A water tank is made by rotating f(x)=2^x-1 between [0,2] about the y-axis. The water tank is initially full, when a hole is opened at the bottom tip so that the water drains at a rate of 4 unit^3/second. How fast is the height of the water decreasing when the height has decreased to 1? I did the integration and I got that the volume was 46.35, but I'm not sure what to do after that.
When the height of the water is y, then the radius of the surface is x=log_2(y+1)
Thus, the volume of the water when the depth is h is
∫[0,h] πx^2 dy
Now,
∫π(log_2(y+1))^2 dy
= π(y+1)(ln^2(y+1)-2ln(y+1)+2)/ln^2(2)
That gives me
∫[0,3]π(log_2(y+1))^2 dy = 30.06
That's somewhat different from your answer.
Anyway, we have v as a function of h.
v = F(h), so dv/dt = f(h)
Then we can say
dv/dt = f(h) dh/dt
-4 = f(1) dh/dt
So, plug in your expression for f(h). In this case, it is (I assert) π(log_2(h+1))^2
To find how fast the height of the water is decreasing when the height has decreased to 1, we can use related rates.
Let's denote the height of the water as h and the volume of the water as V.
We know that the rate of change of volume with respect to time is given by dV/dt = -4, because the water is draining at a rate of 4 unit^3/second.
And we need to find dh/dt when h = 1.
From integration, you found that the volume of the water is 46.35.
So, we have V = 46.35 and we want to find dh/dt when h = 1.
The volume of the water in the tank can be calculated using the formula for the volume of a rotation:
V = ∫[0,2] (π*f(x)^2) dx
Since f(x) = 2^x - 1, we can substitute this into the volume equation:
V = ∫[0,2] (π*(2^x - 1)^2) dx
Now, you can differentiate both sides of this equation with respect to t (time):
dV/dt = d/dt (∫[0,2] (π*(2^x - 1)^2) dx)
Using the chain rule, the derivative dV/dt can also be expressed as dV/dx * dx/dt. However, dx/dt is equal to dh/dt, so we can write:
dV/dt = dV/dx * dx/dt = dV/dx * dh/dt
Now, let's find dV/dx:
To do this, we differentiate the integrand with respect to x:
dV/dx = d/dx (π*(2^x - 1)^2)
= π * 2 * (2^x - 1) * d/dx (2^x - 1)
= π * 2 * (2^x - 1) * (ln(2) * 2^x)
Now, substitute this value back into the equation:
dV/dt = π * 2 * (2^x - 1) * (ln(2) * 2^x) * dh/dt
We know dV/dt = -4 (given), V = 46.35 and h = 1 (what we're trying to find). Substitute these values into the equation:
-4 = π * 2 * (2^x - 1) * (ln(2) * 2^x) * dh/dt
Solving for dh/dt, we have:
dh/dt = -4 / (π * 2 * (2^x - 1) * (ln(2) * 2^x))
When h = 1, substitute this value into the equation:
dh/dt = -4 / (π * 2 * (2^1 - 1) * (ln(2) * 2^1))
Simplifying further:
dh/dt = - 4 / (π * 2 * 1 * ln(2) * 2)
dh/dt = -2 / (π * ln(2))
Therefore, when the height of the water has decreased to 1, the rate at which the height is decreasing is approximately -0.913 units/second.
To find the rate at which the height of the water is decreasing, we can use related rates. We have the volume of the water tank as a function of time, which is given by integrating the function f(x)=2^x-1. From the information given, we know that the volume is decreasing at a constant rate of 4 units^3/second.
Here's what you can do:
1. Differentiate the volume function with respect to time to find the derivative of the volume function. The derivative of the volume function gives us the rate of change of volume with respect to time.
2. Determine the value of x when the height of the water has decreased to 1. Recall that the height is represented by the variable x in this problem.
3. Substitute the value of x into the derivative of the volume function to find the rate at which the volume is changing with respect to time when the height is 1.
4. Finally, since we are interested in the rate at which the height of the water is decreasing, we need to find the rate at which the volume is changing with respect to time when the height is 1. Divide this rate by the rate at which the volume is changing (which is given to be 4 units^3/second) to find the rate at which the height is decreasing.
Let's calculate this step by step:
1. Differentiate the volume function with respect to time:
dV/dt = (d/dt)∫[0,2] 2^x-1 dx
Since the limits of integration are constants, we can differentiate inside the integral:
dV/dt = ∫[0,2] d/dt (2^x-1) dx
Applying the chain rule, we get:
dV/dt = ∫[0,2] ln(2) * 2^x dx
To evaluate this integral, we use the fundamental theorem of calculus:
dV/dt = ln(2) * [2^x / ln(2)] evaluated from x = 0 to x = 2
dV/dt = ln(2) * (2 - 1) = ln(2)
Therefore, the rate of change of volume with respect to time is ln(2) units^3/second.
2. When the height of the water has decreased to 1, we know that x = 1.
3. Substituting x = 1 into the derivative, we have:
dV/dt = ln(2)
4. Finally, to find the rate at which the height is decreasing, we need to divide the rate of volume change by the known rate at which the volume is changing:
Rate of height change = dV/dt / rate of volume change
Rate of height change = ln(2) / 4
Therefore, when the height has decreased to 1, the rate at which the height of the water is decreasing is ln(2) / 4 units/second.