The trough shown in the figure is 5 feet The trough shown in the figure is 5 feet long, and its vertical cross sections are long, and its vertical cross sections are inverted isosceles triangles with base 2 inverted isosceles triangles with base 2 feet and height 3 feet. Water is being feet and height 3 feet. Water is being siphoned out of the trough at the rate of 2 siphoned out of the trough at the rate of 2 cubic feet per minute. At any time cubic feet per minute. At any time tt, let , let hh be the depth and be the depth and VV be the volume of be the volume of water in the trough.water in the trough.

a) Find the volume of water in the trough when it is full.a) Find the volume of water in the trough when it is full.

b) What is the rate of change in b) What is the rate of change in hh at the instant at the instant when the trough is ¼ full by volume?when the trough is ¼ full by volume?

c) What is the rate of change in the area of c) What is the rate of change in the area of the surface of the water (shaded in the figure) the surface of the water (shaded in the figure) at the instant when the trough is ¼ full by at the instant when the trough is ¼ full by volume?volume?

a) Find the volume of water in the trough when it is full.a) Find the volume of water in the trough when it is full.

3ft15

5322

12

1

lhbV

b) What is the rate of change in b) What is the rate of change in hh at the instant when the at the instant when the trough is ¼ full by volume?trough is ¼ full by volume?

2

3

5

53

2

2

1

3

2

2

1

h

hh

lhhV

hb

bhh

b

3

2

323

2

b) What is the rate of change in b) What is the rate of change in hh at the instant when the at the instant when the trough is ¼ full by volume?trough is ¼ full by volume?

min/ft5

33

102-

(given) min/ft2

3

10

3

hdt

dhdt

dhh

dt

dVdt

dhh

dt

dV

h

h

2

33

5

4

15 2

minft/4.15

623

5

3

dt

dh

[ 0 , 3 ][ 0 , 3 ]

[ -2 , 0 ][ -2 , 0 ]

Solving for dh/dt graphically...Solving for dh/dt graphically...

25

3V h

10

3

dV dhh

dt dt

102

3

dhhdt

dt

dh

h

5

3

Solving for dh/dt numerically...Solving for dh/dt numerically...

c) What is the rate of change in the area of the surface of the water c) What is the rate of change in the area of the surface of the water (shaded in the figure) at the instant when the trough is ¼ full by (shaded in the figure) at the instant when the trough is ¼ full by volume?volume?

h

h

b

blS

3

10

3

25

5

min/ft3

415

20

5

2

3

103

10

2

dt

dh

dt

dS