Chapter 4 | Applications of Derivatives
349
Let h denote the height of the water in the funnel, r denote the radius of the water at its surface, and V denote the volume of the water. Step 2: We need to determine dh dt when h = 1 2 ft. We know that dV dt = −0.03 ft/sec. Step 3: The volume of water in the cone is V = 1 3 πr 2 h . From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, r h = 1 2 or r = h 2 . Using this fact, the equation for volume can be simplified to h 3 . Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time t , weobtain dV dt = V = 1 3 π ⎛ ⎝ h 2 ⎞ ⎠ 2 h = π 12 h 2 dh dt .
π 4
Step 5: We want to find dh dt
2 ft. Since water is leaving at the rate of 0.03ft 3 /sec, we know that
when h = 1
dV dt =−0.03ft
3 /sec. Therefore,
⎛ ⎝ 1 2
⎞ ⎠
2 dh
−0.03= π 4
dt ,
which implies
−0.03= π 16
dh dt .
It follows that
dh dt = − 0.48 π = −0.153 ft/sec.
At what rate is the height of the water changing when the height of the water is 1 4 ft?
4.4
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