Lesson 18 - Related Rates, Slides of Calculus for Engineers

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Lesson 7

Related Rates

(Time – Rate Problems)

OBJECTIVES:

• to “know by heart” the different rules of

differentiation;

• to utilize fully these rules and appreciate its

simplicity; and

• to extend these basic rules to other “complex”

algebraic functions.

A Strategy for Solving Related Rates Problems (p. 205)

Example 2

A balloon leaving the ground 60 feet from an observer, rises vertically at the rate

10 ft/sec. How fast is the balloon receding from the observer after 8 seconds?

    t time  sec  sincetheballoonstartstorise fromtheground

L distance ft of theballoon fromtheobserveratanyinstant

Let h height ft of theballoon fromthegroundatanyinstant







Viewer 60 feet

h

L

sec

ft 10 dt

dh 

? dt

dL

t 8 sec

 

 

L h 3600  Working Equation

In the figure: L h 60

2

2 2 2

 

 

h 3600

dt

dh h

dt

dL

2 h 3600

dt

dh 2 h

dt

dL

2

2



 

  







 

  





8sec  80 ft.

sec

ft h 10

and t 8sec sec

ft 10 dt

dh Since,

^  

  

 

 

  

 

sec

ft 8 100

800

dt

dL

10 , 000

800

6400 3600

800

dt

dL

80 3600

80 10

dt

dL 2

 

 







 

 

R length ^ ft  of theropeout at any instant

x ditsnace ft of theboat fromthewharf at anyinstant

Let t time ft sincetheboat startstoapproachthewharf







x R 400 (WorkingEquation)

x R 400

R x 20

when R 25 ft dt

dx Find

2

2 2

2 2 2

 

 

 



R 400

dt

dR R

2 R 400

dt

dR 2 R

x

x R 400

2 2

2



 



 

 

 

sec

ft

3

20

dt

dx

15

25 4

25 400

25 4

dt

dx

sec

ft 4 dt

dR and

When R 25 ft

2

 

 



 

 



 

h height(ft)of thewater at any instant

r radius(ft)of thewater surfaceat anyinstant

Lett time min sincethewater flowsintothereservoir







 r  h 3

1 Bh 3

1 V

whenthewateris 8 ft deep dt

dh Find

2   

h 4

1 r h

r

20

5

By ratioand proportion

  

 

    min

ft 4

15 8

16 15 h

dt

dV 16

dt

dh

dt

dh h dt 16

dh 3 h dt 48

dV h Working Equation 48

V

h h 4

1 3

1 Thus, V

2 2 h 8 ft

3 2 2

2

 ^ 

  



  

 

  

   

  

  

  

   



 





 

  

  

 





Example 5.

Water is flowing into a vertical tank at the rate of 24. If

the radius of the tank is 4 feet, how fast is the surface

rising?

h

4 feet

min

ft 24

3

Example 6

A triangular trough is 10 feet long, 6 feet across the top, and 3 feet

deep. If water flows in at the rate of 12 ft 3 /min, find how fast the

surface is rising when the water is 6 inches deep?

min

ft 12

3

h

6 feet

3 feet

x

6 feet

3 feet

V volume  ft  of thewaterat any instant

end at anyinstant

x horizontalwidth ft of thewateratthetriangular

h height ft of thewater atanyinstant

Let t time min sincethewater flowsintothetrough

3 







when the water is 6 inches deep. dt

dh Find

Thus, V 5xh 5 2h h 10h Working Equation

x 2 h 3

6

h

x By ratio and proportion,

xh 10 5xh 2

1 V

From V Bh

2   

  

  

  

 



Example 7

A train, starting at noon, travels at 40 mph going north. Another train, starting

from the same point at 2 : 00 pm travels east at 50 mph. Find how fast the two

trains are separating at 3 : 00 pm.

80 miles

x

2p m

B

C

D

A

L

y

3pm

3pm

hr

mi 40 dt

dy 

hr

mi 50 dt

dx  12p m

2pm

when t 1hr. dt

dL Find 

 

L x  80 y  Working Equation

From the figure: L x 80 y

2 2

2 2 2

  

  

2 2 2 x ( 80 y)

dt

dy 2 ( 80 y) dt

dx 2 x

dt

dL

 

 



 ^ ^ 





 

   y 40 mph 1 hr 40 miles

x 50mph 1 hr 50 miles After 1 hr

AB  40  2 80 miles

40 mph dt

dy 50 mph and dt

dx Since

 

 