RELATED RATES

What are 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

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. Identify and LABEL all the given info and what you are asked to find. Draw a picture if appropriate.

2. Write an EQUATION relating the variables.

3. Differentiate both sides of the equation with respect to TIME.

4. 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?

Finddh

dt2V r h

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.

Differentiate both sides with respect to t

34

3V r

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?

34

3V r

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 water level rises? Explain.

21

3V r h

Need to get rid of the variable r some how, can we rewrite r in

terms of h? We can if we use similar

triangles.2 31 1

(.4 ) .163 3

V h h h

2

2

1(3)(.16)

3

(.16)

dV dhh

dt dtdV dh

hdt dt

10dV

dt

210 .16 /dh

h solve for dh dtdt

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.

16

h

x

We want to find d/dt (both derivatives with respect to t) of both x and h, so…

2 2 216d d dx h

dt dt dt

2 2 216x h

So… 2 2 0dx dhx hdt dt

2

2

dxxdh dt

dt h

dxxdh dt

dt h

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

2 2 216x h

1

8(3)

13.86t

dh

dt

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 = 5db/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

changeinvolumewith respect to timedt

dhchangein height with respect to time

dt

6

V lwh

V h

6dV dh

dt dt 3 6

dh

dt Solve

now for dh/dt

32

5

10.795775ft /min

(.16)t

dh

dt 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…

tan10

y 2 1sec

10

d dy

dt dt

2

10

cos

dy d

dt dt

3

23

10(.5)

cos ( )

dy

dt

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 line-of-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