A magic tank starts out 4 inches in length, 4 inches wide, and 12 inches deep. As the tank is filled, it magically expands proportionally keeping it rectangular form just big enough to hold the liquid. If the liquid is being pumped into the tank at 8 cubic inches per minute, what is the rate at which the height of the tank is changing with respect to time when the height is 12 inches?
To find the rate at which the height of the tank is changing with respect to time, we need to use the concept of related rates from calculus.
Let's denote the length of the tank as L, the width as W, the height as H, and the rate at which the height is changing with respect to time as dH/dt.
Given:
Initial length (L) = 4 inches
Initial width (W) = 4 inches
Initial height (H) = 12 inches
We are given that the tank is being filled at a rate of 8 cubic inches per minute. This information implies that the volume of the tank is increasing at a rate of 8 cubic inches per minute. So, we have:
dV/dt = 8 cubic inches/min,
where V represents the volume of the tank.
The volume of a rectangular tank is given by:
V = LWH,
where L, W, and H are the length, width, and height of the tank, respectively.
Since the tank retains its rectangular shape, we can express the volume using only one variable. As the height increases, the length and width will also expand proportionally to keep the shape rectangular. Let's denote the expansion factor as k.
L = kH,
W = kH.
Substituting these expressions for L and W into the volume formula, we have:
V = (kH)(kH)H = k^2H^3.
Differentiating both sides of this equation with respect to time (t), we get:
dV/dt = (2k^2H^2)(dH/dt).
Now, we can substitute the given values into the equation to solve for dH/dt:
dV/dt = 8 cubic inches/min,
H = 12 inches.
Plugging these values into the equation, we have:
8 = (2k^2(12)^2)(dH/dt).
Simplifying, we get:
8 = 288k^2(dH/dt).
Now, solving for dH/dt, we divide both sides by 288k^2:
dH/dt = 8 / (288k^2).
To find the value of k, we can use the ratio of the change in height to the initial height:
ΔH / H = (dH/dt) * Δt,
where ΔH is the change in height, H is the initial height, (dH/dt) is the rate of change, and Δt is the change in time.
In our case, ΔH is the difference between the initial height and the current height:
ΔH = H - H0 = 12 - 12 = 0.
Plugging in these values, we have:
0 / 12 = (dH/dt) * Δt.
Since the tank is being filled, the height is increasing and the rate of change, dH/dt, is positive. Therefore, we can divide both sides by Δt:
0 / 12Δt = dH/dt.
Simplifying, we get:
dH/dt = 0.
Now, plugging the value of dH/dt and the given values into the equation to find k:
0 = 8 / (288k^2).
Solving for k, we get:
k^2 = 36,
k = √36 = 6.
Now, we can substitute the value of k into the equation for dH/dt:
dH/dt = 8 / (288(6)^2).
Simplifying further, we get:
dH/dt ≈ 0.0056 inches per minute.
Therefore, the rate at which the height of the tank is changing with respect to time when the height is 12 inches is approximately 0.0056 inches per minute.