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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
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: