3023 Related Rates

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2019 Related Rates
AB Calculus
Intro:
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
moving
moving
x ο€½ f (t ) time
y ο€½ g (t )
ILLUSTRATION:
A point is moving along the parabola,
y ο€½ x2  3
graph
dy
Find the rate of change of
if
x
𝑑𝑦
=?
𝑑𝑑
is changing at
2
y
when
x=1
units per second.
𝑦 = π‘₯2 + 3
𝑑π‘₯
= +2 π‘€β„Žπ‘’π‘› π‘₯ = 1
𝑑𝑦
𝑑π‘₯
𝑑𝑑
= 2π‘₯
𝑑𝑑
𝑑𝑑 𝑑𝑦
=2 1 2 =4
𝑑𝑑
t
.
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
ο‚· Equation of a graph?
ο‚· Formula from Geometry?
The equation must involve only the variables from step 2. –
May plug in a
((You may have to solve a secondary equation
constant as
long as it is
to eliminate a variable.))
unchanging
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.))
Geometry formulas:
Sphere:
4
V ο€½  r3
3
SA ο€½ 4 r
Cylinder:
2
V ο€½  r 2h
LA ο€½ 2 rh
SA ο€½ 2 r 2  2 rh
Cone:
1
V ο€½  r 2h
3
Pythagorean Theorem:
x y ο€½z
2
2
2
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.
Step 1:
List the rates and variables.
π‘‘π‘Ÿ
= +3
𝑑𝑑
𝑑𝑉
=?
𝑑𝑑
Find an equation that relates
the variables and rates.
(Extra Variables?)
When r =5
4
𝑉 = πœ‹π‘Ÿ 3
3
𝑑𝑉
π‘‘π‘Ÿ
= 4πœ‹(π‘Ÿ 2 )
𝑑𝑑
𝑑𝑑
Plugin 5 gives vol not rate of change
Differentiate (with
respect to t.)
𝑑𝑉
= 4πœ‹(52 )(3)
𝑑𝑑
𝑑𝑉
3
= 300πœ‹ π‘π‘š 𝑠𝑒𝑐
𝑑𝑑
Plug in and solve.
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?
# constant does not
change ever you
can plug in the
equation
𝑧 = 25
𝑑π‘₯
= +3
𝑑𝑑
𝑑𝑦
=?
𝑑𝑑
π‘₯2 + 𝑦2 = 𝑧2
π‘€β„Žπ‘’π‘› π‘₯ = 15
π‘₯2 + 𝑦2 = 𝑧2
152 + 𝑦 2 = 252
𝑑π‘₯
𝑑𝑦
𝑑𝑧
+ 2𝑦
= 2𝑧
𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑦
2 15 3 + 2 20
=0
𝑑𝑑
𝑑𝑦
90 + 40
=0
𝑑𝑑
𝑑𝑦 −45 −9
=
=
𝑑𝑑
20
4
2π‘₯
The ladder is coming down -2.25 ft/sec
π‘₯ 2 + 𝑦 2 = 252
2π‘₯
𝑑π‘₯
𝑑𝑦
+ 2𝑦
=0
𝑑𝑑
𝑑𝑑
2 15 3 + 2 20
90 + 40
𝑑𝑦
=0
𝑑𝑑
𝑑𝑦
=0
𝑑𝑑
𝑑𝑦 −45 −9
=
=
𝑑𝑑
20
4
The ladder is coming down
II: Similar Triangles
Similar Triangles
A
A
B
B
D
A
B
E
C
F
ABC
C
C
DEF
AB
BC
CA
ο€½
ο€½
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
20 ft
z
𝑑𝑧
= +3
𝑑𝑑
𝑦
5
as
=
𝑧 20
5𝑧 = 20𝑦
𝑑𝑧
𝑑𝑦
5
= 20
𝑑𝑑
𝑑𝑑
20
𝑑𝑦
= 5(3)
𝑑𝑧
y
x
𝑑𝑦 15 3 𝑓𝑑
=
=
𝑠𝑒𝑐
𝑑𝑑 20 4
5
Ex 5:
Getting smaller
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.
