An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the container is 35pi in^3/sec , how fast is the depth of the water dropping when the height is 5in?
let the height of water be h in, let the radius of the water surface be r in
By similar triangles, r/h = 21/15 = 7/5
so r = 7h/5
V = (1/3)pi(r^2)h
= (1/3)pi(49h^2/25)(h
= (49/75)pi(h^3)
dV/dt = (49/25)pi(h^2)dh/dt
-35pi = (49/25)pi(25)dh/dt
solve for dh/dt
-5/7
To solve this problem, we can apply related rates, which involves taking derivatives with respect to time. Let's denote the depth of the water as "h" (in inches) and the volume of water as "V" (in cubic inches).
First, let's determine the equation that relates the volume and the depth of the water in the inverted cone. The volume of a cone can be given by the formula:
V = (1/3)πr²h,
where r is the radius of the base of the cone.
Since the diameter of the inverted cone is given as 42 inches, the radius can be calculated as half of the diameter, so r = 21 inches.
Substituting the values into the formula for the volume, we have:
V = (1/3)π(21²)h.
Now, let's differentiate both sides of the equation with respect to time:
dV/dt = (1/3)π(2r)(dh/dt).
The left side, dV/dt, represents the rate at which the volume is changing, which is given as 35π in³/sec.
We are interested in finding the rate at which the depth of the water is changing, dh/dt, when the height is 5 inches. So, we need to substitute the known values into the equation and solve for dh/dt.
35π = (1/3)π(2(21))(dh/dt).
Simplifying further:
35 = 14(dh/dt).
Finally, solve for dh/dt:
dh/dt = 35/14 = 5/2 = 2.5 in/sec.
Therefore, the depth of the water is dropping at a rate of 2.5 inches per second when the height is 5 inches.