Related rates - Fay`s Mathematics

Report
RELATED RATES
What are related rates?
 Related rates are found when there are two
or more variables that all depend on
another variable, usually time
 Two or more quantities change as time
changes
 Since the variables are related to each other,
the rates at which they change (their
derivatives) are also related
 In related rate problems, the goal is to calculate an
unknown rate of change in terms of other rates of
change that are known.
 How fast does the top of the ladder move if the
bottom of the ladder is pulled away from the wall at
a constant speed?
 It may be surprising to you that the top and bottom are
moving at different speeds. The top actually speeds up
as the bottom stays at a constant.
 Suppose x and y are both differentiable functions
of t and are related by the equation y = x2+3. Find
dy/dt when x=1, given that dx/dt=2 when x=1.
 Differentiate both sides with respect to t. Will
have to put d__/dt for each variable.
 Then substitute dx/dt with 2 and x with 1.
Procedure:
1.
2.
3.
4.
Identify and LABEL all the given info and what
you are asked to find. Draw a picture if
appropriate.
Write an EQUATION relating the variables.
Differentiate both sides of the equation with
respect to TIME.
Substitute and Solve. Sometimes you will need
to use the original equation or other equations
to solve for missing parts.
BASIC SKILL:
DRAW A SKETCH AND
DIFFERENTIATE BASIC GEOMETRY FORMULAS WITH RESPECT TO TIME.
1. Let A be the Area of a circle of radius r.
How is dA/dt related to dr/dt?
BASIC SKILL:
DRAW A SKETCH AND
DIFFERENTIATE BASIC GEOMETRY FORMULAS WITH RESPECT TO TIME.
2. Let V be the Volume of a cube of side length x.
How is dV/dt related to dx/dt?
BASIC SKILL:
DRAW A SKETCH AND
DIFFERENTIATE BASIC GEOMETRY FORMULAS WITH RESPECT TO TIME.
3. Let V be the Volume of a sphere of radius r.
How is dV/dt related to dr/dt?
A pipe is filling a cylindrical tank at the rate of 2500
cm3 per minute. If the radius of the tank is 25cm,
how fast is the height of the water in the tank
changing?
What is changing?
What is constant?
Water is draining from a cylindrical tank at 3
liters/second. How fast is the surface
dropping when the cylinder’s radius is 20 cm?
Find
dh
dt
V   r 2h

 A pebble is dropped into a calm pond, causing ripples
in the form of cocentric 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?
 Find a model that relates the information.
 A=πr2
 Differentiate each variable with respect to t.
 Substitute dr/dt=1 and r =4
 Air is being pumped into a spherical balloon at a
rate of 4.5 cubic feet per minute. Find the rate of
change of the radius when the radius is 2 feet.
4 3
V  r
3
Differentiate both sides with
respect to t
 A bouillon cube with side lengths 0.8cm is placed into
boiling water. Assuming it roughly resembles a cube as
it dissolves, at approximately what rate is its volume
changing when its side length is 0.25cm and is
decreasing at a rate of 0.12 cm/sec?
 Volume of a cube is V=s3
 A spherical cell is growing at a constant rate of 400
µm3 / day. At what rate is its radius increasing
when the radius is 10 µm?
4 3
V  r
3
 Water pours into a conical tank of height 10 ft and
radius 4 ft at a rate of 10 ft3/min. How fast is the
water level rising when it is 5 ft. high? As time
passes, what happens to the rate at which the
dV
water level rises? Explain.
Need to get rid of the
 10
1 2
V  r h
3
variable r some how, can
we rewrite r in terms of h?
We can if we use similar
triangles.
1
1
2
V   (.4h) h   .16h3
3
3
dV 1
2 dh
  (3)(.16) h
dt 3
dt
dV
2 dh
  (.16) h
dt
dt
dh
10  .16 h
solve for dh / dt
dt
2
dt
 A basin has the shape of an inverted cone with altitude
100 cm and radius at the top of 50 cm. Water is poured
into the basin at the constant rate of 40 cubic
cm/minute. At the instant when the volume of water in
the basin is 486π cubic centimeters, find the rate at
which the level of water is rising.
 A 16 ft. ladder leans against a wall. The bottom of the
ladder is 5 ft. from the wall at time t=0 and slides away
from the wall at a rate of 3 ft/s. Find the velocity of the
top of the ladder at time t=1.
 x=x(t) distance from the bottom of the ladder to the wall
 h=h(t) height of the ladder’s top
 Both x and h are functions of time. The velocity of the bottom is dx/dt=3 ft/s.
The velocity of the top is dh/dt (change in height for some t) and the initial
distance from the bottom to the wall is x(0)=5.
x  h  16
2
16
h
x
2
2
We want to find d/dt (both derivatives with respect
to t) of both x and h, so…
d 2 d 2 d 2
x  h  16
dt
dt
dt
So…
x  h  16
2
dx
2 x
dh
dt

dt
2h
dx
x
dh
dt

dt
h
2
2
dx
dh
2 x  2h
0
dt
dt
We were told that dx/dt was 3 ft/s. Plug
it in, we also know that x(0)=5 (how far
the ladder is away from the base of the
wall at time 0, so to see where it is at
t=1 we would take 5+3=8. Now we know
x @ t=1, find h @ t=1 by the Pythagorean
theorem. h(1) = 13.86
dh
dt
8(3)

