# calculus

posted by on .

A water-trough is 10m long and has a cross-section which is the shape of an isosceles trapezoid
that is 30cm wide at the bottom, 80cm wide at the top, and has height 50cm. If the trough is being fi lled with water at the rate of 0.2 m3/min, how fast is the water level rising when the water is 30cm deep?

The steps I took were as follows

b1 is the width of the water at height 0

b2 is the width of the water at height h

knowing the volume of a trapezoid is

V = 0.5(b1 + b2)hL

where L is the length of the trapezoid.

As the height increases by 1 so does the width so that means dh/dt = dw/dt

Now I tried solving for h in meters.

V = 5h(0.3 +(0.3 + h))
[0.3 + h is the width of the water at the combined height]

V = 3h + 5h^2

differentiate by d/dt

dV/dt = 3dh/dt + 10h dh/dt

(dV/dt)/(3 + 10h) = dh/dt

solve for dV/dt = 0.2 and h = 0.3

1/30 m/min = dh/dt

The answer is 1/3 cm/min but when I convert this to cm I get 10/3 cm/min.

What exactly am I doing wrong here?

• calculus - ,

well first of all think easy way
for small changes (as dh ---> 0)
change in volume = surface area * change in height
or
dV = A dh

Now do the problem (the surface is a rectangle)
we know the length, 10 meters
what is the width when h = 30 ???
linear in depth (straight sides)
w = .30 + [(.80-.30) /.50] h
or
w = .3 + h
so for the area of the surface
A = 10(.3+h)
so
when h = .3, A = 10(.6) = 6 m^2
dV = A dh
so
dV/dt = A dh/dt
.2 m^3/min = 6 m^2 dh/dt
so
dh/dt = (.2/6) meters/min
= .03333 meters/min
= 3 1/3 cm /min