Chapter 4 | Applications of Derivatives
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x dx dt . 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 ds dt when x =3000ft. Since the speed of the plane is 600 ft/sec, we know that dx dt = 600 ft/sec. We are not given an explicit value for s ; however, since we are trying to find ds dt when x =3000ft, we can use the Pythagorean theorem to determine the distance s when x =3000 and the height is 4000ft. Solving the equation 3000 2 +4000 2 = s 2 for s , we have s =5000ft at the time of interest. Using these values, we conclude that ds / dt is a solution of the equation (3000)(600) = (5000) · ds dt . Therefore, ds dt = 3000·600 5000 = 360 ft/sec. 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 x ( t ) and s ( t ) by the equation dt = s ds
⎤ ⎦ 2 +4000 2 = ⎡
⎤ ⎦ 2 .
⎡ ⎣ x ( t )
⎣ s ( t )
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 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted x ( t ) =3000 into the equation before differentiating, our equation would have been 3000 2 +4000 2 = ⎡ ⎣ s ( t ) ⎤ ⎦ 2 . After differentiating, our equation would become 0= s ( t ) ds dt . As a result, we would incorrectly conclude that ds dt =0.
4.2
What is the speed of the plane if the distance between the person and the plane is increasing at the rate of 300 ft/sec?
We now return to the problem involving the rocket launch from the beginning of the chapter.
Example 4.3 Chapter Opener: A Rocket Launch
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