2019 related rates ab calculus. known limits:
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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
ππ¦ππ‘
=?ππ₯ππ‘
=+2 hπ€ πππ₯=1π¦=π₯2+3ππ¦ππ‘
=2π₯ππ₯ππ‘ππ¦ππ‘
=2 (1 ) (2 )=4
PROCEDURE:
1). DRAW A PICTURE! β 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 Β· Equation of a graph?
Β· Formula from Geometry? The equation must involve only the variables from step 2. β
((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:
ππππ‘
=+3
ππππ‘
=?
When r =5
π=43π π3
πππ π‘
=4π (52)(3)
πππ π‘
=4π (π 2)ππππ‘
πππ π‘
=300π ππ3
π ππ
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?
π§=25ππ₯ππ‘
=+3
ππ¦ππ‘
=?
hπ€ πππ₯=15π₯2+π¦2=π§2
2 π₯ππ₯ππ‘
+2 π¦ππ¦ππ‘
=2 π§ππ§ππ‘
π₯2+π¦2=π§2
π₯2+π¦2=252
2 π₯ππ₯ππ‘
+2 π¦ππ¦ππ‘
=0
2 (15 ) (3 )+2 (20 ) ππ¦ππ‘
=0
ππ¦ππ‘
=β 4520
=β 94
90+40ππ¦ππ‘
=0
The ladder is coming down -2.25 ft/sec
2 (15 ) 3+2 (20 ) ππ¦ππ‘
=0
90+40ππ¦ππ‘
=0ππ¦ππ‘
=β 4520
=β 94
# constant does not change ever you can plug in the equation
The ladder is coming down
152+π¦2=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 β 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
ππ§ππ‘
=+3
π¦π§=
520
as
5 π§=20 π¦
5ππ§ππ‘
=20ππ¦ππ‘
20ππ¦ππ‘
=5(3)
ππ¦ππ‘
=1520
=34
ππ‘π ππ
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 Patβs shadow at the moment Pat is 24 ft. from the base of the lamppost.
ππ₯ππ‘
ββ 5
ππ¦ππ‘
=? hπ€ πππ₯=24
20π₯+π¦
=6π¦ 20 π¦=6 π₯+6 π¦
20ππ¦ππ‘
=6ππ₯ππ‘
+6ππ¦ππ‘
14ππ¦ππ‘
=β30
ππ¦ππ‘
=β 3014
=β15
7
How fast is the tip of Patβs shadow changing
The distance of top of shadow from post
6
y-x6
y
6x
y
20
20π¦=
6π¦βπ₯
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
π=13ππ2 h
Three variables
πππ π‘
=+23
π hππ‘
=? hπ€ ππ h=4
26=πh
2 h=6πh3=
2 h6=π
r changesh changes
π=13π( h
3 )2
h
π=13ππ2 h
π=π27
h3
πππ π‘
=π9
h2 hπππ‘
23=π9
42 hπππ‘
+916π
β23=
hπππ‘
38π
=hπ
ππ‘
Too many variables need to find r
Only two variables
III: Angle of Elevation
hyp
opp
adj
sin π=πππhπ¦π
csc π=hπ¦ππππ
sπππ=hπ¦ππππ
cot π=ππππππ
sin π=πππhπ¦π
cosπ=πππhπ¦π
tanπ=ππππππ
Angles of Elevation
ΞΈa
b
c SOH β CAH - TOA
Hint: The problem may not require solving for an angle measure β¦ only a specific trig ratio.
ie. need sec ΞΈ instead of ΞΈ
ΞΈ3
4
5 π=?
secπ=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 β2When y =100
secπ=100 β2100
=β2
ππ¦ππ‘
=40
ππππ‘
=? hπ€ πππ¦=100
tanπ=π¦π₯
orπ¦
100
π ππ2πππ§ππ‘
=1
100ππ¦ππ‘
(β2 )2 π π§ππ‘
=1
100(10 )
2ππππ‘
=1
10ππππ‘
=120
ππππ ππ
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.
ππ§ππ‘
=β1ππ‘π ππ
ππππ‘
=? When z = 25 ft
sin π=15π§
βcscπ=π§
15
π§ sin π=15
π§ (cosπ ππππ‘ )+sinπ ππ§
ππ‘=0
25 ( 45 ( ππππ‘ ))+ 3
5(β1 )=0
20ππ§ππ‘
β35=0
ππππ‘
=35 ( 1
20 )ππππ‘
=3
100ππππ ππ π
z15
πππ15
20
25
cosπ=2025
=45
sin π=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
ππππ‘
=? hπ€ πππ‘=10π ππ
π¦=50 π‘ 2tanπ=
π¦2000
tanπ= 50 π‘2
2000= π‘ 2
400= 1
400π‘2
sec2πππππ‘
=1
200π‘
(β292 )
2 ππππ‘
= 1200
(10 )π¦=50 (10 )2=5000
20002+50002=4000000+25000000=β29000000=ΒΏΒΏ1000β29
secπ=1000 β292000
π y
2000ft
y1000 β29y5000
2000294ππππ‘
=1
204
29 ( 120 )= 1
145ππππ π ππ
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
π₯2+ π¦2=π§2
ππ¦ππ‘
=2
ππ₯ππ‘
=β3
ππ§ππ‘
=? hπ€ πππ¦=72ππππ₯=96
2 π₯ππ₯ππ‘
+2 π¦ππ¦ππ‘
=2 π§ππ§ππ‘
π§ 2=962+722
π§ 2=9216+5184
π§=β14400
π§=120
2 (96 ) (β 3 )+2 (72 ) (2 )=2 (120 )(ππ§ππ‘ )β576+288=240
ππ§ππ‘
ππ§ππ‘
=β288240
=β1.2πππ ππ
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.?
π¦=24π₯=32
ππ¦ππ‘
=+24
ππ₯ππ‘
=β 32
ππ§ππ‘
=? hπ€ πππ§=40
π₯2+ π¦2=π§2
2 π₯ππ₯ππ‘
+2 π¦ππ¦ππ‘
=2 π§ππ§ππ‘
2 (32 ) (β32 )+2 (24 ) (24 )=2 (40 ) ππ§ππ‘
ππ§ππ‘
=β896
80ππ§ππ‘
=β11.2
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
β’ 11/12/11
β’ BC:
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