Asked by Angel
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.
Answers
Answered by
Steve
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.
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.
Answered by
Angel
Thanks Steve for your explanation. The questions never gave how deep the water is, which was why I was a bit thrown off.
Answered by
Steve
That's unusual. Since the rise rate depends on the current depth, it's important to know.
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