A tank for storing water is conical in shape, with maximum radius 20 m and maxi­mum depth 15 m. The large, circular end of the tank points vertically upwards and is open to the air. It is known that the volume of water in the tank will decrease at a rate proportional to the area of the water's surface, due to evaporation. Show that the water-level in the tank decreases at a constant rate. If the tank originally stores 1000 $\mathrm{m}^3$ of water and the water-level drops 1cm/day, how long will it be before the tank is empty?

First, you have draw the figure, and figure volume in the tank as a function of height (radius=1.5*height)

Volume=1/3 PI (1.5h)^2 *h
the rate of volume change (dV/dt) is given as constant (which ignores Bernoulli principle), but anyway

dV/dt= PI/3 ( 2.25h^2 + 3h^2) dh/dt
check that.
Now consider evaporation:
Ve=constant*PIr^2=k PI*(1.5h)^2
dVe/dt=constant*PI*3h dh/dt

so in the end, total change in volume/time is
deltaV/dt=-PI PI/3 ( 2.25h^2 + 3h^2) dh/dt - constant*PI*3h dh/dt

but dh/dt=.01m/day
so now you have some function for dV/dt.

V=INT dV=INT from o to V = INT function dt from zero to time T. Solve for T after you put in the limits.

If the area of the water is A, and its depth is h, then

V = 1/3 Ah
But, due to the conical shape, r = kh, so V = kr^3

Taking derivatives,
dV/dt = kr^2 dh/dt
But, we are told that dV/dt = kA, so

(dV/dt)/A = k = dh/dt

So, dh/dt = k

Note that all the k's used are different, but each is constant.

To show that the water level in the tank decreases at a constant rate, we need to determine the relationship between the volume of water in the tank and the area of the water's surface.

Let's denote the radius of the water surface as r and the depth of water as h. As given, the maximum radius of the tank is 20 m, and the maximum depth is 15 m.

The volume V of a cone can be calculated using the formula:

V = (1/3) * π * r^2 * h

The rate at which the volume of water decreases is directly proportional to the area of the water's surface. If we assume the rate of decrease is k, the rate of change of volume with respect to time (dv/dt) can be expressed as:

dv/dt = -k * A

where A is the surface area of the water.

The surface area A can be calculated using the formula for the lateral surface area of a cone:

A = π * r * s

where s is the slant height of the cone.

Since the large end of the tank points vertically upwards, the slant height s can be calculated using the Pythagorean theorem:

s^2 = r^2 + h^2

Now, let's differentiate the volume equation with respect to time to find an expression for dv/dt:

dv/dt = (1/3) * π * [2r * dr/dt * h + r^2 * dh/dt]

We can rearrange the above equation to solve for dh/dt, which gives us the rate at which the water level decreases:

dh/dt = -(2/3) * (dr/dt) * (r/h)

We know that the initial volume of water in the tank is 1000 m^3, and the water level drops by 1 cm/day. We can convert the drop in water level to meters by dividing by 100. Therefore, dh/dt = -0.01 m/day.

To find the time it takes for the tank to be empty, we need to determine when the water level reaches zero. We can plug in the given values into the rate equation and solve for time:

0 = -(2/3) * (dr/dt) * (r/h)

Since we require the water level to decrease, dr/dt must be negative, and r and h must also be positive. From the equation above, we can see that if dh/dt is constant, dr/dt must also be constant to ensure a constant rate of decrease in water level.

Therefore, the water-level in the tank decreases at a constant rate.

To find how long it will take for the tank to be empty, we need to determine when the water level reaches zero. We need to solve the differential equation:

dh/dt = -(2/3) * (dr/dt) * (r/h)

Given that dh/dt = -0.01 m/day, we can substitute this into the equation:

-0.01 = -(2/3) * (dr/dt) * (r/h)

We know that initially, r = 20 m when the tank stores 1000 m^3 of water. Substituting r = 20 m and h = 15 m, we can solve for dr/dt:

-0.01 = -(2/3) * (dr/dt) * (20/15)

Simplifying the equation:

dr/dt = -0.01 * (15/20) * (3/2)

dr/dt = -0.0075 m/day

Now, we can determine the time it takes for the tank to be empty by dividing the initial volume by the rate of decrease in volume:

Time = 1000 m^3 / (0.0075 m/day) = 133,333.33 days

Therefore, it will take approximately 133,333.33 days for the tank to be empty.