13.86
t 1
NOW WE NEED TO GET RID
OF W (the extra variable)
 A 5-foot girl is walking toward a 20-foot lamppost at a rate of 6
feet per second. How fast is the tip of her shadow (cast by the
lamppost) moving?
 Let x(t) be the distance of the girl to the base of the post, and let
y(t) be the distance of the tip of the shadow to the base of the
post. If you've drawn the right setup, you should see similar
triangles...this question involves asking how fast the tip of her
shadow is moving so we think about the distance between her and
the tip of the shadow (y-x)
 If it asks how fast is the length of the shadow moving, now we
have to think about the distance from the light pole to the tip of
the shadow (y+x).
 A man is 5 ft tall and is walking toward a light post at a
rate of 4 ft/sec. Find the rate that the tip of his shadow
is moving, and determine the rate at which the length of
his shadow is changing if the light post is 22 feet high.
A light shines from the top of a pole 50 ft high. A ball is dropped
from the same height from a point 30 ft away from the light as
shown in the figure. How fast is the shadow of the ball moving
along the ground ½ second later?
 One vehicle starts driving north and one vehicle starts driving
west from an intersection (as indicated by the arrows above). At
the time the first vehicle is .3 miles north the intersection it is
traveling at 20 mph. Simultaneously the second vehicle is .4 miles
west of the intersection and is traveling at 25 mph. At that instant,
how fast are the two vehicles separating?
 Two hikers begin at the same location and travel in
perpendicular directions. Hiker A travels due north at a
rate of 5 miles per hour. Hiker B travels due west at a
rate of 8 miles per hour. At what rate is the distance
between the hikers changing 3 hours into the hike?
da/dt = 5
db/dt = 8
We want to find the rate of change of
the hypotenuse which would be
dd/dt.
The Pythagorean theorem relates the
variables.
 Water pours into a fish tank at a rate of 3 ft3/min.
How fast is the water lever rising if the base of the
tank is a rectangle of dimensions 2 x 3 ft?
dV
 change in volume with respect to time
dt
dh
 change in height with respect to time
dt
V  lwh
V  6h
dV
dh
6
dt
dt
dh Solve now
36
dt for dh/dt
dh
dt
10
3
 .795775ft /min
2
t 5 (.16) h
To think about the second part of the question, one would think about the
volume of a cross section close to the tip of the cone, and a cross section at
the base of the cone. Which is a greater volume?
Or you could think about letting t=1 (close to when you first start filling up)
and compare that to t=5.
 A spy tracks a rocket through a telescope to determine its velocity. The rocket is
traveling vertically from a launching pad located 10 km away. At a certain
moment, the spy’s instruments show that the angle between the telescope and
the ground is equal to pi/3 and is changing at a rate 0f 0.5 rad/min. What is the
rocket’s velocity at that moment?
We want to find dy/dt, the velocity (distance/time) of
the rocket
We know dӨ/dt
A relationship between theta and y would be…
tan  
y
10
d 1 dy
sec

dt 10 dt
2
dy
10 d

dt cos 2  dt
dy
dt
  3
10

(.5)
2 
cos ( 3 )
 A hot air balloon rising vertically is tracked by an
observer located 2 miles from the lift-off point. At a
certain moment, the angle between the observer’s lineof-sight and the horizontal is pi/5, and it is changing at a
rate of 0.2 radians/minute. How fast is the balloon
rising at this moment.
 At a given moment, a plane passes directly above a radar
station at an altitude of 6 miles. If the plane’s speed is
500 mph. Suppose that the line through the radar
station and the plane makes an angle theta with the
horizontal. How fast is theta changing 10 minutes after
the plane passes directly above the station?
 In the engine shown, a 7 inch connecting rod is fastened to
a crank of radius 3 inches. The crankshaft rotates
counterclockwise at a constant of 200 revolutions per
minute. Find the velocity of the piston when Ө=π/3
 dӨ/dt=200(2pi) = 400 π
Find dx/dt when Ө=π/3
The relationship to use is law of cosines.
a2=b2+c2-2bc*cosA

similar documents