Activity: Teacher-Directed Instruction

C CONVERSATION: Voice level 0. No talking!

HHELP: Raise your hand and wait to be called on.

AACTIVITY: Whole class instruction; students in seats.

M MOVEMENT: Remain in seat during instruction.

P PARTICIPATION: Look at teacher or materials being discussed. Raise hand to contribute; respond to questions, write or perform other actions as directed.NO SLEEPING OR PUTTING HEAD DOWN, TEXTING, DOING OTHER WORK.

S

Activity: Teacher-Directed Instruction

Calculus AB

2013 Implicit Differentiation

Objective

• C: The swbat differentiate implicitly equations in more than one variable.

• L: the sw explain to others how to find derivatives of multiple types of problems verbally and demonstratively

Implicit Differentiation

Equation for a line:

Explicit Form

<One variable given explicitly in terms of the other>

Implicit Form

<Function implied by the equation>

Differentiate the Explicit

< Explicit: , y is function of x >

Differentiation taking place with respect to x. The derivative is explicit also.

y mx b

Ax By C

24 3 4y x x

8 3dy xdx


Equation of circle:

To work explicitly; must work two equations

2 2 9y x

29y x

Implicit Differentiation is a Short Cut - A method to handle equations that are not easily written explicitly.

( Usually non-functions)

29y x 𝑑𝑦𝑑𝑥 =1

2(9−𝑥2 )

−12 (−2 𝑥 )

𝑑𝑦𝑑𝑥 =

−𝑥√9−𝑥2

𝑑𝑦𝑑𝑥=− 1

2( 9− 𝑥2 )

− 12 (−2𝑥 )

𝑑𝑦𝑑𝑥 =

𝑥√9−𝑥2

𝑥2𝑦+2 𝑦2𝑥+3 𝑦3=7Don’t want to solve for y


Chain Rule Pretend y is some function like

so becomes

(A)

(B)

(C)

Note: Use the Leibniz form. Leads to Parametric and Related Rates.

2 2 3y x x 2 4( 2 3)x x 4y

Find the derivative with respect to x

< Assuming - y is a differentiable function of x >

32y

4y

2 3x y

=

=

2 𝑑𝑥𝑑𝑥 +3 𝑑𝑦𝑑𝑥=2+3 𝑑𝑦𝑑𝑥

4 𝑦 3(2𝑥+2)

6 𝑦2(2 𝑥+2)


Find the derivative with respect to x

< Assuming - y is a differentiable function of x >

𝑥𝑦=¿ 𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥 ¿ 𝑥 𝑑𝑦𝑑𝑥 + 𝑦

𝑥2+ 𝑦2=¿2 𝑥 𝑑𝑥𝑑𝑥 +2 𝑦 𝑑𝑦𝑑𝑥 ¿2 𝑥+2 𝑦 𝑑𝑦𝑑𝑥

sin (𝑥𝑦) ¿cos (𝑥𝑦 )∗(𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥 )¿ 𝑥cos (𝑥𝑦 ) 𝑑𝑦𝑑𝑥 + 𝑦 cos (𝑥𝑦)


(D) Product Rule

2xy ¿ 𝑥 (2 𝑦 ) 𝑑𝑦𝑑𝑥 + 𝑦2 𝑑𝑥𝑑𝑥

¿2 𝑥𝑦 𝑑𝑦𝑑𝑥 +𝑦2

¿ 𝑥 𝑑𝑦𝑑𝑥+ 𝑦❑ 𝑑𝑥𝑑𝑥

¿ 𝑥 𝑑𝑦𝑑𝑥 +𝑦❑

𝑥𝑦


(E) Chain Rule 3( )xy

Product inside a chain

3 (𝑥𝑦 )2(𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥 )

