Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the rate of 2 cm^3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep?
The trick is that the rate of change of volume is the surface area times the rate of change of depth
dh/dt(pi r^2) = dV/dt = 2 cm^3/min
and r = (4/16)8 = 2 cm
Thank you!
You are welcome.
To find the rate at which the depth of the water is changing, we need to use related rates. Related rates is a method that relates the rates of change of different variables that are related to each other by an equation. In this case, we need to relate the depth of the water (h) and the radius of the water (r).
First, we need to set up the equation that relates the radius and height of the cone. We can use the similarity of triangles to find the relationship.
The larger triangle formed by the height of the cone and the slant height of the cone is similar to the smaller triangle formed by the depth of the water and the radius of the water.
Therefore, we have a proportion:
(r / h) = (R / H)
where r is the radius of the water, h is the depth of the water, R is the radius of the cone, and H is the height of the cone.
We are given that R = 4 cm and H = 16 cm.
We are asked to find the rate at which the depth of the water is changing, dh/dt, when the water in the tank is 8 cm deep.
Now, let's differentiate the equation with respect to time t:
(d(r) / dt) / h - r(dh / dt) / h^2 = (d(R) / dt) / H
We are given that (d(R) / dt) = 0, because the radius of the tank is constant.
Now we can rearrange the equation to solve for dh / dt:
(d(r) / dt) = (r/h) * (dh / dt) + r/h^2 * (dh / dt)
We know that r = 4 cm and h = 8 cm.
Substituting these values into the equation:
(d(r) / dt) = (4/8) * (dh / dt) + 4/(8^2) * (dh / dt)
Simplifying:
(d(r) / dt) = (1/2) * (dh / dt) + 1/16 * (dh / dt)
(d(r) / dt) = (9/16) * (dh / dt)
Now we can substitute the given value for (d(r) / dt):
2 = (9/16) * (dh / dt)
To find the value of dh / dt, we can solve for it:
dh / dt = (2 * 16) / 9
dh / dt = 32/9
Therefore, the rate at which the depth of the water is changing at the instant when the water in the tank is 8 cm deep is 32/9 cm/min.