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
๐๐ฆ๐๐ก
=?๐๐ฅ๐๐ก
=+2 h๐ค ๐๐๐ฅ=1๐ฆ=๐ฅ2+3๐๐ฆ๐๐ก
=2๐ฅ๐๐ฅ๐๐ก๐๐ฆ๐๐ก
=2 (1 ) (2 )=4
PROCEDURE:
1). DRAW A PICTURE! โ 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 ยท Equation of a graph?
ยท Formula from Geometry? The equation must involve only the variables from step 2. โ
((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:
๐๐๐๐ก
=+3
๐๐๐๐ก
=?
When r =5
๐=43๐ ๐3
๐๐๐ ๐ก
=4๐ (52)(3)
๐๐๐ ๐ก
=4๐ (๐ 2)๐๐๐๐ก
๐๐๐ ๐ก
=300๐ ๐๐3
๐ ๐๐
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?
๐ง=25๐๐ฅ๐๐ก
=+3
๐๐ฆ๐๐ก
=?
h๐ค ๐๐๐ฅ=15๐ฅ2+๐ฆ2=๐ง2
2 ๐ฅ๐๐ฅ๐๐ก
+2 ๐ฆ๐๐ฆ๐๐ก
=2 ๐ง๐๐ง๐๐ก
๐ฅ2+๐ฆ2=๐ง2
๐ฅ2+๐ฆ2=252
2 ๐ฅ๐๐ฅ๐๐ก
+2 ๐ฆ๐๐ฆ๐๐ก
=0
2 (15 ) (3 )+2 (20 ) ๐๐ฆ๐๐ก
=0
๐๐ฆ๐๐ก
=โ 4520
=โ 94
90+40๐๐ฆ๐๐ก
=0
The ladder is coming down -2.25 ft/sec
2 (15 ) 3+2 (20 ) ๐๐ฆ๐๐ก
=0
90+40๐๐ฆ๐๐ก
=0๐๐ฆ๐๐ก
=โ 4520
=โ 94
# constant does not change ever you can plug in the equation
The ladder is coming down
152+๐ฆ2=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 โ 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
๐๐ง๐๐ก
=+3
๐ฆ๐ง=
520
as
5 ๐ง=20 ๐ฆ
5๐๐ง๐๐ก
=20๐๐ฆ๐๐ก
20๐๐ฆ๐๐ก
=5(3)
๐๐ฆ๐๐ก
=1520
=34
๐๐ก๐ ๐๐
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 Patโs shadow at the moment Pat is 24 ft. from the base of the lamppost.
๐๐ฅ๐๐ก
โโ 5
๐๐ฆ๐๐ก
=? h๐ค ๐๐๐ฅ=24
20๐ฅ+๐ฆ
=6๐ฆ 20 ๐ฆ=6 ๐ฅ+6 ๐ฆ
20๐๐ฆ๐๐ก
=6๐๐ฅ๐๐ก
+6๐๐ฆ๐๐ก
14๐๐ฆ๐๐ก
=โ30
๐๐ฆ๐๐ก
=โ 3014
=โ15
7
How fast is the tip of Patโs shadow changing
The distance of top of shadow from post
6
y-x6
y
6x
y
20
20๐ฆ=
6๐ฆโ๐ฅ
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
๐=13๐๐2 h
Three variables
๐๐๐ ๐ก
=+23
๐ h๐๐ก
=? h๐ค ๐๐ h=4
26=๐h
2 h=6๐h3=
2 h6=๐
r changesh changes
๐=13๐( h
3 )2
h
๐=13๐๐2 h
๐=๐27
h3
๐๐๐ ๐ก
=๐9
h2 h๐๐๐ก
23=๐9
42 h๐๐๐ก
+916๐
โ23=
h๐๐๐ก
38๐
=h๐
๐๐ก
Too many variables need to find r
Only two variables
III: Angle of Elevation
hyp
opp
adj
sin ๐=๐๐๐h๐ฆ๐
csc ๐=h๐ฆ๐๐๐๐
s๐๐๐=h๐ฆ๐๐๐๐
cot ๐=๐๐๐๐๐๐
sin ๐=๐๐๐h๐ฆ๐
cos๐=๐๐๐h๐ฆ๐
tan๐=๐๐๐๐๐๐
Angles of Elevation
ฮธa
b
c SOH โ CAH - TOA
Hint: The problem may not require solving for an angle measure โฆ only a specific trig ratio.
