Water is being poured into a conical tank that has a height of 4.5m and an upper radius of 2.7m. If the height of the water is increasing at a rate of 11.5cm/s at the moment when the water's height is 1.24m, at what rate, in m^3/min, is the water being added.
I understand that V = π*r^2*h/3, just don't know how to set up the derivative's for a question like this and solve.
Two things:
- we would want to get rid of one of the variables on the right side
- since we are given dh/dt, and we are looking for dV/dt, we have to differentiate with respect to t
Given: dh/dt = 11.5 m/s when h = 1.24
Find: dV/dt at that moment.
At a given time of t seconds, let the height of the water level be h,
let the radius of the water level be r
h/r = 4.5/2.7
h = (5/3)r or r = (3/5)h
then V = (1/3)π r^2 h = (1/3)π (9/25)h^2 (h) = (9/75)π h^3
now in dV/dt = (9/25)π h^2 dh/dt
put in our given:
dV/dt = (9/25)π(1.24)^2 (11.5) m^3/s = appr 20 m^3/s
(I could have replaced the h and have my V = ... in terms of r, but that would
require extra steps to find dr/dt for that moment)
Good work, but dh/dt is cm/s, not m/s.
or, using the product rule,
v = π/3 r^2 h
dv/dt = π/3 (2rh dr/dt + r^2 dh/dt)
r = 3/5 h, so dr/dt = 3/5 dh/dt
3/5 * 1.24 = 0.744
dv/dt = π/3 (2*0.744*1.24*(3/5 * 11.5/100) + 0.744^2 * 11.5/100) = 0.199983
or, 0.20 m^3/s
Another way of looking at it is to first order
dh/dt = dV/dt / surface area
Ah, I see you're trying to pour some brainpower into this conical tank problem. Let's break it down with a splash of humor!
We know that the volume (V) of a cone is given by the formula V = (π * r^2 * h) / 3, where r is the radius and h is the height.
Now, we want to find the rate at which the water is being added, so we're looking for dV/dt. To do that, we need to take the derivative of the volume function with respect to time, which involves using the chain rule. Hang on, folks – it's about to get derivative-y!
Since the radius (r) isn't changing, let's treat it as a constant (if it starts changing, we may have more on our hands than a math problem).
Differentiating the volume equation gives us dV/dt = (π/3)(2r * dh/dt) + (π * r^2 * dh/dt). Notice how we multiplied the derivative of the height (dh/dt) by 2r in the first term – that's the chain rule at play!
Now, let's plug in the values we know. The radius (r) is 2.7m, and the height is 1.24m. We're also given that dh/dt = 11.5cm/s. Remember to convert centimeters to meters before continuing, or we'll be drowning in units!
Plugging everything into the equation, we get: dV/dt = (π/3)(2 * 2.7 * [11.5/100]) + (π * 2.7^2 * [11.5/100]).
Simplifying that expression should give you the rate at which the water is being added – voila!
Remember, keep your math skills afloat, and don't forget to have some fun while solving these problems. Good luck!
To solve this problem, we need to find the rate at which the volume of the water in the tank is changing over time. We are given that the height of the water is increasing at a rate of 11.5 cm/s. We want to find at what rate the volume is changing, in m³/min.
First, let's differentiate the volume formula with respect to time (t):
V = (π * r² * h) / 3
dV/dt = (π/3) * [(2r * dr/dt * h) + (r² * dh/dt)]
In this equation, dV/dt represents the rate of change of volume with respect to time (which is what we want to find), dr/dt represents the rate of change of the radius with respect to time (which is what we don't have), and dh/dt represents the rate of change of the height with respect to time (which is given as 11.5 cm/s).
Now, we need to find dr/dt. Since the conical tank is being filled with water, the radius of the water surface will be changing as well.
Using similar triangles, we can determine that:
h / r = H / R
Where h is the height of the water, r is the radius of the water surface, H is the total height of the cone, and R is the total radius of the cone.
Substituting the given values, we have:
1.24 m / r = 4.5 m / 2.7 m
r = (2.7 m * 1.24 m) / 4.5 m
r ≈ 0.7456 m
Now, we can differentiate the equation with respect to time:
dh/dt = 11.5 cm/s
dr/dt = (dh/dt * r) / h
Substituting the known values, we get:
dr/dt = (0.115 m/s * 0.7456 m) / 1.24 m
dr/dt ≈ 0.0693 m/s
Now we have dr/dt, we can substitute all the values back into the first equation to find dV/dt:
dV/dt = (π/3) * [(2 * 0.7456 m * 0.0693 m/s * 1.24 m) + (0.7456 m² * 0.115 m/s)]
Evaluating this expression will give the rate at which the volume is changing in m³/s. To find the rate in m³/min, multiply the result by 60 since there are 60 seconds in a minute.