Transcript Aim: How do we find related rates when we have more than

Aim: How do we find related rates when we
have more than two variables?
Do Now:
Find the points on the curve
x2 + y2 = 2x +2y where the
tangent is parallel to the x-axis.
(1, 1 + 21/2), (1, 1 – 21/2)
Aim: Related Rates
Course: Calculus
Related Rates involving TIME
Most quantities encountered in science or
everyday life vary with time.
If two such quantities are related to each
other by an equation, and if we know the
rate at which one of them changes, then, by
differentiating the equation with respect to
time, we can find the rate at which the other
quantity changes.
Implicit differentiation
and the Chain Rule
dy dy dt


dx dt dx
Aim: Related Rates
Course: Calculus
Related Rates involving TIME
Suppose a particle starts from the origin and
moves along the curve y = x2.
gx  = x 2
8
at (3, 9)
dy
3
dt
•as it moves both x and y values change
Suppose:
6
•the rate of change for x is ½ unit/sec.
at (2, 4)
dx 1
 unit/sec
dy
dt 2
2
dy
dt
•How can we find the rate of change of y?
dt
dy
dy dy dx
 2x


dx
dt dx dt
dy
dx
1
 2x 
 2x    x
dt
dt
 2
Related Rates
Course: Calculus
rate ofAim:change
of y equals x at
that point.
4
2
Related Rates involving TIME
Suppose that the diameter D and the area of a
circle are differentiable functions of t. Write
an equation that relates dA/dt and dD/dt.
A  r
2
dA
dr
 ?2 r
dt
dt
D  2r
dD
dr
 ?2
dt
dt
dr 1 dD

dt 2 dt
dA
1 dD
dD
D dD
 2 r
 r

dt
2 dt
dt
2 dt
1
2
Dr
D  2r
A  r
2
2
2
dA 2 dD
D dD
1 
 D

D

A    D    
dt
4
dt
2 dt
2 
2
Aim: Related Rates
Course: Calculus
Model Problem
A particle moves along the curve y = 3x2 – 6x
so that the rate of change of the x coord. is
dx
 2 unit/sec
dt
Find the rate of change of the y-coord. when
the particle is at the origin. @ x = 0
dy
?
dt
dy
 6x  6
dx
dy dy dx


dt dx dt
@x=0
dx
 2 units/sec
dt
dy
 6 x  6   2   12 x  12
dt
dy
 12  0   12  12 units/sec
dt
Aim: Related Rates
Course: Calculus
Model Problem
A particle moves along a circle 25 = x2 + y2.
If the particle is at (-4, 3) and the y-coord. is
increasing so that dy  2units/sec.
dt
Find the rate of change of the x-coord.
dx
dy
?
 2 units/sec
dt
dt
dx
dy d
2x
 2y

 25   0
dt
dt dx
Take derivative of x2 +
y2 = 25 with respect to t
dx
2  4 
 2  3  2   0
dt
Substitute and solve for
dx/dt.
dx 3
 units/sec.
dt 2
Aim: Related Rates
Course: Calculus
Model Verbal Problem Guidelines
1. Identify all given quantities and
quantities to be determined. Make a
sketch and label quantities.
2. Write an equation involving the variables
whose rates of change either are given or
are to be determined.
3. Using the Chain Rule, implicitly
differentiate both sides of the equation
with respect to time t.
4. After completing Step 3, substitute into
the resulting equation all known values
for the variable and their rates of change
and then solve.
Aim: Related Rates
Course: Calculus
Related Rates involving TIME
V

r 2h
3
volume in dependent
on measurements of
radius and height
If water were draining
out of this cone, the
volume V, the height h,
and the radius r, of the
water level would all be
functions of time t.
Aim: Related Rates
Course: Calculus
The Draining Cone
V

r 2h
3
implicitly differentiate the
Volume formula in terms
of time t.
  2 dh
dV
 dr  
 r
 h  2r  
3  dt
dt
 dt  

dh
dr 
 r
 2rh 
3  dt
dt 
2
this reinforces the fact
that the rate of Volume
change is related to the
changes in h and r.
Aim: Related Rates
Course: Calculus
The Draining Cone
If the water height is changing at a rate of
-0.2 foot per minute and the radius is
changing at the rate of -0.1 foot per
minute, what is the rate of change in the
volume when the radius r = 1 foot and the
height h = 2 feet?
dh
dr
dV
 2
 0.2
 0.1 V  r h
?
dt
dt
dt
3
dV   2 dh
dr 
 r
 h2 r 
dt
3  dt
dt 
Chain Rule
dV  2
 1  0.2   2  1 2  0.1 
dt
3
= -0.628 . . . cubic feet per min.
Aim: Related Rates
Course: Calculus
Model Problem
x and y are both differentiable functions of t
and related by the equation y = x2 + 3. Find
dy/dt when x = 1, given that dx/dt = 2 when
x = 1.
y = x2 + 3
original equation
d
d
2

y

x
 3 



dt
dt
differentiate with
respect to t
dy
dx
 2x
dt
dt
Chain Rule
When x = 1 and dx/dt = 2, you have
dy
 2  1 2   4
dt
Aim: Related Rates
Course: Calculus
Aim: How do we find related rates when we
have more than two variables?
Do Now:
Suppose that the diameter D and the area of a
circle are differentiable functions of t. Write
an equation that relates dA/dt and dD/dt.
Aim: Related Rates
Course: Calculus
Aim: How do we find related rates when we
have more than two variables?
Do Now:
A pebble is dropped into a calm pond,
causing ripples in the form of concentric
circles. The radius r of the outer ripple is
increasing at a constant rate of 1 foot per
second. When the radius is 4 feet, at what
rate is the total area A of the disturbed water
changing?
Aim: Related Rates
Course: Calculus
Model Verbal Problem
A pebble is dropped into a calm pond,
causing ripples in the form of concentric
circles. The radius r of the outer ripple is
increasing at a constant rate of 1 foot per
second. When the radius is 4 feet, at what
rate is the total area A of the disturbed water
changing?
A   r2
dr
1
dt
Area formula for circle
rate of change
what to find?
dA
 ?, when r  4
dt
Aim: Related Rates
Course: Calculus
Model Verbal Problem
A pebble is dropped into a calm pond,
causing ripples in the form of concentric
circles. The radius r of the outer ripple is
increasing at a constant rate of 1 foot per
second. When the radius is 4 feet, at what
rate is the total area A of the disturbed water
2
A


r
changing?
dA d
differentiate with
  r 2 
respect to t
dt dt
dA
d 2
dr
   r    2r
Chain Rule
dt
dt
dt
dA
Substitute
  2  1 4   8
dt
When r = 4, the area is changing
a rate
of 8 sq.ft./sec.
Aim: at
Related
Rates
Course: Calculus
Model Verbal Problem
Air is being pumped into a spherical balloon
at a rate of 4.5 cubic inches per minute.
Find the rate of change of the radius
when the radius is 2 inches.
4 3
Volume formula for sphere
V  r
3
dV
 4.5
dt
rate of change
what to find?
dr
 ?, when r  2
dt
Aim: Related Rates
Course: Calculus
Model Problem (con’t)
Air is being pumped into a spherical balloon at a rate of 4.5
cubic inches per minute. Find the rate of change of the
radius when the radius is 2 inches.
4 3
V  r
3
dV 4
2 dr
   3r 
dt
3
dt
2 dr
 4 r
dt
dr
1 dV

dt 4 r 2 dt
dr
1

4.5  
2 
dt 4  2 
Volume formula for sphere
differentiate with respect to t
solve for dr/dt
substitute
0.09 inches per min.
Aim: Related Rates
Course: Calculus
Model Problem
A airplane is flying at an altitude of 6 mi. on a
flight path that will take it directly over a
radar tracking station. If s is decreasing at
a rate of 400 miles per hour when s = 10
miles, what is the speed of the plane?
rate of change
a2 + b2 = c2
ds
 400
dt
when s = 10, x = 8 miles
what to find?
dx
 ?,
dt
when s  10 and x = ?8
10s
8x
Aim: Related Rates
Course: Calculus
6 mi.
Model Problem
A airplane is flying on a flight path that will take it directly
over a radar tracking station. If s is decreasing at a rate of
400 miles per hour when s = 10 miles, what is the speed of
the plane?
x2 + 6 2 = s2
Pythagorean Theorem
dx
ds
2x
 2s
dt
dt
differentiate with respect to t
dx 2 s ds

dt 2 x dt
solve for dr/dt
dx  10 

 400 
dt
 8
s
substitute
 500 mi/hr
x
Aim: Related Rates
Course: Calculus
6 mi.
Model Problem
Find the rate of change in the angle of
elevation of the camera at 10 seconds
after liftoff.
 is the angle of elevation
s = 50t2
ds
 100t
dt
Position Equation
Velocity of rocket
when t = 10,
s = 5000 feet
Substitution &
Evaluation
what to find?
d
 ?, when t  10
dt
and s  5000
Aim: Related Rates
Course: Calculus
Model Problem
Find the rate of change in the angle of elevation of the camera
s
at 10 seconds after liftoff.
tan 
2000

d
1  ds 
sec 

dt 2000  dt 
2

differentiate with
respect to t
d
1
2
 cos 
 100t 
dt
2000
substitute ds/st = 100t
and solve for d/dt
2
 100t
d 
2000


