Water is being poured into an inverted cone(vertex down) of radius 4 in and height 10 in at a rate of 3 in^3/sec. Find the rate at which the height of the water is increasing when the water is 5 in high.
at water depth y, the radius of the surface is 2y/5, so y = 5r/2
v = 1/3 pi r^2 h = 1/3 pi (25y^2/4)(y) = 25pi/12 y^3
dv/dt = 25pi/12 * 3y^2 dy/dt = 25pi/4 y^2 dy/dt
3 = 25pi/4 * 25 dy/dt
dy/dt = 12 / 625pi
Seems kinda slow - better double check my algebra
Ahh. I see - I did the wrong substitution
r = 2y/5
Make that change and follow it through
To find the rate at which the height of the water is increasing, we need to use related rates, specifically the chain rule from calculus.
Let's denote the height of the water in the cone as h (measured in inches) and the radius of the water surface as r (measured in inches). We are given that the radius of the cone is always 4 inches, and the height of the cone is always 10 inches.
We want to find the rate at which the height of the water, dh/dt, is changing when the water is 5 inches high, which means h = 5.
First, we need to find an equation relating the height and the radius of the water surface in the cone. Since the cone is inverted with its vertex down, similar triangles can be used to establish a relationship between h and r.
The smaller cone formed by the water has a similar shape to the original cone. Thus, we can write the equation for the similar triangles as:
r/h = 4/10 or r = 2h/5
Differentiate both sides of this equation with respect to t (time) to get the rate of change of r with respect to t:
dr/dt = (2/5)(dh/dt)
Now, we know that the rate of change of volume (dV/dt) is given as 3 in^3/sec. The volume of a cone is given by V = (1/3)πr^2h. Substituting the equation for r obtained earlier, we can express the volume as a function of h:
V = (1/3)π(2h/5)^2h = (4/75)πh^3
Take the derivative of V with respect to t (time) to get the rate of change of volume:
dV/dt = (4/75)π(3h^2)(dh/dt)
Now plug in the known rate of change of volume (dV/dt = 3 in^3/sec) and the given height (h = 5) to solve for the rate of change of the height of water (dh/dt):
3 = (4/75)π(3(5)^2)(dh/dt)
Simplify and solve for dh/dt:
dh/dt = (3*75)/(4π(3(5)^2))
Evaluate this expression to find the rate at which the height of the water is increasing when the water level is 5 inches high.
Note: Don't forget to convert the units to simplify the expression and get the final answer.