A boat is pulled into a dock by a rope attached to the bow of the boat and passi

Question

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 7 m from the dock? (Round your answer to two decimal places.) m/s Read It Watch Master It Talk to a Tutor 10. 1 points SCalcET8 3.9.025. My Notes Ask Your Tea Water is leaking out of an inverted conical tank at a rate of 8,500 cm /min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.) cm3/min Watch

Explanation / Answer

Solution:

Let x be the length of rope.
Let d be distance of boat from the dock.

Now the bow of the boat (bottom of rope) the pulley (top of rope) and the dock (where it is level with bow of boat) form a right triangle.
height of pulley above level of boat's bow = 1m

Using Pythagorean Theorem, we get:

x2 = d2 + 1

Differentiating both sides with respect to t (time in seconds), we get:
2x dx/dt = 2d dd/dt + 0
x dx/dt = d dd/dt .............(*)

Rope is pulled in at rate of 1 m/s;   dx/dt = 1
Boat is 7m from the dock;   d = 7

x2 = d2 + 1 => x = ?(49 + 1) = ?50

Substituting these values into equation (*), we get

?50 * 1 = 7 dd/dt
dd/dt = (?50) / 7
dd/dt =1.01015 ? 1.01

Boat is approaching dock at a rate of about 1.01 m/s

Solution:(10)

Let all linear measures be in centimeters and time be measured in minutes.

"Water is leaking out of an inverted conical tank at a rate of 8,500 cm^3/min at the same time that water is being pumped into the tank at a constant rate."

dV/dt = R - 8500

Where V is the volume of the water in the tank at time t and R is the rate at which water is being pumped into the tank. R is the quantity we are asked to find.

Using the volume of a cone, we know the volume of the water in the tank may be given by:

V = (1/3) ? r2 h

where r is the radius of the surface of the water, and h is the depth of the water. We know that at any given time or volume of water, the ratio of the radius of the water at the surface to its depth will remain constant, and in fact will be in the same proportions as the tank itself.

Because we are given information regarding the time rate of change of the depth and the depth itself, we need to place the radius with a function of the depth. Hence, we may use:

r/h = 2/6 = 1/3 ==> r = h/3

And so the volume as a function of h is:

V = (1/3) ? (h/3)2 h = ?h3/27

Now, differentiating with respect to t, we obtain:

dV/dt = ?/9 * h2 * dh/dt

Equating the two expressions for dV/dt, we find:

?/9 * h2 * dh/dt = R - 8500

R = ?/9 * h2 * dh/dt + 8500

Now, using the given data (making sure all of our units match):

h = 200cm, dh/dt = 20cm/min

we have:

R = ?/9 * (200 cm)2 * (20 cm/min) + 8500 cm3/min

R = (500/9) (1600? + 153) cm3/min

Rounded to the nearest integer this is:

R ? 287753 cm3/min

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