water drips into an upside down cone, whose diameter at the base is 10 cm , and whose height is 15 cm. If the water is dripping int a rate of 2cm^3 per minute, how fast is the height rising.
Thanks Steve for your explanation. The questions never gave how deep the water is, which was why I was a bit thrown off.
That's unusual. Since the rise rate depends on the current depth, it's important to know.
using similar triangles, it is clear that r = h/3
So, with water at depth h, the volume of water
v = 1/3 πr^2h = 1/3 π(h/3)^2*h = π/27 h^3
dv/dt = π/9 h^2 dh/dt
You don't say how deep the water is, but when you figure it out, just plug in your numbers and solve for dh/dt.
To find how fast the height is rising, we need to find the rate of change of the height with respect to time. Let's denote the height of the cone as h(t), where t is the time.
Given that the water is dripping at a constant rate of 2 cm³ per minute, we can say that the volume of water in the cone, V(t), is increasing at a rate of 2 cm³/min.
The volume of a cone can be calculated using the formula:
V = (1/3)πr²h,
where r is the radius of the base and h is the height of the cone.
In this case, the radius of the base is half of the diameter, so it is 10 cm / 2 = 5 cm.
The volume of the cone can be expressed in terms of the height as:
V(h) = (1/3)π(5²)h.
Now, let's differentiate both sides of the equation with respect to time:
dV/dt = d((1/3)π(5²)h)/dt.
The left side represents the rate of change of volume, which is 2 cm³/min.
2 = (1/3)π(5²)*(dh/dt),
Simplifying further:
2 = (25/3)π*(dh/dt).
Finally, we can solve for dh/dt:
dh/dt = 2 / ((25/3)π).
Calculating this, we find that the rate at which the height is increasing is approximately 0.076 cm/min.