2
2
dt  s  2000  2000
cos   2000 /
 100  10 
d 
2000


dt  50002  20002  2000
s  2000
2
2
d
2

radians per second
dt 29 Aim: Related Rates
substitute s, and t
and simplify
Course: Calculus
2
Aim: How do we find related rates when we
have more than two variables?
Do Now:
At a sand and grave plant, sand is falling off a
conveyor and onto a conical pile at a rate of 10
cubic feet per minute. The diameter of the
base of the cone is approximately three times
the altitude. At what rate is the height of the
pile changing when the pile is 15 feet high?
Aim: Related Rates
Course: Calculus
Model Problem
In an engine, a 7-inch connecting rod is
fastened to a crank of radius 3 inches. The
crankshaft rotates counterclockwise at a
constant rate of 200 revolutions per minute.
Find the velocity of the piston when  = /3.
7
3
 = /3
x
the velocity of a piston is related
to the angle of the crankshaft.
what to find?
dx

 ?, when  
dt
3
1 revolution
2 radians
rev  2 radians 
rad
d 
 400
  200
?


min  1 revolution 
min
dt 
How do we relate x to ?
Aim: Related Rates
Course: Calculus
Model Problem
In an engine, a 7-inch connecting rod is fastened to a crank of radius
3 inches. The crankshaft rotates counterclockwise at a constant rate
of 200 revolutions per minute. Find the velocity of the piston when
 = /3.
How do we relate x to ?
Law of Cosines: b2 = a2 + c2 – 2ac cos 
a

b
x
72 = 32 + x2 – 2(3)(x) cos 
dx
d
dx  differentiate with

0  2x
 6   x sin
 cos

respect to t
dt
dt
dt


dx
dx
d
isolate dx/dt
0  2x
 6cos
 6 x sin
dt
dt
dt
dx
d
  2 x  6cos 
 6 x sin 
dt
dt
dx
6 x sin  d 



Aim:
Related
Rates
dt 6cos  2 x  dt 
Course: Calculus
Model Problem
In an engine, a 7-inch connecting rod is fastened to a crank of radius
3 inches. The crankshaft rotates counterclockwise at a constant rate
of 200 revolutions per minute. Find the velocity of the piston when
 = /3.
72 = 32 + x2 – 2(3)(x) cos /3
49 = 9 + x2 – 6(x)(1/2)
0 = x2 – 3x - 40
0 = (x – 8)(x + 5)
solve for x by
substituting  = /3
x=8
 3
6 8 

2 
dx


400 

dt
1
6    16
 2
solve for dx/dt by
substituting x = 8 and
 = /3
dx
6 x sin  d 

dt 6cos  2 x  dt 
9600 3

 4018 inches per minute
13 Aim: Related Rates
Course: Calculus
Model Problem
A circular pool of water is expanding at the
rate of 16π in2/sec. At what rate is the
radius expanding when the radius is 4
inches?
Aim: Related Rates
Course: Calculus
Model Problem
A 25-foot ladder is leaning against a wall
and sliding toward the floor. If the foot of
the ladder is sliding away from the base of
the wall at a rate of 15 ft/sec, how fast is the
top of the ladder sliding down the wall
when the top of the ladder is 7 feet from the
ground?
Aim: Related Rates
Course: Calculus
Model Problem
A spherical balloon is expanding at a rate of
60π in3/sec. How fast is the surface area of
the balloon expanding when the radius of
the balloon is 4 in.
Aim: Related Rates
Course: Calculus
Model Problem
An underground conical tank, standing on
its vertex, is being filled with water at the
rate of 18π ft3/min. If the tank has a height
of 30 feet and a radius of 15 feet, how fast is
the water level rising when the water is 12
feet deep?
Aim: Related Rates
Course: Calculus
Model Problem
A rocket is rising vertically at a rate of 5400
miles per hour. An observer on the ground
is standing 20 miles from the rocket’s
launch point. How fast (in radians per
second) is the angle of elevation between the
ground and the observer’s line of sight of
the rocket increasing when the rocket is at
an elevation of 40 miles?
Aim: Related Rates
Course: Calculus
Model Problem
A man 2 meters tall walks at the rate of 2
meters per second toward a streetlight
that’s 5 meters above the ground. At what
rate is the tip of his shadow moving?
Aim: Related Rates
Course: Calculus