A cylindrical jar of radius 5 cm contains water to a depth of 8 cm. the water from jar is poured at a constant rate into a hemispherical bowl. After t seconds, the depth of the water in the bowl is x cm and the volume, V of water is given as 1/3πh^2(18-h). Given that all the water is transferred in π seconds, find
I) dV/do
I I) the rate of change of x when x=8, giving your answer in terms of 1/π
Assuming you meant x instead of h,
V = 1/3πx^2(18-x)
= 6πx^2 - 1/3πx^3
dV/dx = 12πx - πx^2
the jar's volume is 400π cm^3. If it drains in π seconds, that's 400cm^3/s. That is dV/dt
dV/dt = (12π - 2πx) dx/dt
-400 = (12π-16π) dx/dt
dx/dt = 100(1/π)
To find the rate of change of volume, dV/do, we need to find the derivative of the volume V with respect to time t.
The given volume formula is V = (1/3)πh^2(18 - h).
Now, we need to express the depth h in terms of time t. We know that the water is poured from the cylindrical jar into the hemispherical bowl at a constant rate. Therefore, the volume of water poured into the bowl depends on the rate of pouring, which we'll represent as r.
The volume of water poured into the bowl is equal to the volume of water initially in the jar minus the volume of water left in the jar at time t.
The volume of water initially in the jar is the volume of a cylinder, V_cylinder = πr^2h_initial, where h_initial is the initial depth of water in the cylindrical jar.
Since the water is poured at a constant rate, the volume of water left in the jar at time t is given by V_left = πr^2(h_initial - r*t).
Now, we can express the depth h in terms of time t as h = h_initial - r*t.
Substituting this expression for h into the volume formula, we get:
V = (1/3)π(h_initial - r*t)^2(18 - (h_initial - r*t)).
Next, we differentiate the volume V with respect to time t, using the chain rule:
dV/do = dV/dt * (dt/do).
Since all the water is transferred in π seconds, dt/do = 1/π.
So, we need to find dV/dt and substitute dt/do = 1/π.
Differentiating V = (1/3)π(h_initial - r*t)^2(18 - (h_initial - r*t)) with respect to t, we get:
dV/dt = (1/3)π(2(h_initial - r*t)(-r)(18 - (h_initial - r*t)) + (h_initial - r*t)^2(-r)).
= -πr(h_initial - r*t)(18 - (h_initial - r*t)) - (1/3)π(h_initial - r*t)^2.
Now, substitute dt/do = 1/π:
dV/do = dV/dt * (dt/do) = [-πr(h_initial - r*t)(18 - (h_initial - r*t)) - (1/3)π(h_initial - r*t)^2] * (1/π).
= -r(h_initial - r*t)(18 - (h_initial - r*t)) - (1/3)(h_initial - r*t)^2.
So, the rate of change of volume, dV/do, is given by the expression -r(h_initial - r*t)(18 - (h_initial - r*t)) - (1/3)(h_initial - r*t)^2.
Now, for the second question, we need to find the rate of change of x (depth of the water in the bowl) when x = 8.
From the given information, the depth of the water in the bowl is related to the depth of the water in the cylindrical jar by x = h_initial - r*t.
When we plug in x = 8, we get:
8 = h_initial - r*t.
Solving for t, we have:
t = (h_initial - 8) / r.
Now, we can differentiate x = h_initial - r*t with respect to t to find dx/dt:
dx/dt = -r.
Since dx/dt = -r, the rate of change of x when x = 8 is simply equal to the rate at which the water is poured into the bowl, given by dx/dt = -r.