3 𝑥3 𝑦2 𝑑𝑦𝑑𝑥+3 𝑥2 𝑦3

¿3 𝑥2 𝑦2(𝑥 𝑑𝑦𝑑𝑥 + 𝑦 𝑑𝑥𝑑𝑥 )

sin (𝑥𝑦)

cos (𝑥𝑦 )∗(𝑥 𝑑𝑦𝑑𝑥 + 𝑦 𝑑𝑥𝑑𝑥 )

𝑥cos (𝑥𝑦 ) 𝑑𝑦𝑑𝑥 + 𝑦 cos(𝑥𝑦 )


(E) Chain Rule

Product inside a chain

𝑢=𝑥𝑦 𝑑𝑢=𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥

sin𝑢 (𝑑𝑢)

cos𝑢(𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥 )

cos 𝑥𝑦 (𝑥 𝑑𝑦𝑑𝑥 )+cos 𝑥𝑦 (𝑦 𝑑𝑥𝑑𝑥 )


To find implicitly.

EX: Diff Both Sides of equation with respect to x

Solve for

dydx

2 2 9x y dydx2 𝑥 𝑑𝑥𝑑𝑥+2 𝑦 𝑑𝑦𝑑𝑥=0

2 𝑥+2 𝑦 𝑑𝑦𝑑𝑥=0

𝑑𝑦𝑑𝑥=

−2𝑥2 𝑦 =

−𝑥𝑦

29y x

29y x

Need both x and y to find the slope.

C CONVERSATION: Voice level 0. No talking!

HHELP: Raise your hand and wait to be called on.

AACTIVITY: Whole class instruction; students in seats.

M MOVEMENT: Remain in seat during instruction.

P PARTICIPATION: Look at teacher or materials being discussed. Raise hand to contribute; respond to questions, write or perform other actions as directed.NO SLEEPING OR PUTTING HEAD DOWN, TEXTING, DOING OTHER WORK.

S

Activity: Teacher-Directed Instruction

Objective

• C: The swbat differentiate implicitly equations in more than one variable.

• L: the sw explain to others how to find derivatives of multiple types of problems verbally and demonstratively

EX 1:3 2 25 4y y y x

(a) Find the derivative at the point ( 5, 3 ) , at ( -1,-3 )

(b) Find where the curve has a horizontal tangent.

(c) Find where the curve has vertical tangents.

3 𝑦2 𝑑 𝑦𝑑𝑥 +2 𝑦 𝑑𝑦𝑑𝑥 −5 𝑑𝑦𝑑𝑥 −2𝑥 𝑑𝑥𝑑𝑥=0

(3 𝑦¿¿2+2 𝑦−5)𝑑𝑦𝑑𝑥=2 𝑥¿

𝑑𝑦𝑑𝑥 =

2𝑥3 𝑦2+2 𝑦−5

𝑑𝑦𝑑𝑥 |¿ (5,3)=10

28𝑑𝑦𝑑𝑥 |¿ (−1 ,−3)=−2

16

EX 1:3 2 25 4y y y x

(b) Find where the curve has a horizontal tangent. Horizontal tangent has a 0 slope

𝑎𝑏=0∴𝑎=0

2 𝑥=0𝑥=0

EX 1:3 2 25 4y y y x

(c) Find where the curve has vertical tangents. Vertical tangent has an undefined slope𝑎

𝑏𝑢𝑛𝑑𝑒𝑓 𝑏=0

3 𝑦2+2 𝑦−5=0(3 𝑦+5)(𝑦−1)

3 𝑦+5=0𝑦=

−53

𝑦−1=0𝑦=1

Ex 2:

3 3 2x y xy

< Folium of Descartes >

3 𝑥2 𝑑𝑥𝑑𝑥 +3 𝑦2 𝑑𝑦

𝑑𝑥=2(𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥 )3 𝑥2+3 𝑦2 𝑑𝑦

𝑑𝑥=2𝑥 𝑑𝑦𝑑𝑥 +2 𝑦

3 𝑦2 𝑑𝑦𝑑𝑥 −2 𝑥 𝑑𝑦𝑑𝑥=2 𝑦−3 𝑥2

𝑑𝑦𝑑𝑥 ( (3 𝑦

2−2 𝑥)(3 𝑦 2−2 𝑥))= 2 𝑦−3𝑥2

(3 𝑦2−2𝑥 )