ie. need sec ฮธ instead of ฮธ
ฮธ3
4
5 ๐=?
sec๐=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 โ2When y =100
sec๐=100 โ2100
=โ2
๐๐ฆ๐๐ก
=40
๐๐๐๐ก
=? h๐ค ๐๐๐ฆ=100
tan๐=๐ฆ๐ฅ
or๐ฆ
100
๐ ๐๐2๐๐๐ง๐๐ก
=1
100๐๐ฆ๐๐ก
(โ2 )2 ๐ ๐ง๐๐ก
=1
100(10 )
2๐๐๐๐ก
=1
10๐๐๐๐ก
=120
๐๐๐๐ ๐๐
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.
๐๐ง๐๐ก
=โ1๐๐ก๐ ๐๐
๐๐๐๐ก
=? When z = 25 ft
sin ๐=15๐ง
โcsc๐=๐ง
15
๐ง sin ๐=15
๐ง (cos๐ ๐๐๐๐ก )+sin๐ ๐๐ง
๐๐ก=0
25 ( 45 ( ๐๐๐๐ก ))+ 3
5(โ1 )=0
20๐๐ง๐๐ก
โ35=0
๐๐๐๐ก
=35 ( 1
20 )๐๐๐๐ก
=3
100๐๐๐๐ ๐๐ ๐
z15
๐๐๐15
20
25
cos๐=2025
=45
sin ๐=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
๐๐๐๐ก
=? h๐ค ๐๐๐ก=10๐ ๐๐
๐ฆ=50 ๐ก 2tan๐=
๐ฆ2000
tan๐= 50 ๐ก2
2000= ๐ก 2
400= 1
400๐ก2
sec2๐๐๐๐๐ก
=1
200๐ก
(โ292 )
2 ๐๐๐๐ก
= 1200
(10 )๐ฆ=50 (10 )2=5000
20002+50002=4000000+25000000=โ29000000=ยฟยฟ1000โ29
sec๐=1000 โ292000
๐ y
2000ft
y1000 โ29y5000
2000294๐๐๐๐ก
=1
204
29 ( 120 )= 1
145๐๐๐๐ ๐ ๐๐
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
๐ฅ2+ ๐ฆ2=๐ง2
๐๐ฆ๐๐ก
=2
๐๐ฅ๐๐ก
=โ3
๐๐ง๐๐ก
=? h๐ค ๐๐๐ฆ=72๐๐๐๐ฅ=96
2 ๐ฅ๐๐ฅ๐๐ก
+2 ๐ฆ๐๐ฆ๐๐ก
=2 ๐ง๐๐ง๐๐ก
๐ง 2=962+722
๐ง 2=9216+5184
๐ง=โ14400
๐ง=120
2 (96 ) (โ 3 )+2 (72 ) (2 )=2 (120 )(๐๐ง๐๐ก )โ576+288=240
๐๐ง๐๐ก
๐๐ง๐๐ก
=โ288240
=โ1.2๐๐๐ ๐๐
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.?
๐ฆ=24๐ฅ=32
๐๐ฆ๐๐ก
=+24
๐๐ฅ๐๐ก
=โ 32
๐๐ง๐๐ก
=? h๐ค ๐๐๐ง=40
๐ฅ2+ ๐ฆ2=๐ง2
2 ๐ฅ๐๐ฅ๐๐ก
+2 ๐ฆ๐๐ฆ๐๐ก
=2 ๐ง๐๐ง๐๐ก
2 (32 ) (โ32 )+2 (24 ) (24 )=2 (40 ) ๐๐ง๐๐ก
๐๐ง๐๐ก
=โ896
80๐๐ง๐๐ก
=โ11.2
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
โข 11/12/11
โข BC: