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He's kind of right, however the answer s flawed because a cone's volume formula is (((pi)r^2)h))/3. He solved the problem for a cylinder. Do the work again put with the afforementioned formula from the top. I'm not gonna give you the answer put you should be able to follow this guy's work with the new formula.

Relations:

From similar triangles, h=4r [Since a full pool has h=4r, so does any partially full pool.]

V= (pi/3) r^2h

Diff:

dv/dt= (pi/3)*(h/4)^2 h

dv/dt= pi/16h^2 dh/dt

Answer:

dh/dt= 12/pi

Part 2 How fast the radius changing.

We have 4r=h

v=1/3pir^2 h

v=(1/3)pi r^2 (4r)

v=(1/3) pi4 =r^3

dv/dt =(4/3) pi 3r^2 dr/dt h=4r so at h=2 r=1/2

so dv/dt =(4/3) pi 3/4 dr /dt

3=pi dr /dt

dr/dt =3/pi meter/minute

Or take h=4r

differentiate dh/dt=4 dr/dt we know that dh/dt =12/pi

12/pi=4dr/dt

dr/dt=3/pi meter/minute

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Calculus

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.**Education/Credentials**

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