2019 Related Rates AB Calculus. Known Limits:

2019 Related Rates

AB Calculus

Known Limits:0

, 0, 0,0 0

c c

Intro:

( )

( )

x f t

y 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

=?𝑑𝑥𝑑𝑡

=+2 h𝑤 𝑒𝑛𝑥=1𝑦=𝑥2+3𝑑𝑦𝑑𝑡

=2𝑥𝑑𝑥𝑑𝑡𝑑𝑦𝑑𝑡

=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:  

34

3V r

24SA r

2V r h2LA rh

22 2SA r rh

21

3V r h

2 2 2x y z

METHOD: Inflating a Balloon - 1

A 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=43𝜋 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 equation

A 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 𝑥𝑑𝑥𝑑𝑡

+2 𝑦𝑑𝑦𝑑𝑡

=2 𝑧𝑑𝑧𝑑𝑡

𝑥2+𝑦2=𝑧2

𝑥2+𝑦2=252

2 𝑥𝑑𝑥𝑑𝑡

+2 𝑦𝑑𝑦𝑑𝑡

=0

2 (15 ) (3 )+2 (20 ) 𝑑𝑦𝑑𝑡

=0

𝑑𝑦𝑑𝑡

=− 4520

=− 94

90+40𝑑𝑦𝑑𝑡

=0

The ladder is coming down -2.25 ft/sec

2 (15 ) 3+2 (20 ) 𝑑𝑦𝑑𝑡

=0

90+40𝑑𝑦𝑑𝑡

=0𝑑𝑦𝑑𝑡

=− 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 Triangles

A

B

C

A

B

C

A B

C

D E

F

ABC DEF AB BC CA

DE 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

z

xy

20 ft

5

𝑑𝑧𝑑𝑡

=+3

𝑦𝑧=

520

as

5 𝑧=20 𝑦

5𝑑𝑧𝑑𝑡

=20𝑑𝑦𝑑𝑡

20𝑑𝑦𝑑𝑡

=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 𝑦

20𝑑𝑦𝑑𝑡

=6𝑑𝑥𝑑𝑡

+6𝑑𝑦𝑑𝑡

14𝑑𝑦𝑑𝑡

=−30

𝑑𝑦𝑑𝑡

=− 3014

=−15

7

How fast is the tip of Pat’s shadow changing

The distance of top of shadow from post

6

y-x6

y

6x

y

20

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?

2

3

𝑉=13𝜋𝑟2 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=13𝜋𝑟2 h

𝑉=𝜋27

h3

𝑑𝑉𝑑 𝑡

=𝜋9

h2 h𝑑𝑑𝑡

23=𝜋9

42 h𝑑𝑑𝑡

+916𝜋

∗23=

h𝑑𝑑𝑡

38𝜋

=h𝑑

𝑑𝑡

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

𝑠𝑒𝑐2𝜃𝑑𝑧𝑑𝑡

=1

100𝑑𝑦𝑑𝑡

(√2 )2 𝑑 𝑧𝑑𝑡

=1

100(10 )

2𝑑𝜃𝑑𝑡

=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.

𝑑𝑧𝑑𝑡

=−1𝑓𝑡𝑠𝑒𝑐

𝑑𝜃𝑑𝑡

=? When z = 25 ft

sin 𝜃=15𝑧

↔csc𝜃=𝑧

15

𝑧 sin 𝜃=15

𝑧 (cos𝜃 𝑑𝜃𝑑𝑡 )+sin𝜃 𝑑𝑧

𝑑𝑡=0

25 ( 45 ( 𝑑𝜃𝑑𝑡 ))+ 3

5(−1 )=0

20𝑑𝑧𝑑𝑡

−35=0

𝑑𝜃𝑑𝑡

=35 ( 1

20 )𝑑𝜃𝑑𝑡

=3

100𝑟𝑎𝑑𝑠𝑒𝑐 𝜃

z15

𝜃𝜃𝜃15

20

25

cos𝜃=2025

=45

sin 𝜃=1525

=35

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 y is measured in feet and t in seconds. The camera is

is 2000 ft. from the launch pad. Find the rate of change of the angle of

elevation of the camera 10 sec. after lift off.

250y t

𝑑𝜃𝑑𝑡

=? h𝑤 𝑒𝑛𝑡=10𝑠𝑒𝑐

𝑦=50 𝑡 2tan𝜃=

𝑦2000

tan𝜃= 50 𝑡2

2000= 𝑡 2

400= 1

400𝑡2

sec2𝜃𝑑𝜃𝑑𝑡

=1

200𝑡

(√292 )

2 𝑑𝜃𝑑𝑡

= 1200

(10 )𝑦=50 (10 )2=5000

20002+50002=4000000+25000000=√29000000=¿¿1000√29

sec𝜃=1000 √292000

𝜃 y

2000ft

y1000 √29y5000

2000294𝑑𝜃𝑑𝑡

=1

204

29 ( 120 )= 1

145𝑟𝑎𝑑𝑠𝑠𝑒𝑐

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

zy

x

𝑥2+ 𝑦2=𝑧2

𝑑𝑦𝑑𝑡

=2

𝑑𝑥𝑑𝑡

=−3

𝑑𝑧𝑑𝑡

=? h𝑤 𝑒𝑛𝑦=72𝑎𝑛𝑑𝑥=96

2 𝑥𝑑𝑥𝑑𝑡

+2 𝑦𝑑𝑦𝑑𝑡

=2 𝑧𝑑𝑧𝑑𝑡

𝑧 2=962+722

𝑧 2=9216+5184

𝑧=√14400

𝑧=120

2 (96 ) (− 3 )+2 (72 ) (2 )=2 (120 )(𝑑𝑧𝑑𝑡 )−576+288=240

𝑑𝑧𝑑𝑡

𝑑𝑧𝑑𝑡

=−288240

=−1.2𝑖𝑛𝑠𝑒𝑐

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.?

𝑦=24𝑥=32

𝑑𝑦𝑑𝑡

=+24

𝑑𝑥𝑑𝑡

=− 32

𝑑𝑧𝑑𝑡

=? h𝑤 𝑒𝑛𝑧=40

𝑥2+ 𝑦2=𝑧2

2 𝑥𝑑𝑥𝑑𝑡

+2 𝑦𝑑𝑦𝑑𝑡

=2 𝑧𝑑𝑧𝑑𝑡

2 (32 ) (−32 )+2 (24 ) (24 )=2 (40 ) 𝑑𝑧𝑑𝑡

𝑑𝑧𝑑𝑡

=−896

80𝑑𝑧𝑑𝑡

=−11.2

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.

Last Update

• 11/12/11

• BC: