![]() We know that water is flowing into the tank at a rate of 3.The large cylinder is the tank, and the small cylinder is the water in the tank. In order to do this, we should consider that this example essentially has two cylinders in it. First let’s think about what was directly given. This question didn’t give us a lot of information but we can figure out a little extra. To find this, we need the height of the cylinder to be in the equation. The question is asking us to find the rate of change of the height of one of the cylinders. ![]() Remember in the last section I said we may want to create an equation that didn’t contain the height of either cylinder because we have no information about them? Well now we know that’s not really an option. Once we take the derivative of our equation, that will introduce the variable representing “how fast things are changing.” Therefore, we know that we will need the height of the water to be in our equation. The question is asking us to find how fast the height of the water is increasing. To do this, we need to sort out what information we know and what we are looking for. Now that we have a drawing of the situation being described, we need to come up with our equation. We can also see that the water level is rising, but we don’t know the height of the water at this instant. The height of the tank is unknown, but we know the radius is 5 m. Looking at the above drawing, you can see that water is being poured into a cylindrical tank at a rate of 3. This is a relatively simple situation being described, so we can go ahead and draw it. Draw a sketchĪs with any related rates problem, the first thing we need to do is draw the situation being described to us. Since we have a related rates problem here, we will want to follow the same four steps as all other related rates problems. One of these options actually won’t work, but we will explore them further later in this example. If this one works we should be able to plug in any height and get the same answer for all of them. or plug in some random value for the height and see what we get.Create an equation that doesn’t include the height of the tank or the water,.So there’s two possible ways to deal with this: We actually don’t need it! Because the surface of the water in this tank will always be a circle with radius 5 m as this tank fills up, the height of the water or the tank won’t impact our answer. Step 3.How do we deal with this missing information? Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find ds/dt when x=3000 ft. We are told the speed of the plane is 600 ft/sec. Since x denotes the horizontal distance between the man and the point on the ground below the plane, dx/dt represents the speed of the plane. Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length 4000 ft is perpendicular to the line segment of length x feet, creating a right triangle. We do not introduce a variable for the height of the plane because it remains at a constant elevation of 4000 ft. Note that both x and s are functions of time. The variable s denotes the distance between the man and the plane. We denote those quantities with the variables s and x, respectively.Īs shown, x denotes the distance between the man and the position on the ground directly below the airplane. The distance between the person and the airplane and the person and the place on the ground directly below the airplane are changing. ![]() An airplane is flying at a constant height of 4000 ft.
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