Suppose that sand is collecting in the shape of a cone in such a way that the base radius of the cone is always one-third of its height. If 3cm^3/min is the rate at which the sand is being added to the cone, how fast is the height of the cone changing when it is 7cm tall?
v = 1/3 π r^2 h ... r = h / 3
v = 1/3 π (h / 3)^2 h = π/27 h^3
dv/dt = π/9 h^2 dh/dt
3 cm^3/min = π/9 * 49 cm^2 * dh/dt
To solve this problem, we need to use related rates and the given information about the cone. Let's define the variables:
Let h represent the height of the cone.
Let r represent the base radius of the cone.
The problem states that the base radius is always one-third of the height:
r = (1/3)h
We are given that the sand is being added to the cone at a rate of 3 cm³/min. This means we need to find the rate at which the height of the cone is changing, or dh/dt, when the height is 7 cm.
To find the rate of change of the height with respect to time (dh/dt), we need to express h and r in terms of t (time) and differentiate both sides of the equation with respect to t.
First, let's differentiate the equation r = (1/3)h with respect to t:
dr/dt = (1/3)(dh/dt)
Now, let's express r in terms of h to eliminate one variable:
Using the equation r = (1/3)h, we can substitute (1/3)h for r in the formula for the volume of a cone:
V = (1/3)πr²h
Substituting (1/3)h for r:
V = (1/3)π((1/3)h)²h
V = (1/27)πh³
We know that the volume of the cone is changing at a rate of 3 cm³/min, so we have:
dV/dt = 3
Next, differentiate the volume equation with respect to t to relate the rates of change:
dV/dt = (1/27)π(3h²)(dh/dt)
Substituting dV/dt = 3 and simplifying the equation:
3 = (1/9)πh²(dh/dt)
Now, solve for dh/dt:
3 / ((1/9)πh²) = dh/dt
(27/πh²) = dh/dt
Finally, substitute h = 7 cm into the equation to find the rate at which the height is changing when the height is 7 cm:
dh/dt = (27/π(7)²) = 27/49π
So, the height of the cone is changing at a rate of 27/49π cm/min when it is 7 cm tall.