An inverted cylindrical cone, \(20 \mathrm{ft}\) deep and \(10 \mathrm{ft}\) across at the top, is being filled wi water at a rate of \(10 \mathrm{ft}^3 / \mathrm{min}\). At what rate is the water rising in the tank when the the water is:
How long will the tank take to fill when starting at empty?
To find dh/dt
\(\mathrm{We}\) know dV \(/ \mathrm{dt}=10 \mathrm{ft}^3 / \mathrm{min}\). Volume of a right circular cone \(=(1 / 3)(\pi) r^2 \mathrm{~h}\)
We need to get the volume in terms of one variable, \(h\), since we’re looking for \(\mathrm{dh} / \mathrm{dt}\)
The diameter at the top of the cone is \(10^{\prime}\) so the radius is \(5^{\prime}\) We can find \(\mathbf{h}\) by similar ratios using the height and radius of the water in the tank
\[ 5 / 20=r / h \Rightarrow 1 / 4=r / h \Rightarrow r=h / 4 \]
Substituting into the volume we get \(V=(1 / 3)(\pi)(h / 4)^2 h=(1 / 3)(\pi) h^3 /(16)\) \(\mathrm{dV} / \mathrm{dt}=\left[(1 / 3)(3)(\pi)(\mathrm{h})^2 / 16\right](\mathrm{dh} / \mathrm{dt})=\left[(\pi)(\mathrm{h})^2 / 16\right](\mathrm{dh} / \mathrm{dt})\)
Solving for \((\mathrm{dh} / \mathrm{dt})\) we get \(\mathrm{dh} / \mathrm{dt}=(\mathrm{dV} / \mathrm{dt})(16) /\left[\left((\pi)(\mathrm{h})^2\right]\right.\)
We know that the change in volume is \(\mathrm{dV} / \mathrm{dt}=10 \mathrm{ft}^3 / \mathrm{min} \mathrm{dh} / \mathrm{dt}=\left(10 \mathrm{ft}^3 / \mathrm{min}\right)(16\) \(\left[\left((\pi)(h)^2\right])\right.\)
When \(\mathrm{h}=1 \mathrm{ft}, \mathrm{dh} / \mathrm{dt}=\left(10 \mathrm{ft}^3 / \mathrm{min}\right)(16) /\left[\left((\pi)(1 \mathrm{ft})^2\right] \approx 50.93 \mathrm{ft} / \mathrm{min}\right.\)
When \(\mathrm{h}=10 \mathrm{ft}, \mathrm{dh} / \mathrm{dt}=\left(10 \mathrm{ft}^3 / \mathrm{min}\right)(16) /\left[\left((\pi)(10 \mathrm{ft})^2\right]=\left(160 \mathrm{ft}^3\right) /\left(100 \mathrm{ft}^2 \pi\right) \approx\right.\) \(0.51 \mathrm{ft} / \mathrm{min}\)
When \(\mathrm{h}=51 \mathrm{ft}, \mathrm{dh} / \mathrm{dt}=\left(10 \mathrm{ft}^3 / \mathrm{min}\right)(16) /\left[\left((\pi)(51 \mathrm{ft})^2\right]=\left(160 \mathrm{ft}^3\right) /\left(2601 \mathrm{ft}^2 \pi\right) \approx\right.\) \(0.02 \mathrm{ft} / \mathrm{min}\)
The total volume of the inverted cylindrical cone when it’s full ( \(h=20\) feet) is: \[ V=\frac{\pi \mathrm{h}^3}{48} \quad \frac{\pi(20)^3}{48}=\frac{8000 \pi}{48}=\frac{500 \pi}{3} \mathrm{ft}^3 \]
To find the time it takes to fill the tank: \[ \text { Time to fill }=\frac{\frac{500 \pi}{3} \mathrm{ft}^3}{10 \mathrm{ft}^3 / \mathrm{min}}=\frac{500 \pi}{30} \approx 52.36 \text { minutes. } \]
# Given values
dV_dt <- 10 # ft^3/min
h <- 20 # feet
# Function to calculate dh/dt at a given height h
calculate_dh_dt <- function(h) {
dh_dt <- (dV_dt * 16) / (pi * h^2)
return(dh_dt)
}
# Calculate dh/dt at different depths
depths <- c(1, 10, 51) # Depths of the water in feet
dh_dt_values <- sapply(depths, calculate_dh_dt)
# Output results for dh/dt at different depths
for (i in seq_along(depths)) {
cat("Rate of water rising when depth is", depths[i], "feet:", dh_dt_values[i], "ft/min\n")
}## Rate of water rising when depth is 1 feet: 50.92958 ft/min
## Rate of water rising when depth is 10 feet: 0.5092958 ft/min
## Rate of water rising when depth is 51 feet: 0.01958077 ft/min# Given values
total_volume <- (pi * h^3) / 48 # ft^3
rate_of_filling <- 10 # ft^3/min
# Calculate time to fill the tank
time_to_fill <- total_volume / rate_of_filling
# Output result
cat("Time to fill the tank starting from empty:", time_to_fill, "minutes\n")## Time to fill the tank starting from empty: 52.35988 minutes