + 1/x]6 6 [3 – 1/x[3 – 1/x22]]

D.D. 7 [3 – 1/x]7 [3 – 1/x]66

Replace g(x) with y.Replace g(x) with y.

Instead of ([g(x)]Instead of ([g(x)]n n )’ = n [g(x)])’ = n [g(x)]n-1n-1g’(x)g’(x)

We get (y We get (y n n )’ = n [ y ])’ = n [ y ]n-1 n-1 y’y’

Replace g(x) with y.Replace g(x) with y.

We get (y We get (y n n )’ = n [ y ])’ = n [ y ]n-1 n-1 y’y’

Recall [(x+1)(xRecall [(x+1)(x22-3)]’ -3)]’ = (x+1)(2x)+ (x= (x+1)(2x)+ (x22-3)-3)

So [(x+1)y]’ = (x+1)y’ + y So [(x+1)y]’ = (x+1)y’ + y

[3 y[3 y3 3 ]’ = 9 y]’ = 9 y2 2 y’y’

' 3 2 22

23 3

2 3 4 ' 2 (6 3)2

2 3 2 3

x x yy y xy

x x x x

So [(2x+1)y]’ = ?So [(2x+1)y]’ = ?

A.A. 2y’2y’

B.B. (2x+1) y’ + 2y(2x+1) y’ + 2y

C.C. (2x+1) y’ + 2y’(2x+1) y’ + 2y’

D.D. (2x+1) y’(2x+1) y’

So [(2x+1)y]’ = ?So [(2x+1)y]’ = ?

A.A. 2y’2y’

B.B. (2x+1) y’ + 2y(2x+1) y’ + 2y

C.C. (2x+1) y’ + 2y’(2x+1) y’ + 2y’

D.D. (2x+1) y’(2x+1) y’

Find [2yFind [2y66 + 2y]’ + 2y]’

A.A. 12y12y55 y’ – 2 y’ – 2

B.B. 12y12y55 y’ + 2y’ y’ + 2y’

C.C. 12y12y55 y’ + 2yy’ y’ + 2yy’

D.D. 12y12y55 + 2y’ + 2y’

Find [2yFind [2y66 + 2y]’ + 2y]’

A.A. 12y12y55 y’ – 2 y’ – 2

B.B. 12y12y55 y’ + 2y’ y’ + 2y’

C.C. 12y12y55 y’ + 2yy’ y’ + 2yy’

D.D. 12y12y55 + 2y’ + 2y’

Guidelines for Guidelines for Finding the Finding the Derivative ImplicitlyDerivative ImplicitlyLet y stand for one of any number of Let y stand for one of any number of

functions.functions.

Differentiate both sides of the equation, Differentiate both sides of the equation, using chain rule, power rule, product using chain rule, power rule, product rule, quotient rule.rule, quotient rule.

If xIf x22 + y + y22 = 36 find y’. = 36 find y’.

What is the derivative of xWhat is the derivative of x22 ? ?

2x2x

What is the derivative of yWhat is the derivative of y22 ? ?

2yy’2yy’

What is the derivative of 36 ?What is the derivative of 36 ?

What is the derivative of What is the derivative of 36?36?

0.00.0

0.10.1

Differentiate both Differentiate both sidessides x x22 + y + y22 = 36 = 36A.A. 2x + 2yy’ = 02x + 2yy’ = 0

B.B. 2x + 2yy’ = 362x + 2yy’ = 36

C.C. 2x + 2y = 02x + 2y = 0

Differentiate both Differentiate both sidessides x x22 + y + y22 = 36 = 36A.A. 2x + 2yy’ = 02x + 2yy’ = 0

B.B. 2x + 2yy’ = 362x + 2yy’ = 36

C.C. 2x + 2y = 02x + 2y = 0

Guidelines for Guidelines for Finding the Finding the Derivative ImplicitlyDerivative ImplicitlyLet Y stand for one of any number of Let Y stand for one of any number of

functions.functions.


