A right circular cylinder through a constant volume is diminish in elevation at a price of 0.2 in/sec. In ~ the minute that the elevation is 4 inches and the radius is 3 inches, what is the rate at i beg your pardon the radius is decreasing?

After the test ns still don"t know the really answer, yet I am quite curious around what the answer should have actually been.

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What ns tried to be the following.

Givens:

$V=(\pi)r^2(h)$ $V=(\pi)(3)^2(4)$ $V=36 \pi$

$r=3$ $h=4$

$dh/dt=-0.2$ in /sec

$dr/dt=?$

Work:

$V=(\pi)r^2(h)$ $36\pi=(\pi)r^2(dh/dt)+h(2\pi(r)(dr/dt))$ $36\pi=(\pi)(3)^2(-0.2)+4(2\pi(3)(dr/dt))$ $36\pi=9(\pi)(-0.2)+24(\pi)(4dr/dt)$ $36\pi=-1.8(\pi)+24(\pi)(4dr/dt)$ $37.8\pi=24(\pi)(4dr/dt)$ $37.8\pi/24(\pi)=(24(\pi)(4dr/dt))/24(\pi)$ $1.6 \pi=4(dr/dt)$ $dr/dr=0.4\pi$ in/sec

Could you tell me what i did was correct or did ns make any mistakes? All help is considerably appreciated.

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edited Mar 15 "15 at 1:22

Pedro

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asked Mar 15 "15 at 0:43

Adam SilversteinAdam Silverstein

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Not quite. Friend missed a vital word in the problem. The very first sentence speak you the cylinder is to decrease in height, however with a

**constant**volume. If miscellaneous is constant, then it is no changing. If that is not changing, its rate of change (i.e. Derivative) is zero.

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But wait, how have the right to the cylinder be decreasing in height, yet remain consistent in volume? This indicates that the radius need to be increasing, to counter the shed volume native a to decrease height.

You are correct in separating $V=\pi r^2 h$ v respect to time. It"s simply implicit differentiation. So we have$$\fracdVdt = \pi r^2 \fracdhdt + 2\pi r h \fracdrdt.$$

However, in ~ this suggest we must recognize the the problem just told us the volume is constant! If other is constant, it doesn"t change, so its derivative is zero. The is,

$$0 = \pi r^2 \fracdhdt + 2\pi r h \fracdrdt.$$

Plugging in the provided rate $dh/dt$, and evaluating at $r=3$ inches, and $h=4$ inches, we have

$$0 = \pi (3 \text in)^2 \left(\frac-1 \text in5 \text sec\right) + 2\pi (3 \text in) (4 \text in) \fracdrdt$$

$$0 = -\frac95\pi\ \frac\textin^3\textsec + 24\pi \ \textin^2\fracdrdt.$$$$\frac95\pi\ \frac\textin^3\textsec = 24\pi \textin^2\fracdrdt$$

Now we divide by $24\pi\ \textin^2$. An alert how i deliberately involve the units in my calculations. It"s not necessary, yet it renders our work simpler to check. Because that example, if the end result is no in the units we expect, then we understand we do an error in ours algebra.$$\frac9120\frac\textin\textsec = \fracdrdt.$$