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A spherical balloon is being filled with air at the constant rate of Figure. How fast is the radius increasing when the radius is? The volume of a sphere of radius centimeters is. Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, seconds after beginning to fill the balloon with air, the volume of air in the balloon is. Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation.
When the radius ,. What is the instantaneous rate of change of the radius when? Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4.
We examine this potential error in the following example.
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The first example involves a plane flying overhead. The relationship we are studying is between the speed of the plane and the rate at which the distance between the plane and a person on the ground is changing. An airplane is flying overhead at a constant elevation of ft. A man is viewing the plane from a position ft from the base of a radio tower. The airplane is flying horizontally away from the man. Step 1.
Draw a picture, introducing variables to represent the different quantities involved. As shown, denotes the distance between the man and the position on the ground directly below the airplane. The variable denotes the distance between the man and the plane. Note that both and are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of ft. Step 2. Since denotes the horizontal distance between the man and the point on the ground below the plane, represents the speed of the plane.
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 when ft. Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating and :. Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation.
Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find when ft. We are not given an explicit value for ; however, since we are trying to find when ft, we can use the Pythagorean theorem to determine the distance when and the height is ft. Solving the equation. Using these values, we conclude that is a solution of the equation.
Note : When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities and by the equation. Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant to denote that quantity. However, the other two quantities are changing.
If we mistakenly substituted into the equation before differentiating, our equation would have been.
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As a result, we would incorrectly conclude that. We now return to the problem involving the rocket launch from the beginning of the chapter. Let denote the height of the rocket above the launch pad and be the angle between the camera lens and the ground. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is ft off the ground. That is, we need to find when ft. Now we need to find an equation relating the two quantities that are changing with respect to time: and. How can we create such an equation?
Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have. Differentiating this equation with respect to time , we obtain. We want to find when ft.
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We need to determine. Recall that is the ratio of the length of the hypotenuse to the length of the adjacent side.
Opus Majus, Volumes 1 and 2 - Roger Bacon - Google Книги
We know the length of the adjacent side is ft. To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is ft, the length of the other leg is ft, and the length of the hypotenuse is feet as shown in the following figure. Therefore, when , we have. Recall from step 4 that the equation relating to our known values is. Substituting these values into the previous equation, we arrive at the equation.
Find when ft. In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.
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Water is draining from the bottom of a cone-shaped funnel at the rate of. The height of the funnel is ft and the radius at the top of the funnel is ft. At what rate is the height of the water in the funnel changing when the height of the water is ft? Let denote the height of the water in the funnel, denote the radius of the water at its surface, and denote the volume of the water. Step 2: We need to determine when ft. We know that. From the figure, we see that we have similar triangles.
Therefore, the ratio of the sides in the two triangles is the same.
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Therefore, or. Using this fact, the equation for volume can be simplified to. Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time , we obtain. Step 5: We want to find when ft. Since water is leaving at the rate of , we know that. At what rate is the height of the water changing when the height of the water is ft?
We need to find when. Find at and if.