Place all of the terms containing Y' on Place all of the terms containing Y' on one side and the other terms on the one side and the other terms on the other.other.


2x + 2yy’ = 02x + 2yy’ = 0

2yy’ = -2x2yy’ = -2x

Guidelines for Guidelines for Finding the Finding the Derivative ImplicitlyDerivative Implicitly


Place all of the terms containing y' on Place all of the terms containing y' on one side and the other terms on the one side and the other terms on the other.other.

Factor the y' out if necessary and solve Factor the y' out if necessary and solve for y’. for y’.


2yy’ = -2x2yy’ = -2x

yy’ = -xyy’ = -x

y’ = y’ = x

y

Guidelines for Guidelines for Finding the Finding the Derivative ImplicitlyDerivative Implicitly

Place all of the terms containing Y' on Place all of the terms containing Y' on one side and the other terms on the one side and the other terms on the other.other.

Factor the Y ' out if necessary and solve Factor the Y ' out if necessary and solve for Y '.for Y '.

Replace x and y by the values given. Replace x and y by the values given.

y’ =y’ = x x22 + y + y22 = 36 = 36

Top point only!Top point only!

Find the slope when x = 2.Find the slope when x = 2.

When x = 2, y = When x = 2, y = oror

Y = Y =

Thus y’ = Thus y’ =

x

y

32

236 x2

32

y’ = y’ =

Y’ = for top pointY’ = for top point

Y’ = bottom Y’ = bottom pointpoint

2 2

32 32

x

y

2

32

If y’ =If y’ =

find the slope at (-3 , find the slope at (-3 , 3)3)

1.7321.732

0.10.1

x

y

3

Thus if the dolphins Thus if the dolphins forehead could be forehead could be approximated by a circle,approximated by a circle,

we could calculate we could calculate

the slope there ifthe slope there if

we knew the x andwe knew the x and

y coordinates. y coordinates.

y’ = y’ = x

y

And if the dolphins throat And if the dolphins throat could be approximated by a could be approximated by a circle,circle,

we could calculate we could calculate

the slope there ifthe slope there if

we knew the x andwe knew the x and

y coordinates. y coordinates.

y’ = y’ = x

y

Guidelines for Finding the Derivative Guidelines for Finding the Derivative ImplicitlyImplicitly

Let Y stand for one of any number of Let Y stand for one of any number of functions.functions.

Differentiate both sides of the equation, Differentiate both sides of the equation, usingusing

chain rule, power rule, product rule, quotient chain rule, power rule, product rule, quotient rule.rule.

Place all of the terms containing y' on one Place all of the terms containing y' on one side side and the other terms on and the other terms on the other. the other.

Factor the y’ out if necessary and solve for Factor the y’ out if necessary and solve for y’. y’.

Replace x and y by the values given. Replace x and y by the values given.

Let y stand for one of 5 Let y stand for one of 5 functions. functions.

If 3xIf 3x22 + xy + xy55 = 16x find y’. = 16x find y’.

What is the derivative of 3xWhat is the derivative of 3x22 ? ?

6x6x

What is the derivative of xyWhat is the derivative of xy55 ? ?

x5yx5y44y’+yy’+y55

What is the derivative of 16x ?What is the derivative of 16x ?

Differentiate xyDifferentiate xy55 using the using the product rule and power product rule and power rulerule

x5yx5y44y’ + yy’ + y55

3x3x22 + xy + xy55 = 16x = 16xDifferentiate both Differentiate both sidessidesA.A. 6x + 5y6x + 5y44 y’= 16 y’= 16

B.B. 6x +5xy6x +5xy4 4 y’= 16y’= 16

C.C. 6x +5xy6x +5xy4 4 y’+yy’+y5 5 = 16= 16

D.D. 6 + 6x +5xy6 + 6x +5xy4 4 y’+yy’+y5 5 = 0= 0

3x3x22 + xy + xy55 = 16x = 16xDifferentiate both Differentiate both sidessidesA.A. 6x + 5y6x + 5y44 y’= 16 y’= 16

B.B. 6x +5xy6x +5xy4 4 y’= 16y’= 16

C.C. 6x +5xy6x +5xy4 4 y’+yy’+y5 5 = 16= 16

D.D. 6 + 6x +5xy6 + 6x +5xy4 4 y’+yy’+y5 5 = 0= 0

Guidelines for finding Guidelines for finding the derivative the derivative implicitlyimplicitlyLet y stand for one of 5 functions.Let y stand for one of 5 functions.


