Question

Water is flowing into a trough at a rate of 100 cubic centimeters per second. The trough has a length of 3 meters and cross sections in the form of a trapezoid, whose height is 50 cm, whose lower base is 25 cm and whose upper base is 1 meter. At what rate is the water rising when the depth of the water is 25 cm?

Answer 1

when the water has depth y cm, the width of the surface has width

25+(75/50)y cm

So, at depth y, the area of the trapezoidal cross-section is

1/2 (25 + (75/50)y)(y) = 3/4 y^2 + 25/2 y cm^2

So, the volume is
v = 9/4 y^2 + 75/2 y cm^3

So, we want dy/dt when y=25, and we know that dv/dt=100:

9/4 (2y dy/dt) + 75/2 dy/dt = 100
9/4 (50 + 75/2) dy/dt = 100
dy/dt = 32/63 cm/s

An unusual answer; better check my math.

Answer 2

I tried a slightly different approach

I looked at the trapezoid, and drew two verticals upwards from the base,
creating a rectangle of 25 by 50 cm and two right-angled triangles ,
each of height 50 and base 75/2

I drew a water line, of height h and width of r in each of the triangles
by ratios: r/h = (75/2) / 50 = 3/4
r = (3/4)h

V = 300( 25h + 2(1/2)rh)
= 300(25h + rh)
= 300(25h + 3/4)h^2)

dV/dt = 300( 25dh/dt + (3/2) h dh/dt)
= 300(dh/dt)(25 + (3/2)h)

when h = 25 , and dV/dt = 100

100 = 300 (dh/dt) (25 + 37.5)

dh/dt = 100/(300(62.5)) = 100/18750 = 2/375 cm/s or appr .00533 cm/s

As usual, my arithmetic should be checked also

Answer 3

Well, my most obvious mistake was using 3 as the length, not 300 (cm).