A 3-dimensional structure is obtained from rotating the parabola y=x^2 about the y-axis. Each second, 2π units^3 of water is being poured into the structure from the top. When 8π units^3 of water has been poured in the structure, the instantaneous change in water height level is ab, where a and b are coprime positive integers. What is the value of a+b?

Question

A 3-dimensional structure is obtained from rotating the parabola y=x^2 about the y-axis. Each second, 2π units^3 of water is being poured into the structure from the top. When 8π units^3 of water has been poured in the structure, the instantaneous change in water height level is ab, where a and b are coprime positive integers. What is the value of a+b?

Answer 1

the volume of water when y=k is

∫[0,k] πx^2 dy
= ∫[0,k] πy dy
= π/2 k^2

So, at depth y,

v = π/2 y^2
when v=8π, y=4

dv/dt = πy dy/dt
2π = π(4) dy/dt
dy/dt = 1/2

You sure you want ab, and not a/b?
a+b = 1+2 = 3

Answer 2

Yes sorry, it's a/b.

Answer 3

To find the instantaneous change in water height level, we need to find the derivative of the volume function with respect to time.

Let's start by finding the volume function of the 3-dimensional structure. Since the parabola y = x^2 is being rotated about the y-axis, the cross-sectional area at any height y is given by A = πx^2, where x represents the radius of the circular cross-section at height y.

In this case, x = √y (since the parabola equation is y = x^2), so the cross-sectional area is A = π(√y)^2 = πy. Now, to find the volume, we integrate the cross-sectional area from y = 0 to y = h (height of the water level):

V(h) = ∫[0,h] A dy = ∫[0,h] πy dy = π∫[0,h] y dy = π/2 * y^2 |[0,h] = π/2 * h^2

Now, we have the volume V as a function of the water level height h: V(h) = π/2 * h^2.

Next, we need to find the expression for the derivative of the volume function with respect to time (dV/dt). Since the water is being poured into the structure at a constant rate of 2π units^3/s, we have dV/dt = 2π.

Now, we need to find the value of h when the volume V(h) is equal to 8π units^3. Substituting this into the volume function, we get:

8π = π/2 * h^2
16 = h^2
h = 4

Therefore, when 8π units^3 of water has been poured in the structure, the water level height is h = 4.

Finally, to find the instantaneous change in water level, we take the derivative of the volume function with respect to time:

dV/dt = 2π

So, the instantaneous change in water height level is 2π units/s (ab = 2).

Therefore, the value of a+b is 2.