3. A man 6 feet tall walks directly away from a lamp post 12 feet high at the rate of 4 miles an hour. How fast does the shadow of his head move? 4. A man is walking at the rate of 3 miles an hour along a sidewalk which is 8 feet from the buildings beside it. A light on the other side of the street, 30 feet from the walk, casts a shadow of the man on the buildings. How fast does his shadow move? 5. A cruiser is moving parallel to a straight coast at a distance of 6 miles from it, at the rate of 27 miles an hour. How fast is the cruiser approaching a fort on the shore, 15 miles from the point directly opposite the ship? 6. The pressure and volume of a gas at a constant temperature satisfy the relation pv = k. At a certain time the volume is 3 cubic feet, the pressure is 10 pounds per square foot, and the volume is increasing at the rate of 0.2 of a cubic foot per second. Find the rate of change of the pressure. 7. Two ships start from the same place at the same time. east at 12 knots an hour, the other south at 18 knots an hour. will they be separating after half an hour? = One sails 8. The height of a ball thrown vertically upward with a velocity of 64 feet per second is given by s 64t16t2. The rays of the sun make an angle of 30° with the horizontal. How fast is the shadow of the ball moving along the ground just before the ball hits the ground? 103. Small Errors. In measuring any quantity directly, we usually have some idea of the accuracy of the measurement. Thus we may measure the length of the edge of a cubical box and feel confident that the length is 10 inches with an error of not more than of an inch. This is indicated by saying that the length is 10 inches. Frequently the size of the error is of less importance than the relative error, which is the ratio of the error to the quantity measured. Thus the relative error in the edge of the cube above is the ratio of to 10, that is, = 0.0125 or 1.25%. g The relative error, rather than the magnitude of the error itself, indicates the degree of accuracy of a measurement. Nothing can be said of the comparative accuracy of the measurements of the lengths of two lines if it is known that the error in one case is 2 feet and that in the other it is only half a foot. If the first error occurred in measuring a side of a farm half a mile long, the relative error would be 40 = 0.000758, which is less than 0.08 of one per cent, while if the second error was made in measuring the frontage of a city lot 50 feet wide the relative error would be 0.5/50 = 0.01, or one per cent. Under these circumstances the first measurement would be by far the more accurate. Many quantities are measured indirectly. Thus to find the volume of a cubical box we measure the length of an edge, and compute the volume. If the length of the edge is found to be 10 inches, then the volume is V 1000 cubic inches. If an error of of an inch is made in measuring the edge, what is the error in the volume? In the example below we shall develop a method of answering such a question. = EXAMPLE. The side of a square was found by measuring to be 3 inches. If the measurement is 0.1 of an inch too small, what is the approximate error in the computed area? First solution. It is obvious that the computed area and = 9 the true area = (3 + 0.1)2 = 9+ 0.6 +0.01. Since 0.01 is small in comparison with the other terms, it may be disregarded, and the approximate error is 0.6. Not only is the term disregarded much smaller than the others, but it is smaller than the change in the approximate error due to a slight change in the estimated error in the measurement of the side. For if the measurement of the side is 0.11 of an inch too small, then the true area = (3 + 0.11)2 = 9 + 0.66 + 0.0121, and the approximate error is now 0.66. Thus a slight change in the estimated error in the original measurement of the side produces a change of 0.06 in the original approximate error in the area, and this change is 6 times the term originally neglected. The approximate relative error is the ratio of the approximate error to the computed area, i.e., 0.6/9, = 0.07 or 7 %. It is instructive to first solve the problem for any square and then substitute the given numbers. Let x be the value of the side found by measurement, and Ax the estimated error. Let y denote the computed value of the area, and Ay the error in y. Then the computed area is y = x2 (1) and the true area is y + Ay = (x + ▲x)2 = x2 + 2xAx + Ax2 (2) Subtracting (1) from (2), the error in the computed area is seen to be When Ax is small, Ax2 is very small compared to 2xAx, so that Ar2 may be neglected, and the approximate error in y is 2xAx. Ax FIG. 169. Equation (3) is readily interpreted from the figure. The product xAx is the area of either of the rectangles, and Ar2 is the area of the small square. Obviously, if Ax is small compared to r, the small square is negligible as compared with the two rectangles. It is also clear that a slight change in Ax produces a change in the two rectangles which is greater than the small square disregarded. Denoting the approximate error by dy, we obtain from the above dy = 2x▲x. The relative error is therefore so that the relative error in y is twice the relative error in x. Substituting the given values of x and Ax we get, as before, = dy = 2 × 3 × 0.1 0.6, and dy/y = 2 × 0.1/3 = 0.07, or 7 %. Second solution. The first solution is useful for clarifying the ideas involved, but the method now to be considered is more convenient in all but the simplest examples. = Letting x be the side of any square, the computed area is given by (1), whose graph is shown in Fig. 170. The ordinate y MP of any point represents the computed area of a square whose side as measured is x = OM. If the error in the measurement is Ax = MN, then the ordinate NQ represents the true area and Ay = RQ the true error in y. Construct the line tangent at P, and let it cut the ordinate NQ at S. Let dy = RS. Then dy is an approximate value of the true error Ay, and if Ax is very small this gives a very good approximation. For if MN were less than one-eighth of an inch in this figure, S would be practically coincident with Q. In order to have a more consistent notation set Ax = dx, so that dx represents the error in x. Then in the right triangle PRS, we have PR = dx, and From (1) we have Day 2x, and hence, from (4), dy = 2x dx, the same result that was obtained in the first solution. The error for the given y P dx R У х M N FIG. 170. square, and the relative error, are found as before. If a magnitude x be measured, and a second magnitude y be computed from it, then y is a function of x, y = f(x). The reasoning in the second solution applies equally well to the graph of this function, if we regard the curve in the figure as the graph of any function. Hence: If y = f(x), an approximate value of the error in y due to an error of dx in the measurement of x is the prod uct of the derivative of y with respect to x and dx; i.e., An approximate value of the relative error is given by dy /y. In what follows we shall use the terms error and relative error to denote these approximations unless the contrary is explicitly mentioned. х In applying this method, it is necessary first to compute y as a function of x without using the given numerical value of x. For example, in the illustration above. we must first find the area of any square. EXERCISES 1. The edge of a cube is found by measurement to be 8.43 inches with a probable error of 0.005 inches. What is the error in the volume?. The relative error? 2. The side of a square is almost exactly 4 inches. What is the allowable error in the measurement of a side if the error in the area is to be determined to within one-tenth of one per cent? 3. A surveyor measures a square field with a 66-foot chain and finds the area to be 40 acres. If it is found that his chain has stretched and is one inch too long, find the correct area approximately. 4. Find the side of a cubical box to hold one quart (57.75 cubic inches). How much variation may be allowed on a side if the error in the capacity is not to exceed one per cent? 5. What is the allowable error in the measurement of the largest dimension of a rectangular block, 10 × 6 × 4 inches, if the volume is to be determined to within one-fifth of one per cent? What is the allowable error in the smallest dimension? (Assume in each part of the problem that the other two dimensions are exact.) 6. The intensity of a certain light at a distance x from the source is I = 1/2. What is the error in I if x = 12, with a probable error of 0.1? 7. The quantity of heat necessary to raise the temperature of a certain substance from 0° C. to 0° C. is Q = .5290 +.0003 02. If in an experiment is found to be 20°, with an estimated error of not more than 0°.3, what is the error in Q? The relative error? 8. The relation between Fahrenheit and Centigrade thermometer readings is F 1.8C +32. A reading of 19°.2 is made on a Centigrade thermometer, and the temperature Fahrenheit computed from it. Find the percentage of error in the computed value, if the error in the reading is 0°.1. = 9. The depth of the Colorado Canyon at a certain point was found from the relation s 16ť2 by dropping a stone and observing the time of falling. Is the percentage of error in the computed depth less than or greater than the percentage of error in the measurement of the time? # 10. The distance to the sun as determined by recent measurements has been stated to be 92,780,000 ± 500,000 miles (the notation indicating that the error is probably not greater than 500,000 miles). 90° r PE FIG. 171. subtended by the radius of the earth. What is the allowable percentage of error in the radius to give the above result for the distance to the sun? The radius of the earth is found by measuring an arc of one degree, or 0.017453 radians, on the equatorial or on a meridian circle, and applying the theorem on page 171. Assuming that the angular measurement is exact, what is the allowable percentage of error in the measurement of the arc? An arc of one degree has been measured recently by using a brass rod |