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 a/b, where a and b are coprime positive integers. What is the value of a+b?
when the water depth is a, the volume is
v = ∫[0,a] π x^2 dy
= ∫[0,a] πy dy
= π/2 a^2
dv/dt = πa da/dt
when v=8π, a=4, so since dv/dt=2π,
2π = π(4) da/dt
da/dt = 1/2
1+2=3
To find the value of "a + b," you need to determine the value of "a" and "b" first. Let's break down the problem step by step.
1. Visualize the 3-dimensional structure: When the parabola y = x^2 is rotated around the y-axis, it forms a symmetrical shape known as a "solid of revolution." In this case, it creates a rotational solid shaped like a bowl or a vase.
2. Determine the volume of the solid: To find the volume of the solid, you can use the formula for the volume of a solid of revolution, which is given by:
V = ∫[a,b] A(x)dx
In this case, the limits of integration can be determined by finding the x-values where the parabola intersects the y-axis. Since the equation is y = x^2, when y = 0, x = 0. So the limit of integration is from x = 0 to some positive value x = b.
Therefore, the volume of the solid is:
V = ∫[0,b] A(x)dx = ∫[0,b] πy^2dx = π∫[0,b] x^4dx
Integrating with respect to x gives:
V = π[x^5/5] from 0 to b = π(b^5/5)
3. Calculate the rate of change of the water level: We know that 2π units^3 of water is being poured into the solid each second. Therefore, the change in volume with respect to time is:
dV/dt = 2π
Since V = π(b^5/5), we can differentiate both sides of the equation with respect to time (t) using implicit differentiation:
dV/dt = d/dt (π(b^5/5))
dV/dt = (π/5)(5b^4)(db/dt)
Simplifying, we get:
2π = (π/5)(5b^4)(db/dt)
Canceling out π and 5, we are left with:
2 = b^4 * (db/dt)
4. Find the instantaneous change in water height level: The instantaneous change in water height level is the rate at which the water level is changing with respect to time. In this case, it is represented by db/dt. From the equation obtained in step 3, we can solve for db/dt:
2 = b^4 * (db/dt)
db/dt = 2 / b^4
Note that when 8π units^3 of water has been poured, the volume V is equal to 8π. Therefore:
8π = π(b^5/5)
Solving for b, we get:
b^5/5 = 8
b^5 = 40
b = (40)^(1/5) ← Taking the fifth root of both sides
5. Calculate the value of a + b: Now that we have the value of "b," we can substitute it back into the equation for db/dt:
db/dt = 2 / b^4
db/dt = 2 / (40^(4/5))
To simplify the expression further, we can rewrite 40^(4/5) as (2^3 * 5)^(4/5) = 2^(12/5) * 5^(4/5).
db/dt = 2 / (2^(12/5) * 5^(4/5))
Simplifying, we get:
db/dt = 2^(3/5) * 5^(-1/5)
Therefore, the value of "a" is 2^(3/5) and "b" is 5^(-1/5). Since "a" and "b" are coprime positive integers, the value of a + b is:
a + b = 2^(3/5) + 5^(-1/5)