a) the dh/dt when h=2cm.Originally Posted byjacs

V = (pi/3)(h^2)[3r -h] ---the r is not changing, is a constant.

V = (pi/3)[3*10*h^2 -h^3]

V = (pi/3)[30h^2 -h^3]

Differentiate both sides with respect to time t,

dV/dt = (pi/3)[60h*dh/dt -(3h^2)*dh/dt]

Since dV/dt is 6cc/sec, and h=2cm at the time in question, then,

6 = (pi/3)[60*2*dh/dt -3*2^2 *dh/dt]

6 = (pi/3)[120 -12]dh/dt

6 = (108pi/3)dh/dt

dh/dt = 6 /(108pi/3) = 18/(108pi) = 1/(6pi) cm/sec ---------answer.

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b.1) The radius of the water surface when depth of water is h.

Imagine, or draw the figure on paper.

It is a semicircle, with the water level being below the diameter. Call the radius of the water level as R, to differentiate or make it different from the radius r of the semicircle. Draw a radius r to one end of the water level. A right triangle is formed, with these:

---hypotenuse = r

---vertical leg = r-h

---horizontal leg = R

By Pythagorean Theorem,

r^2 = (r-h)^2 +R^2

R^2 = r^2 -(r-h)^2

The righthand side is in the form a^2 -b^2 which is (a+b)(a-b),so,

R^2 = (r +(r-h))(r -(r-h))

R^2 = (2r -h)(h)

R^2 = 2rh -h^2 ---------------***

R = sqrt(2rh -h^2) --------------------------answer.

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b.2) The dA/dt of the water surface when h=2cm.

A = pi(R^2)

A = pi(2rh -h^2) ----where r is 10cm constantly.

A = pi(20h -h^2)

Differentiate both sides with respect to time t,

dA/dt = pi(20*dh/dt -2h*dh/dt)

Since h=2cm, and dh/dt = 1/(6pi) cm/sec, at the time in question, then,

dA/dt = pi[20*1/(6pi) -2*2*1/(6pi)]

dA/dt = pi[20/(6pi) -4/(6pi)]

dA/dt = pi[16/(6pi)]

dA/dt = 8/3 sq.cm/sec ------------answer.