If 3xy + x + 5yIf 3xy + x + 5y22 = 16 find = 16 find y’.y’.

3xy’ + 3y + 1 + 10yy’ = 03xy’ + 3y + 1 + 10yy’ = 0

Place all (y’)’s on the left and Place all (y’)’s on the left and factorfactor

3xy’ + 10yy’ = -1 - 3y3xy’ + 10yy’ = -1 - 3y

(3x+10y)y’= -1 - 3y(3x+10y)y’= -1 - 3y

y’ = y’ = 1 3

3 10

y

x y

If xIf x22 + 5y + 5y22 = 6 find y’. = 6 find y’.

2x + 10yy’ = 02x + 10yy’ = 0

y’ = -2x/10y = -x/5yy’ = -2x/10y = -x/5y

y’ = -x/5yy’ = -x/5yFind y’ when x = y = 1Find y’ when x = y = 1

-0.2-0.2

0.10.1

If xIf x22 + 5y + 5y22 = 6 find y’’. = 6 find y’’.

y’ = -x/5yy’ = -x/5y

5y(-1)-(-x)5y’ 5xy’-5y5y(-1)-(-x)5y’ 5xy’-5y

y’’ = ----------------- = -------------y’’ = ----------------- = -------------

25 y25 y2 2 25 y25 y22

Find y’’ when x = y = 1Find y’’ when x = y = 1and y’ = -0.2 and y’ = -0.2

-0.24-0.24

0.10.1

2

5 ' 5''

25

xy yy

y

Find y’’ when x = y = 1Find y’’ when x = y = 1and y’ = -0.2 and y’ = -0.2

-0.24-0.24

0.10.1

2

5 ' 5 1 5''

25 25

xy yy

y

Find dy/dxFind dy/dx

1. x1. x2 2 y + yy + y2 2 x = 6x = 6 6. (3xy + y)6. (3xy + y)3 3 = 6y = 6y

Find the equation of Find the equation of the tangent and the the tangent and the normal normal 29. x29. x2 2 + xy - y+ xy - y2 2 = 1 at (2, 3)= 1 at (2, 3) 31. x31. x2 2 yy2 2 = 9 at (1, 3)= 9 at (1, 3) 35. 2xy + 35. 2xy + x = 2 x = 2 at (1, at (1, /2)/2)

xx22 – 2xy + y – 2xy + y22sin(xy) > 4sin(xy) > 4

Find y’ at the point (-2, 0)Find y’ at the point (-2, 0)

2x -2xy’-2y+y2cos(xy)[xy’+y]2x -2xy’-2y+y2cos(xy)[xy’+y]+2sin(xy)yy’=0+2sin(xy)yy’=0

Example OnlyExample Only

Implicit Implicit DifferentiationDifferentiation

Related RatesRelated Rates

x’ = dx/dx = 1x’ = dx/dx = 1

x’ = dx/dt x’ = dx/dt

The edge of an ice The edge of an ice cube is decreasing at cube is decreasing at 2mm/sec when x = 2mm/sec when x = 400mm, 400mm, find the rate the volume is find the rate the volume is

changing at that instant.changing at that instant.

The edge of a cube is The edge of a cube is decreasing at 2mm/sec decreasing at 2mm/sec when x = 400mm, find when x = 400mm, find v’.v’. v = xv = x3 3

v’ = 3xv’ = 3x22

The edge of a cube is The edge of a cube is decreasing at 2mm/sec decreasing at 2mm/sec when x = 400mm, find when x = 400mm, find v’.v’. v = xv = x3 3

v’ = 3xv’ = 3x2 2 x’x’

The edge of a cube is The edge of a cube is decreasing at 2mm/sec decreasing at 2mm/sec when x =400mm, find when x =400mm, find v’.v’. v = xv = x3 3

v’ = 3xv’ = 3x2 2 x’x’ v’ = 3(160000)(-2) = v’ = 3(160000)(-2) = -480000 mm-480000 mm33/ sec/ sec

Related RatesRelated RatesRead the problem, drawing a pictureRead the problem, drawing a picture

No non-constants on the pictureNo non-constants on the picture

Write an equationWrite an equation

Differentiate implicitlyDifferentiate implicitly

Enter non-constants and solveEnter non-constants and solve

The edge of a cube is The edge of a cube is decreasing at 2mm/sec decreasing at 2mm/sec when x = 400mm, find when x = 400mm, find v’.v’. find the rate the volume is find the rate the volume is

changing at that instant.changing at that instant.

400 does not appear on the 400 does not appear on the drawing drawing

Set x=400 after v’ = 3xSet x=400 after v’ = 3x2 2 x’x’

Related RatesRelated RatesSuppose a painter is standing on a Suppose a painter is standing on a 13 foot ladder and Joe ties a rope 13 foot ladder and Joe ties a rope to the bottom of the ladder and to the bottom of the ladder and walks away at the rate of 2 feet walks away at the rate of 2 feet per second.per second.

Related RatesRelated RatesSuppose a painter is standing on a Suppose a painter is standing on a 13 foot ladder and Joe ties a rope 13 foot ladder and Joe ties a rope to the bottom of the ladder and to the bottom of the ladder and walks away at the rate of 2 feet walks away at the rate of 2 feet per second.per second.

How fast is the painter How fast is the painter

falling when x = 5 feet?falling when x = 5 feet?

Related RatesRelated Rates13 is a constant and appears 13 is a constant and appears

the rate of 2 feet per second [x’ = the rate of 2 feet per second [x’ = 2]2]



Related RatesRelated Rates13 is a constant and appears 13 is a constant and appears

the rate of 2 feet per second [x’ = the rate of 2 feet per second [x’ = 2]2]



NOT ON THE DRAWINGNOT ON THE DRAWING

What does the What does the Pythagorean Theorem Pythagorean Theorem say ?say ?

A.A. xx22 + y + y22 = 13 = 13

B.B. 2x2x22 + y + y22 = 169 = 169

C.C. xx22 + 2y + 2y22 =13 =13

D.D. xx22 + y + y22 = 169 = 169

What does the What does the Pythagorean Theorem Pythagorean Theorem say?say?

A.A. xx22 + y + y22 = 13 = 13

B.B. 2x2x22 + y + y22 = 169 = 169

C.C. xx22 + 2y + 2y22 =13 =13

D.D. xx22 + y + y22 = 169 = 169

Related RatesRelated RatesWrite an equationWrite an equation

Differentiate the equationDifferentiate the equation

implicitlyimplicitly

2x x’ + 2y y’ = 0 or xx’ + yy’ = 02x x’ + 2y y’ = 0 or xx’ + yy’ = 0

If Joe pulls at 2 ft./sec., find the If Joe pulls at 2 ft./sec., find the speedspeed

of the painter when x = 5.of the painter when x = 5.

2 2 213x y

Find y when x = 5Find y when x = 5

A.A. 55

B.B. 88

C.C. 1212

D.D. 1313

2 2 213x y

Find y when x = 5Find y when x = 5

A.A. 55

B.B. 88

C.C. 1212

D.D. 1313

2 2 213x y

Related RatesRelated RatesUse algebra to find y.Use algebra to find y.

xx22 + y + y22 = 169 = 169

5522 + y + y22 = 169 = 169

yy22 = 169 – 25 = 144 = 169 – 25 = 144

y = 12y = 12

2x x’ + 2y y’ = 0 or 2x x’ + 2y y’ = 0 or xx’ + yy’ = 0xx’ + yy’ = 0Back to the calculus with y = 12,Back to the calculus with y = 12,

x = 5, and x’ = 2 ft/secx = 5, and x’ = 2 ft/sec

xx’ + yy’ =0xx’ + yy’ =0

5(2) + 12(y’) = 05(2) + 12(y’) = 0

y’ = -10/12 = -5/6 ft./sec.y’ = -10/12 = -5/6 ft./sec.

2 2 213x y

Related RatesRelated RatesSummarySummary

Even though Joe is walking 2 ft/sec,Even though Joe is walking 2 ft/sec,

the painter is only falling -5/6 the painter is only falling -5/6 ft/sec. ft/sec.

If the x and y values were If the x and y values were reversed,reversed,

xx’ + yy’ =0 Find xx’ + yy’ =0 Find y’ if x’=2, y=5, and x = y’ if x’=2, y=5, and x = 1212A.A. -24/5-24/5

B.B. 5/245/24

C.C. -5/24-5/24

D.D. -5-5

xx’ + yy’ =0 Find xx’ + yy’ =0 Find y’ if x’=2, y=5, and x = y’ if x’=2, y=5, and x = 1212A.A. -24/5-24/5

B.B. 5/245/24

C.C. -5/24-5/24

D.D. -5-5

Related RatesRelated RatesSummarySummary

0lim ' ? when 5 ' 0yy x yy

0

5limy

x

y

0

5lim ? since x is approaching 13y

x

y


Suppose air is entering a balloon at Suppose air is entering a balloon at the rate of 25 cubic feet per the rate of 25 cubic feet per minute. How fast is the radius minute. How fast is the radius changing when r = 30 feet?changing when r = 30 feet?


r’ = 0.00221r’ = 0.00221

ft per minft per min

34

3v r

24dv dr

rdt dt

225 4 30dr

dt

25

4 900

dr

dt


Suppose a 6 ft tall person walks Suppose a 6 ft tall person walks away from a 13 ft lamp post at a away from a 13 ft lamp post at a speed of 5 ft per sec. How fast is speed of 5 ft per sec. How fast is the tip of his shadow moving when the tip of his shadow moving when 12 ft from the post? 12 ft from the post?

6

s


Suppose a 6 ft tall person walks Suppose a 6 ft tall person walks away from a 13 ft lamp post at a away from a 13 ft lamp post at a speed of 5 ft per sec. How fast is speed of 5 ft per sec. How fast is the tip of his shadow moving when the tip of his shadow moving when 12 ft from the post? 12 ft from the post?

6 13

s

Find ?Find ?

A.A. ss

B.B. xx

C.C. s + xs + x

D.D. 33

6 13

?s

Find ?Find ?

A.A. ss

B.B. xx

C.C. s + xs + x

D.D. 33

6 13

?s


The tip of the shadow has a speed The tip of the shadow has a speed of of

(s+x)’, not s’. What is s’ ?(s+x)’, not s’. What is s’ ?

s’ is the growth of the shadow and s’ is the growth of the shadow and includes getting shorter on the includes getting shorter on the right.right.6 13

s s x


The tip of the shadow has a speed The tip of the shadow has a speed of of

(s+x)’, not s’. What is s’ ?(s+x)’, not s’. What is s’ ?

s’ is the growth of the shadow and s’ is the growth of the shadow and includes getting shorter on the includes getting shorter on the right.right.

Cross multiplyingCross multiplying

6 13

s s x

6( ) 13s x s


6( ) 13s x s 6 6 13s x s


Since x’ = 5Since x’ = 5

6( ) 13s x s 6 6 13

6 7

6 ' 7 '

s x s

x s

x s

7 '5

6'

s

s


Thus s’ isThus s’ is

andand

Note that the tip is moving almostNote that the tip is moving almost

twice as fast as the walker, and more twice as fast as the walker, and more than twice as fast as the shadow than twice as fast as the shadow regardless of x.regardless of x.

30'7

s 30 65' ' 5

7 7s x

Related RatesRelated RatesSuppose a radar gun on first base Suppose a radar gun on first base catches a baseball 30 feet away catches a baseball 30 feet away from the pitcher and registers 50 from the pitcher and registers 50 feet per second. How fast is the feet per second. How fast is the ball really traveling?ball really traveling?

a baseball 30 feet away a baseball 30 feet away from the pitcher and from the pitcher and registers 50 registers 50 . feet per . feet per second.second.

A.A. x=30 y’=50x=30 y’=50

B.B. x=30 y =50x=30 y =50

C.C. x’=30 y’=50x’=30 y’=50

D.D. x’=30 y=50x’=30 y=50


The calculus.The calculus.

x = 30 y’ = 50 y = ?x = 30 y’ = 50 y = ?

The algebra.The algebra.

2 2 22 45 x y

2 ' 2 'xx yy

2 2 22 45 30 y

4950 y

DifferentiateDifferentiateimplicitlyimplicitly

A.A. 2x = 2y y’2x = 2y y’

B.B. 2x = y y’2x = y y’

C.C. 2x x’ = y y’2x x’ = y y’

D.D. 2x x’ = 2y y’2x x’ = 2y y’

2 2 22 45 x y

DifferentiateDifferentiateimplicitlyimplicitly

A.A. 2x = 2y y’2x = 2y y’

B.B. 2x = y y’2x = y y’

C.C. 2x x’ = y y’2x x’ = y y’

D.D. 2x x’ = 2y y’2x x’ = 2y y’

2 2 22 45 x y

Related RatesRelated Ratesx = 30 y’ = 50 y = ?x = 30 y’ = 50 y = ?

Back to the calculus.Back to the calculus.

2 ' 2 'xx yy

4950 y

2(30) 'x

Related RatesRelated RatesBack to the calculus.Back to the calculus.

x = 30 y’ = 50 y = ?x = 30 y’ = 50 y = ?

2 ' 2 'xx yy4950 y

2(30) ' 2 4950 'x y

Related RatesRelated RatesBack to the calculus.Back to the calculus.

x = 30 y’ = 50 y = ?x = 30 y’ = 50 y = ?

2 ' 2 'xx yy4950 y

2(30) ' 2 4950(50)x

a baseball 30 feet away a baseball 30 feet away from the pitcher and from the pitcher and registers 50 registers 50 . feet per . feet per second.second.A.A. x’ = 0.6 root(4950)x’ = 0.6 root(4950)

B.B. x’ = 50 root(4950)x’ = 50 root(4950)

C.C. x’ = 30 root(4950)x’ = 30 root(4950)

D.D. x’ = 5/3 root(4950)x’ = 5/3 root(4950)

2 ' 2 4950 'xx y

a baseball 30 feet away a baseball 30 feet away from the pitcher and from the pitcher and registers 50 registers 50 . feet per . feet per second.second.A.A. x’ = 0.6 root(4950)x’ = 0.6 root(4950)

B.B. x’ = 50 root(4950)x’ = 50 root(4950)

C.C. x’ = 30 root(4950)x’ = 30 root(4950)

D.D. x’ = 5/3 root(4950)x’ = 5/3 root(4950)

2 ' 2 4950 'xx y

Related RatesRelated Ratesx’ = 117.260 feet/sec.x’ = 117.260 feet/sec.

X = 30 y’ = 50 y = ?X = 30 y’ = 50 y = ?

2 ' 2 'xx yy 4950 y

4950(50)'

30x

2(30) ' 2 4950(50)x

Related RatesRelated RatesRead the problem, drawing a pictureRead the problem, drawing a picture

No non-constants on the pictureNo non-constants on the picture

Write an equationWrite an equation

Differentiate implicitlyDifferentiate implicitly

Enter non-constants and solveEnter non-constants and solve

Date post:	14-Dec-2015
Category:	Documents
Upload:	darren-clinton-daniels
View:	216 times
Download:	0 times

Implicit Differentiation. Number of heart beats per minute, t seconds after the beginning of a race...

Documents