2019 related rates

2019 Related Rates

AB Calculus

Intro:

( )( )

x f ty g t

2 3y x

 GOAL: to find the rates of change of two (or more) variables with respect to a third variable (the parameter)

This is a adaptation of IMPLICIT functions  x and y are implicit functions of t .

ILLUSTRATION: 

A point is moving along the parabola,

Find the rate of change of y when x = 1if x is changing at 2 units per second. 

moving

moving time

graphdy

𝑑𝑦𝑑𝑡=? 𝑑𝑥

𝑑𝑡 =+2 h𝑤 𝑒𝑛𝑥=1𝑦=𝑥2+3𝑑𝑦𝑑𝑡=2𝑥 𝑑𝑥𝑑𝑡𝑑𝑦

𝑑𝑡=2 (1 ) (2 )=4

PROCEDURE:

1). DRAW A PICTURE! – Determine what rates are being compared.

 2). Assign variables to all given and unknown quantities and rates.

 3). Write an equation involving the variables whose rates are given or are to be found  ·       Equation of a graph?

·       Formula from Geometry?  The equation must involve only the variables from step 2. –

((You may have to solve a secondary equation to eliminate a variable.))

 4). Use Implicit Differentiation (with respect to the parameter t).

 5). AFTER DIFFERENTIATION, substitute in all known values   (( You may have to solve a secondary equation

to find the value of a variable.))

May plug in a constant as long as it is unchanging

Geometry formulas:

Sphere:   

Cylinder:   

Cone:  

Pythagorean Theorem:  

343

V r

24SA r

2V r h2LA rh

22 2SA r rh

213

V r h

2 2 2x y z

METHOD: Inflating a Balloon - 1A spherical balloon is inflated so that the radius is changing at a rate of 3 cm/sec. How fast is the volume changing when the radius is 5 cm.?

Draw and label a picture.

List the rates and variables.

Find an equation that relates the variables and rates. (Extra Variables?)

Differentiate (with respect to t.)

Plug in and solve.

Step 1:

𝑑𝑟𝑑𝑡 =+3

𝑑𝑉𝑑𝑡 =?

When r =5

𝑉=43 𝜋 𝑟3

𝑑𝑉𝑑 𝑡 =4𝜋 (52)(3)

𝑑𝑉𝑑 𝑡 =4𝜋 (𝑟 2)

𝑑𝑟𝑑𝑡

𝑑𝑉𝑑 𝑡 =300𝜋 𝑐𝑚3

𝑠𝑒𝑐

Plugin 5 gives vol not rate of change

Ex 2: Ladder w/ secondary equationA 25 ft. ladder is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at a rate of 3 ft./sec., how fast is the top of the ladder sliding down the wall when the bottom is 15 ft. from the wall?

𝑧=25𝑑𝑥𝑑𝑡 =+3

𝑑𝑦𝑑𝑡=?

h𝑤 𝑒𝑛𝑥=15𝑥2+ 𝑦2=𝑧2

2 𝑥 𝑑𝑥𝑑𝑡 +2 𝑦 𝑑𝑦𝑑𝑡 =2 𝑧 𝑑𝑧𝑑𝑡

𝑥2+ 𝑦2=𝑧2

𝑥2+ 𝑦2=252

2 𝑥 𝑑𝑥𝑑𝑡 +2 𝑦 𝑑𝑦𝑑𝑡 =0

2 (15 ) (3 )+2 (20 ) 𝑑𝑦𝑑𝑡 =0

𝑑𝑦𝑑𝑡=

− 4520 =

− 94

90+40 𝑑𝑦𝑑𝑡 =0

The ladder is coming down -2.25 ft/sec

2 (15 ) 3+2 (20 ) 𝑑𝑦𝑑𝑡 =0

90+40 𝑑𝑦𝑑𝑡 =0

𝑑𝑦𝑑𝑡=

− 4520 =

− 94

# constant does not change ever you can plug in the equation

The ladder is coming down

152+ 𝑦2=252

II: Similar Triangles

Similar TrianglesA

B

C

A

B

C

A B

C

D E

F

ABC DEF AB BC CADE EF FD

Similar Triangles may be the whole set up.

Similar Triangles may be required to to eliminate an extra variable – or- to find a missing value

Ex 4:A person is pushing a box up a 20 ft. ramp with a 5 ft. incline at a rate of 3 ft.per sec.. How fast is the box rising?

derivative

zx

y

20 ft

5

𝑑 𝑧𝑑𝑡 =+3

𝑦𝑧=

520

as

5 𝑧=20 𝑦5 𝑑𝑧𝑑𝑡 =20 𝑑𝑦𝑑𝑡

20 𝑑𝑦𝑑𝑧=5(3)

𝑑𝑦𝑑𝑡=

1520 =

34

𝑓𝑡𝑠𝑒𝑐

Ex 5:Pat is walking at a rate of 5 ft. per sec. toward a street light whose lamp is 20 ft. above the base of the light. If Pat is 6 ft. tall, determine the rate of change of the length of Pat’s shadow at the moment Pat is 24 ft. from the base of the lamppost.

𝑑𝑥𝑑𝑡 −− 5

𝑑𝑦𝑑𝑡=? h𝑤 𝑒𝑛𝑥=24

20𝑥+𝑦=

6𝑦 20 𝑦=6 𝑥+6 𝑦20 𝑑𝑦𝑑𝑡 =6 𝑑𝑥𝑑𝑡 +6 𝑑𝑦𝑑𝑡

14 𝑑𝑦𝑑𝑡 =−30

𝑑𝑦𝑑𝑡=

− 3014 =

−157

How fast is the tip of Pat’s shadow changing

The distance of top of shadow from post

6x

y

20

66

y-xy

20𝑦 =

6𝑦−𝑥

Getting smaller

Ex 6: Cone w/ extra equation

Water is being poured into a conical paper cup at a rate of

cubic inches per second. If the cup is 6 in. tall and the top of the cup has a radius of 2 in., how fast is the water level rising when the water is 4 in. deep?

