Related Rates

Review Problem:Use implicit differentiation to find If

dx

dy

.104)cos( 32 yyxx

What is a Related Rate?

When working related-rate problems, instead of finding a derivative of an equation y with respect to the variable x, you are finding the derivative of equations with respect to time, a hidden variable t. This means that variables are tied together in their relationship with time.

1. Suppose x and y are functions of .t

Differentiate each side of the following equation with respect to t: 422 yx

a. If ,2dt

dx find

dt

dy when 1x and .3y

1. If xxy 53 3 and dx/dt = 4, find dy/dt when x = 4.

2. If 40022 yx and dy/dt = 6, find dx/dt when y = 16.

Rates of Change

• You have studied the derivative as the slope of a tangent line. Now lets look at it as a rate of change of one variable with respect to another.

• It’s common to use rate of change too describe motion of an object and until now we dealt only with motion in a straight line. But motion is not always in a straight line, more often it is along a curved path. We will study both in this section.

• Remember that average change is written as: _ _ tan s

_ _

change in dis ce

change in time t

Falling Objects

If a billiard ball is dropped from a height of 100 feet, its height s at a time t is given by the position function:S= -16t2 + 100

Where s is measured in feet and t is measured in seconds. Find the average velocity over the interval [1,2]. This procedure uses pre-calculus techniques.

_ _ tan s (2) (1) 36 8448 / sec

_ _ 2 1 2 1

change in dis ce s sft

change in time t

Velocity of a Falling Object

The general formula for a free-falling object is the following: s(t) = (½)gt2 + v0t + s0 where – v0 is the initial velocity of the object

– s0 is the initial position of the object

– g is the acceleration due to gravity. g is either -9.8 m/s or-32 ft/s

Rates of Change with Calculus

Now that we’ve looked at average changes in rates, lets look at instantaneous rates of change.

Suppose we want to find out the exact velocity of an object dropping from a given height at any given time. If we have the position function (location of the object along the straight path over time) then we can easily find the velocity by using calculus. We just take the derivative.

Using the Derivative to Find Velocity

At time t=0, a diver jumps from a platform diving board that is 32 feet above the water. The position of the diver is given by

s(t)= -16t2 + 16t + 32

where s is measured in feet

and t is measured in seconds

1. When does the diver hit the water?

2. What is the diver’s velocity at impact?

Using the Derivative to Find Velocity – continued…

1. To find the time t when the diver hits the water, let s = 0 and solve for t.

-16t2 + 16t + 32 = 0

-16(t+1)(t-2)=0

so t=-1 or t=2. since we can’t go back in time, we’ll use t=2

Using the Derivative to Find Velocity – continued…

2. The velocity at time t is given by the derivative s’(t) = -32t + 16

so the velocity at time t=2 is:

s’(2) = -32(2) + 16 =-48 ft/s

Vertical Motion Practice Problems

1. A silver dollar is dropped from the top of a building that is 1362 feet tall.

a. Determine the position and velocity functions for the coin.s(t)= -16t2 + 1362 {initial velocity is 0 since it was just dropped}

and we have v(t) = s’(t) = -32tb. Determine the average velocity on the interval [1,2]

Using the formula from a previous slide: [s(2)-s(1)]/2-1 = 1298 -1346 = -48 ft/sec

c. Find the instantaneous velocities when t=1 and t=2.v(t) = s’(t) = -32t v(1)= -32ft/sec and v(2)= -64ft/sec

d. Find the time required for the coin to reach ground level.s(t)= -16t2 + 1362 = 0 t = 9.226 sec

e. Find the velocity of the coin at impact.v(9.226)= -295.242 ft/sec

1. Given that the volume of a cone is ,3

1 2hrV find dt

dr when 2r in., 3h in.,

,in. 4 3V sec,/in. 2

1 3dt

dV and in./sec.

4

3

dt

dh

For your notes:

Consider a sphere of radius 10cm. Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec.(Possible if the sphere is a soap bubble or a balloon.)

34

3V r

24dV dr

rdt dt

2 cm4 10cm 0.1

sec

dV

dt

3cm

40sec

dV

dt

The sphere is growing at a rate of .340 cm / sec

Note: This is an exact answer, not an approximation like we got with the differential problems.

Water is draining from a cylindrical tank at 3 liters/second. Where the radius = 3cm. How fast is the surface dropping?

Convert:

L3

sec

dV

dt

3cm3000

sec

Finddh

dt2V r h

Steps for Related Rates Problems:

1. Draw a picture (sketch).

2. Write down known information.

3. Write down what you are looking for.

4. Write an equation to relate the variables. Plug in the constants

5. Differentiate both sides with respect to t.6. Evaluate.7. Check to make sure you answered the correct question.

Hot Air Balloon Problem:

Given:4

rad

0.14min

d

dt

How fast is the balloon rising?

Finddh

dt

tan500

h

2 1sec

500

d dh

dt dt

2

1sec 0.14

4 500

dh

dt

h

500ft

Hot Air Balloon Problem:

Given:4

rad

0.14min

d

dt

How fast is the balloon rising?

Finddh

dt

tan500

h

2 1sec

500

d dh

dt dt

2

1sec 0.14

4 500

dh

dt

h

500ft

2

2 0.14 500dh

dt

1

12

4

sec 24

ft140

min

dh

dt

4x

3y

B

A

5z

Truck Problem:Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

How fast is the distance between the trucks changing 6 minutes later?

r t d 1

40 410

130 3

10

2 2 23 4 z 29 16 z

225 z5 z

4x

3y

30dy

dt

40dx

dt

B

A

5z

Truck Problem:

How fast is the distance between the trucks changing 6 minutes later?

r t d 1

40 410

130 3

10

2 2 23 4 z 29 16 z

225 z5 z

2 2 2x y z

2 2 2dx dy dz

x y zdt dt dt

4 40 3 30 5dz

dt

250 5dz

dt

50dz

dt

miles50

hour

Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

p