A storage tank used to hold sand is leaking. The sand forms a conical pile whose height is twice the radius of the base. The radius of the pile increases at the rate of 2 inches per minute. Find the rate of change of volume when the radius is 5 inches.
since h = 2r
v = 1/3 pi r^2 (2r) = 2/3 pi r^3
dv/dt = 2 pi r^2 dr/dt = 2*pi*25*2 = 314.16
This is an unrealistic problem. If sand is running out, it's likely that the flow gradually slows down. But this setup has the radius of the pile steadily growing, meaning that the volume increases at an increasing rate. They must be piling more sand into the tank, so it comes out faster and faster to keep the radius steadily growing.
Find the absolute max and min of f(x,y)=4xy^2-x^2y^2-xy^2 on the region D
where D is bounded by x=0, y=0 and y=-x+6
To find the rate of change of volume, we need to use calculus.
Let's start by finding an equation that relates the height and radius of the conical pile.
We are given that the height of the conical pile is twice the radius of the base. Let's call the radius "r" and the height "h".
Since the height is twice the radius, we can write h = 2r.
We also know that the radius of the pile is increasing at a rate of 2 inches per minute. So, dr/dt = 2 inches/min.
To find the rate of change of volume, we need to find an equation that relates the volume and the radius.
The volume of a cone is given by the formula: V = (1/3)πr^2h.
Since we know that h = 2r, we can substitute this into the volume equation:
V = (1/3)πr^2(2r) = (2/3)πr^3.
Now, we can differentiate both sides of the equation with respect to time (t) to find the rate of change of volume:
dV/dt = d/dt [(2/3)πr^3].
Using the power rule of differentiation, the derivative of r^3 is 3r^2.
dV/dt = (2/3)π * 3r^2 * (dr/dt)
Now, substitute the given value for dr/dt (2 inches/min) and the given radius (5 inches) into the equation:
dV/dt = (2/3)π * 3(5^2) * 2 = (2/3)π * 3 * 25 * 2 = 300π.
So, the rate of change of volume when the radius is 5 inches is 300π cubic inches per minute.
To find the rate of change of volume, we need to determine the expressions for the radius, height, and volume of the sand pile.
Let's denote:
- r as the radius of the base of the cone
- h as the height of the cone
Given that the height is twice the radius of the base, we can express this as h = 2r.
We know that the radius of the pile is increasing at a rate of 2 inches per minute. Thus, the rate of change of the radius (dr/dt) is constant and given by 2 inches/minute.
The formula for the volume of a cone is V = (1/3)πr²h.
Since we want to find the rate of change of volume with respect to time (dV/dt), we need to express the volume in terms of r and find the derivative.
Substituting the given relationship between h and r into the volume formula, we have V = (1/3)πr²(2r) = (2/3)πr³.
Now, differentiate the volume formula with respect to time (t):
dV/dt = d/dt [(2/3)πr³]
To find this derivative, we use the power rule. The derivative of r³ is 3r² multiplied by the derivative of r with respect to t, which is dr/dt:
dV/dt = (2/3)π * 3r² * dr/dt
Simplifying further, we have:
dV/dt = 2πr² * dr/dt
Now, we can substitute the given value for the radius r when the rate of change of radius (dr/dt) is 2 inches per minute and find the rate of change of volume when r = 5 inches.
Using r = 5 inches and dr/dt = 2 inches/minute in the equation, we have:
dV/dt = 2π(5)² * 2
Simplifying further:
dV/dt = 2π(25) * 2
dV/dt = 100π cubic inches/minute
Therefore, the rate of change of volume when the radius is 5 inches is 100π cubic inches per minute.