23

𝑉=13 𝜋𝑟2 h

Three variables

𝑑𝑉𝑑 𝑡 =

+23

𝑑 h𝑑𝑡=? h𝑤 𝑒𝑛 h=4

26=

𝑟h

2 h=6𝑟h3=

2 h6 =𝑟

r changesh changes

𝑉=13𝜋( h

3 )2

h

𝑉=13 𝜋𝑟 2 h

𝑉=𝜋27 h3

𝑑𝑉𝑑 𝑡 =

𝜋9 h2 h𝑑

𝑑𝑡23=

𝜋9 42 h𝑑

𝑑𝑡+3

16𝜋 ∗ 23=

h𝑑𝑑𝑡

18𝜋=

h𝑑𝑑𝑡

Too many variables need to find r

Only two variables

III: Angle of Elevation

hyp

opp

adj

sin 𝜃=𝑜𝑝𝑝h𝑦𝑝

csc 𝜃= h𝑦𝑝𝑜𝑝𝑝

s𝑒𝑐𝜃=h𝑦𝑝𝑎𝑑𝑗

cot 𝜃= 𝑎𝑑𝑗𝑜𝑝𝑝

sin 𝜃=𝑜𝑝𝑝h𝑦𝑝

cos𝜃= 𝑎𝑑𝑗h𝑦𝑝

tan𝜃=𝑜𝑝𝑝𝑎𝑑𝑗

Angles of Elevation

θa

b

c SOH – CAH - TOA

Hint: The problem may not require solving for an angle measure … only a specific trig ratio.

ie. need sec θ instead of θ

θ3

4

5 𝜃=?

sec𝜃=54

Ex 7:A balloon rises at a rate of 10 ft/sec from a point on the ground 100 ft from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 100 ft. above the ground.

100

100

100 √2When y =100

sec𝜃=100 √2100

=√2

𝑑𝑦𝑑𝑡=40

𝑑𝜃𝑑𝑡 =? h𝑤 𝑒𝑛𝑦=100

tan𝜃= 𝑦𝑥

or𝑦

100

𝑠𝑒𝑐2𝜃 𝑑 𝑧𝑑𝑡 =1

100𝑑𝑦𝑑𝑡

(√2 )2 𝑑 𝑧𝑑𝑡 =1

100(10 )

2 𝑑𝜃𝑑𝑡 =1

10𝑑𝜃𝑑𝑡 =

120

𝑟𝑎𝑑𝑠𝑒𝑐

Ex 8:A fishing line is being reeled in at a rate of 1 ft/sec from a bridge 15 ft above the water. At what rate is the angle between the line and the water changing when 25 ft of line is out.

𝑑 𝑧𝑑𝑡 =−1 𝑓𝑡

𝑠𝑒𝑐𝑑𝜃𝑑𝑡 =? When z = 25 ft

sin 𝜃=15𝑧 ↔ csc𝜃= 𝑧

15𝑧 sin 𝜃=15

𝑧 ()

Ex 9:A television camera at ground level is filming the lift off of a space shuttle that is rising vertically according to the position function

, where s is measured in feet and t in seconds. The camera is is 2000 ft. from the launch pad. Find the rate of change of tin the angle of elevation of the camera 10 sec. after lift off.

250s t

IV: Using multiple rates

Ex 11:If one leg, AB, of a right triangle increases at a rate of 2 in/sec while the other leg, AC, decreases at 3 in/sec, find how fast the hypotenuse is changing when AB is 72 in. and AC is 96 in.

AC

B

Ex 12:A metal rod has the shape of a right circular cylinder. As it is being heated, its length is increasing at a rate of 0.005 cm/min and its diameter is increasing at 0.002 cm/min. At what rate is the volume changing when the rod has a length 40 cm and diameter 3 cm.?

5: AP Questions

Example 12: AP Type

At 8 a.m. a ship is sailing due north at 24 knots(nautical miles per hour) is a point P. At 10 a.m. a second ship sailing due east at 32 knots is a P. At what rate is the distance between the two ships changing at (a) 9 a.m. and (b) 11 a.m.?

Ex 13: AP TypeA right triangle has height 7 cm and the hypotenuse is increasing at a rate of 2 cm/sec. When the hypotenuse is 25 cm, find:

a). the rate of change of the base.

b). The rate of change of the acute angle at the base,

c). The rate of change of the area of the triangle.

AP QuestionA coffee pot is made up of a conical

filter that is 6 in. tall and has a diameter

of 6 in. and a cylindrical pot that is 6 in.

in diameter.

Coffee is draining from the filter into the coffeepot at a rate of 10 in3/sec.

a). How fast is the level in the pot rising when the coffee in the

cone is 5 in. deep?

b). How fast is the level in the cone falling at that instant?

AP Question: Combined Example

At noon a sailboat is 20 km south of a freighter. The sailboat is traveling east at 20 km per hour, and the freighter is traveling south at 40 km per hour.

When is the distance between the boats a minimum?

If the visibility is 10 kilometers, could the people see each other?

At what rate is the distance between the two boats changing at that instant?

Last Update

• 11/12/11

• BC: