Solutions to Examples from Related Rates Notes. ds 2 mm/s. da when s 100 mm

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1 Solutions to Examples from Relate Rates Notes 1. A square metal plate is place in a furnace. The quick temperature change causes the metal plate to expan so that its surface area increases an its thickness ecreases. If the sie length of the plate is increasing at 2 mm per secon, fin the rate the area of the plate is increasing when the sie length is 100 mm. a. Assign smbols to all quantities that are functions of time. Let t be the time in secons. Let s be the sie length of the square plate in mm. Let A be the area of the plate in mm 2. b. Write the given information & the rate ou are tring to fin as appropriate erivatives. s 2 mm/s Fin: A when s 100 mm c. Write an equation that relates the various quantities of the problem. A s 2. Implicitl ifferentiate both sies of equation with respect to time, t. A 2s s e. Substitute given information into resulting equation an solve for the unknown rate. A mm 2 /s s A pebble is thrown into a pon causing concentric circular ripples. The outermost ripple is increasing at a constant rate of 2 feet per secon. When the raius is 4 feet, at what rate is the total area of the isturbe water changing? a. Assign smbols to all quantities that are functions of time. Let t be the time in secons. Let r be the raius of the ripple an A be the enclose area. b. Write the given information & the rate ou are tring to fin as appropriate erivatives. r 2 ft/s Fin: A when r 4ft c. Write an equation that relates the various quantities of the problem. A r 2. Implicitl ifferentiate both sies of equation with respect to time, t. A 2r r 2 r r e. Substitute given information into resulting equation an solve for the unknown rate. A ft 2 /s r 4 3. At a san an gravel plant, san is falling off a conveor an onto a conical (cone-shape) pile at a rate 1

2 of 10 cubic feet per minute. The iameter of the base of the cone is approximatel 3 times the altitue (height). At what rate is the height of the pile changing when the pile is 15 feet high? a. Draw a iagram & Assign smbols to all quantities (there are THREE in this case: The height, base raius, & volume of the conical pile) that are functions of time. Let t be the time in minutes. Let h be the height in feet an r be the base raius in feet (see iagram below). Let V be the volume of the san in cubic feet. h b. Write the given information & the rate ou are tring to fin as appropriate erivatives. V 10 ft 3 /min Fin: h when h 15 ft c. Write an equation that relates the various quantities of the problem. Use information provie to eliminate one of the variables b substitution so that equation has onl two variables (HINT: onl volume & height shoul be left). Equation that relates V, h, anr: V 1 3 r2 h Since the iameter of the pile is three times the height an since the iameter is twice the raius we have 2r 3h or r 3 2 h substituting for r we obtain an equation relating onl V an h: V h 2 h 3 4 h3. Implicitl ifferentiate both sies of equation with respect to time, t. V 3 h 3h2 4 9 h h2 4 or solving for h/ : h 4 9 h 2 V r 2

3 e. Substitute given information into resulting equation an solve for the unknown rate. h 4 h ft/min ft/min A trough is 12 feet long an 3 feet across the top (see figure below). Its ens are isosceles triangles with altitues of 3 feet. If water is being pumpe into the trough at 2 cubic feet per minute, how fast is the water level rising when the water in the trough is 1 foot eep? 3 feet 3 feet 12 feet a. Assign smbols to all quantities (there are THREE in this case) that are functions of time. Let t be the time in minutes. The water will take the shape of the trough. Let h be the height of the water in the trough an let w be the wih across the top of the water (see iagram below). Let V be the volume of the water. h w 12 feet b. Write the given information & the rate ou are tring to fin as appropriate erivatives. V 2ft 3 /min Fin: h when h 1ft c. Write an equation that relates the various quantities of the problem. Use the geometr of the situation to eliminate one of the variables b substitution so that equation has onl two variables. (HINT: Use similar triangles to get a relation ship between two of our variables & then substitute to eliminate one of the three variables.) First we have V 1 wh 12 6wh 2 But, b similar triangles we have w 3 h 3 w h so that V 6h 2. Implicitl ifferentiate both sies of equation with respect to time, t. V 12h h or h 1 12h V 3

4 e. Substitute given information into resulting equation an solve for the unknown rate. h 1 h ft/min ft/min 6 5. Carefull review each solution. Ask if ou have questions. a. Let be the height of the top of the laer above groun an x be the istance from the house to the bottom of the laer. Then we are given that x 2anaskeofin when 12, 5, & 1. The laer an house form a right triangle, so b the Pthagorean Theorem we have x We ifferentiate implicitl with respect to t x x2 2 0 Solving for iels Plugging in our given value iels 2x x 2 2 x 0 2x x x x 2 2x Now when 12, b the Pthagorean Theorem we have x x x 2 25 x 5 Thus when the top of the laer is 12 feet above the groun, ft/s 6 Note that the negative sign inicates that is ecreasing (i.e. the top of the laer is moving ownwar). We can similarl fin for when 5 & 1. The table below summarizes the finings. (in feet) (in ft/s) Note that the rate at which the laer is sliing own the sie of the house is NOT constant. It appears to be speeing up. To express the rate at which the laer is sliing own the sie of the house as a function of the height the top of the laer, we note that at an instant x x x Thus 2x Since 0, 2 an we can further simplif this as 4

5 Note that lim lim since 169 as 0 2 Thus, as the top of the laer gets closer an closer to the groun, the spee at which the laer is falling approaches (negative) infinit. Note that in real life friction will limit the spee of the falling laer. Our moel oes not inclue the effects of friction. b. Let A be the enclose area. Using the same variables from part (a), we want to fin A when 12, 5, & 1. Note that the region enclose is a right triangle so that A 1 2 x Differentiating implicitl with respect to t, wehave A 1 2 x A 1 2 x 1 2 x Using the results from part (a), we have x 12 ft 5ft 1ft x (in feet) (in ft/s) (from part (a)) A (in ft 2 /s) c. Let be the angle between the laer an the groun. We are aske to fin when 12, 5, & 1. Using the variables introuce above we have sin 13 Differentiating implicitl with respect to t, wehave sin 13 cos cos sec 13 Note that So that Thus sec 13 x sec 13 1 x 5

6 12 ft 5ft 1ft x (in feet) (in ft/s) (from part (a)) (in ra/s) Let x be the istance from the blue car to where the re car was at noon. Let the istance the re car has travele since noon. x Blue Car s Re Car If s is the istance between the cars then we are aske to fin s at 2 & 4 p.m. given that x 40 mph (notice that the istance x is getting smaller) an 25 mph. The Pthagorean Theorem equates all our quantities: x 2 2 s 2 Differentiating implicitl with respect to t, wehave x2 2 Thus 2x x 2 s2 2s s s 1 s x x At 2 p.m. x 120 (blue car has travele 80 miles, leaving 120 miles to the location of re car at noon) an 50 so that s Hence s mph 130 At 4 p.m. x 40 (blue car has travele 160 miles, leaving 40 miles to the location of re car at noon) an 100 so that s Hence 6

7 s mph Note that the istance between the cars was ecreasing at 2 p.m. (s/ was negative) an increasing at 4 p.m. (s/ positive). 7