Determine the corresponding rate of increase of the water level height in the cylinder How do I go about doing this?
An ice cube of $16\,\text{cm}^3$ is melting on a mesh as shown below. Each side of the ice cube is
melting at a constant rate of $0.02\,\text{cm/s}$. Determine the corresponding rate of increase of the
water level height in the cylinder, when half of the volume of ice has melted, considering that
water expands by about $10\%$ when frozen. The cylinder has a radius of $2$ cm.
How do I start this? I'm quite bad at math so I'm not sure what formulas or rules to use.
 A: I can’t see the diagram but here is how you can get started:
First, an ice cube of initial volume 16 cm^3 has 6 sides, each melting at an equal rate of 0.02 cm/sec. This means that each face of the cube is approaching the center point of the cube at 0.02 cm/s, and therefore the ice cube is shrinking.
If S_t is the surface area of each side of the cube (in cm^2) at time “t” then the flow rate of melting ice $F_t$ is $6*0.02*S_t/1.1 $(accounting for water being 10% denser than ice).
If V_t ( in cm^3) of ice has melted (we assume equally from all sides) then the surface area at time “t” is given by:
$$ S_t = (16 -V_t)^{2/3} \implies F_t =  6*0.02*(16-V_t)^{2/3}/1.1$$
Now you just divide $F_t$ by the area of the cylinder to get the rate of rise.
A: 
For a cylinder, such as a cylinder of water held in a cylindrical glass, volume $V,$ radius $r,$ and height $h$:
$$V_\text{cylinder} = \pi r^2 h$$
Let's find the derivative with respect to $h.$ Since the formula is first order with respect to $h,$ the derivative is easy. It's just:
$$\frac {dV_\text{cylinder}}{dt} = \pi r^2 \frac{dh}{dt}$$
For a cube, volume $V,$ sidelengths $s:$
$$V_\text{cube} = s^3$$
The problem is a bit unclear on if s is decreasing at a rate of $0.02$ cm/s, or $2\cdot0.02$ cm/s. I'll solve as $0.02$ cm/s.
Take the derivative of the second equation:
$$\frac{dV_\text{cube}}{dt} = V_\text{cube}' = 3s^2 \frac {ds}{dt}$$
Use the above equation along with the rate of decrease from the problem statement ($0.02$ cm/s) to find the general equation for the change in volume of the cube:
$$\frac{dV_\text{cube}}{dt} = 3s^2 (0.02\text{ cm}/\text{s})$$
Now find the side length when half the volume of the cube has melted. When half the volume has melted, $V_\text{half} = (0.5)(16) = 8\text{ cm}^3 = s_\text{half}^3$
$$\implies s_\text{half} = \sqrt[3]{8} = 2$$
Now substitute this into the first equation we found:
$$\frac{dV_\text{cube}}{dt} = 3s^2 (0.02\text{ cm}/\text{s})$$
$$\implies \frac{dV_\text{half}}{dt} = 3(2)^2 (0.02\text{ cm}/\text{s}) = 0.24\text{ cm}^3/\text{s}$$
Now we have the rate of ice melting. multiply that by the scaling factor:
$$0.24\text{ cm}^3/\text{s of ice} = \frac 1 {1.1} 0.24\text{ cm}^3/\text{s} \text{ of water} = \frac {24}{110}\text{ cm}^3/\text{s of water}$$
Now we have the rate of volume of water appearing from the melted ice. Use the formula for the volume of a cylinder to find the rate of change of height:
$$\frac {dV_\text{cylinder}}{dt}= \frac {24}{110} \text{ cm}^3/\text{s} = \pi r^2 \frac{dh}{dt}$$
$$\implies \frac{dh}{dt} = \frac {24}{440\pi} \text{ cm}/\text{s}$$
Simplifying,
$$\frac {24}{440\pi} \text{ cm}/\text{s} = \frac {3}{55\pi} \text{ cm}/\text{s}$$