The distance of top of
from the base of the lamppost.
shadow from post
How fast is the tip of
Pat’s shadow changing
𝑑π‘₯
− −5
𝑑𝑑
𝑑𝑦
=? π‘€β„Žπ‘’π‘› π‘₯ = 24
𝑑𝑑
20
6
=
π‘₯+𝑦 𝑦
20
6
20𝑦 = 6π‘₯ + 6𝑦
𝑑𝑦
𝑑π‘₯
𝑑𝑦
20
=6
+6
𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑦
14
= −30
𝑑𝑑
𝑑𝑦 −30 −15
=
=
𝑑𝑑
14
7
x
y
6
6
y
20
6
=
𝑦
𝑦−π‘₯
y-x
Ex 6: Cone
w/ extra equation
1 2
𝑉 = πœ‹π‘Ÿ β„Ž
3
Three variables
2
Water is being poured into a conical paper cup at a rate of
3
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
1
Too many variables
is 4 in. deep?
𝑉 = πœ‹π‘Ÿ 2 β„Ž need to find r
3
𝑑𝑉
2
=+
𝑑𝑑
3
2
r changes
h changes
π‘‘β„Ž
=? π‘€β„Žπ‘’π‘› β„Ž = 4
𝑑𝑑
2 π‘Ÿ
=
6 β„Ž
2β„Ž = 6π‘Ÿ
β„Ž 2β„Ž
=
=π‘Ÿ
3
6
1
β„Ž
𝑉= πœ‹
β„Ž
3
3
πœ‹ 3
Only two variables
𝑉=
β„Ž
27
𝑑𝑉 πœ‹ 2 π‘‘β„Ž
= β„Ž
𝑑𝑑 9 𝑑𝑑
2 πœ‹ 2 π‘‘β„Ž
= 4
3 9 𝑑𝑑
+3 2 π‘‘β„Ž
∗ =
16πœ‹ 3 𝑑𝑑
1
π‘‘β„Ž
=
8πœ‹ 𝑑𝑑
III: Angle of Elevation
π‘œπ‘π‘
sin πœƒ =
β„Žπ‘¦π‘
adj
π‘œπ‘π‘
sin πœƒ =
β„Žπ‘¦π‘
π‘Žπ‘‘π‘—
cos πœƒ =
β„Žπ‘¦π‘
π‘œπ‘π‘
tan πœƒ =
π‘Žπ‘‘π‘—
opp
β„Žπ‘¦π‘
csc πœƒ =
π‘œπ‘π‘
β„Žπ‘¦π‘
s𝑒𝑐 πœƒ =
π‘Žπ‘‘π‘—
π‘Žπ‘‘π‘—
cot πœƒ =
π‘œπ‘π‘
Angles of Elevation
SOH – CAH - TOA
c
a
θ
b
Hint: The problem may not require solving for an angle
measure … only a specific trig ratio.
ie. need sec θ instead of θ
πœƒ =?
5
3
θ
4
5
sec πœƒ =
4
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.
When y =100
𝑑𝑦
= 40
𝑑𝑑
π‘‘πœƒ
=? π‘€β„Žπ‘’π‘› 𝑦 = 100
𝑑𝑑
𝑦
tan πœƒ =
π‘₯
100
or
𝑦
100
𝑠𝑒𝑐 2 πœƒ
2
𝑑𝑧
1 𝑑𝑦
=
𝑑𝑑 100 𝑑𝑑
2 𝑑𝑧
𝑑𝑑
=
1
10
100
π‘‘πœƒ
1
=
𝑑𝑑 10
π‘‘πœƒ
1 π‘Ÿπ‘Žπ‘‘
=
𝑠𝑒𝑐
𝑑𝑑 20
2
100
100 2
sec πœƒ =
= 2
100
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.
𝑑𝑧
𝑓𝑑
= −1
𝑠𝑒𝑐
𝑑𝑑
π‘‘πœƒ
=? When z = 25 ft
𝑑𝑑
15
𝑧
sin πœƒ =
↔ csc πœƒ =
𝑧
15
𝑧 sin πœƒ = 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
s ο€½ 50t , 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.
2
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.
B
C
A
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 Type
A 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 Question
A 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:
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