𝑑𝑦𝑑𝑥 =

2 𝑦−3𝑥2

3 𝑦2−2𝑥

3 𝑥2−2 𝑦=2 𝑥 𝑑𝑦𝑑𝑥 −3 𝑦2 𝑑𝑦𝑑𝑥

3 𝑥2−2 𝑦=(2 𝑥−3 𝑦2)𝑑𝑦𝑑𝑥

3 𝑥2−2 𝑦(2 𝑥−3 𝑦2)

=𝑑𝑦𝑑𝑥

Why Implicit?

3 3 2x y xy

< Folium of Descartes > Explicit Form:

3 6 3 3 6 33 31

1 1 1 18 82 4 2 4

y x x x x x x

3 6 3 3 6 33 32 1

1 1 1 1 13 8 82 2 4 2 4

y y x x x x x x

3 6 3 3 6 33 33 1

1 1 1 1 13 8 82 2 4 2 4

y y x x x x x x

2nd Derivatives

NOTICE:The second derivative is in terms of x , y , AND dy /dx.

The final step will be to substitute back the value of dy / dx into the second derivative.

EX: Our friendly circle. Find the 2nd Derivative.2 2 9x y

2 𝑥 𝑑𝑥𝑑𝑥 +2 𝑦 𝑑𝑦𝑑𝑥=0

2 𝑥+2 𝑦 𝑑𝑦𝑑𝑥=0

2 𝑦 𝑑𝑦𝑑𝑥 =−2 𝑥

𝑑𝑦𝑑𝑥=

−2𝑥2 𝑦

𝑑𝑦𝑑𝑥=

−𝑥𝑦

𝑑2 𝑦𝑑𝑥2 =¿

𝑦 (−1 𝑑𝑥𝑑𝑥 )−(− 𝑥) 𝑑𝑦𝑑𝑥𝑦2

𝑑2 𝑦𝑑𝑥2 =

−𝑦+𝑥 𝑑𝑦𝑑𝑥𝑦 2

𝑑2 𝑦𝑑𝑥2 =

−𝑦+𝑥 (−𝑥𝑦 )𝑦2 ( 𝑦𝑦 )

−𝑦 2−𝑥2

𝑦3𝑑2 𝑦𝑑𝑥2 =¿

2nd DerivativesEX: Find the 2nd Derivative.

23 5xy

0−(𝑥 2 𝑦 𝑑𝑦𝑑𝑥 + 𝑦2 𝑑𝑥𝑑𝑥 )=0

−2 𝑥𝑦 𝑑𝑦𝑑𝑥 − 𝑦2=0

−2𝑥 𝑑𝑦𝑑𝑥 − 𝑦 (−2 𝑑𝑥𝑑𝑥 )(−2𝑥)2




−2𝑥 𝑑𝑦𝑑𝑥 +2 𝑦

4 𝑥2

−2𝑥 ( 𝑦−2𝑥 )+2 𝑦

4 𝑥2𝑑𝑦𝑑𝑥 =

𝑦 2

−2𝑥𝑦=𝑦−2𝑥


𝑦+2 𝑦4 𝑥2

3 𝑦4 𝑥2


Higher DerivativesEX: Find the Third Derivative.

sin( )y x

cos (𝑦 ) 𝑑𝑦𝑑𝑥=𝑑𝑥𝑑𝑥

𝑑𝑦𝑑𝑥 =

1cos(𝑦 )

𝑑𝑦𝑑𝑥 =sec (𝑦 )




sec (𝑦 ) tan (𝑦 )𝑑𝑦𝑑𝑥

sec (𝑦 ) tan ( 𝑦 ) sec (𝑦 )

𝑠𝑒𝑐2 (𝑦 ) tan (𝑦 )

Last update

• 10/19/10

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