Math 131. Related Rates Larson Section 2.6

Transcription

1 Math 131. Related Rates Larson Section 2.6 There are many natural situations when there are related variables that are changing with respect to time. For example, a spherical balloon is being inflated with air. The volume and radius are related by the volume equation V = 4 3 πr3, and both the volume and radius are changing with respect to time. The volume and radius are typically implicit functions of t since we may not know their explicit expressions. Then we can implicitly differentiate this volume equation with respect to time (the right side will require the chain rule); this will allow us to determine how the rates of change and are related: d [V ] = d [ ] 4 3 πr3 = d [ 4 3 πr3 ] = 4πr2 Example 1. (a) A spherical balloon is being inflated with air so that its radius is increasing at a constant rate of 2 centimiters per second. Find the rate of change of the volume when the radius is 10 centimeters. (b) Another spherical balloon is being inflated at a constant rate of 100 cubic centimeters per second. Find the rate at which the radius is increasing when the radius is 5 centimeters. Solution: (a) From the above, we know that = 2cm/s so when r = 10 cm, we find = 4πr2 = 4π(10cm)2 2 cm s = 4πr2 and the problem gives us that = 800π cm3 s (b) For this part, we also use when r = 5 cm, we have = 4πr2 and we are given that = 100 cm 3 /s. Then 100 = 4π(5 2 ) and so = π = 1 cm π s In a typical related rates problem you will: identify the given quantities and relevant quantities that are changing and label them with variables; write an equation that relates the variables; use the chain rule to implicitly differentiate both sides of the equation with respect to t; use the information given to solve for the required rate of change.

2 The remainder of these notes present several classic examples of related rates problems. Example 2. A stone opped into a still pond sends out a circular ripple whose radius increases at a constant rate of 3.2 feet per second. (a) How rapidly is the area enclosed by the ripple increasing when the radius is 3 feet? (b) How rapidly is the area enclosed by the ripple increasing at the end of 6.2 seconds? Solution: The relevant quantiies are the area of the circle which we call A and the radius of the circle which we call r. These quantities are related by the geometric formula A = πr 2. Then differentiating this with respect to t yields change of the area of the circle, and With this, we can answer both (a) and (b). = 2πr where is the rate of is the rate of change of the radius of the circle. (a) When r = 3 we have = 2 π = 19.2π ft2 /s (b) When t = 6.2 seconds we have r = feet (since it started at 0 and increased at a rate of 3.2 feet per second for 6.2 seconds so r = (3.2)(6.2) = 19.84) and thus = 2 π = π ft2 /s Example 3. A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground (see the figure below). At what rate is the tip of his shadow moving when he is 10 feet from the base of the light? Solution: Let the man be x feet from the base of the lamp post and let his shadow be s feet long. Then s 6 = x + s Then 15 1 ds 6 = 1 dx ds 15

3 and so 1 ds = 1 dx and so ds = 2 dx and we know dx = 5. Thus the length of his shadow is changing at 10 feet per second, and the tip of his shadow is moving away from the lamp 3 post at a rate of 25 feet per second. 3 Example 4. An airplane flies at an altitude of 5 miles toward a point directly over an observer (see the figure below). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation θ is changing when the angle is (i) θ = 30, and (ii) θ = 60. Express answer in radians per minute. Solution: Let x represent the length of the base of the triangle in the figure. cot θ = x ; differentiating with respect to t yields 5 Then we know dx csc 2 θ dθ = 1 5 dx = 10 miles per minute because the jet is traveling 600 miles per hour. Thus (i) When θ = 30, sin θ = 1/2 and so dθ = 1 5 sin2 θ dx radians per minute. dθ = ( 10) = (ii) When θ = 60, sin θ = 3/2 and so radians per minute. dθ = 1 5 ( 3) 2 ( 10) = Example 5. A ladder 15 feet long is leaning against the wall of a house (see the figure below). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. How fast

4 is the top of the ladder moving down the wall when its base is 4 feet from the wall? See the diagram below where L = 15 and x = 2 Solution: Let b be the distance of the base of the ladder from the house, and let h be the height of the top of the ladder above the ground. Then the Pythagorean theorem implies b 2 + h 2 = 225. Differentiating both sides with respect to time yields and then dh = b h db. 2b db + 2hdh = 0 and so 2hdh = 2bdb When b = 4, we use the relation b 2 + h 2 = 225 to find h 2 = 225 and so We now plug this into h 2 = = 209 or h = 209 dh = b h db = Thus top of the ladder is moving down the wall at approximately feet per second Example 6. Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are equal. How fast is the height of the pile increasing when the pile is 11 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V = π 3 r2 h. Solution: In this case r = h since the height and diameter are equal. Therefore V = π 2 12 h3. Then = 3πh2 12 dh = π 4 h2 dh Solving for dh we have dh = 4 = 40 πh 2 121π when the height is 11 feet. feet per minute

5 Example 7. A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 11 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 12 ft/min, at what rate will the boat be approaching the dock when 107 ft of rope is out? Solution: Let x be the horizontal distance of the front of the boat from the dock and let r be the length of rope that is out. Then by the pythagorean theorem x 2 = r 2. Differentiating this equation with respect to t we find 2x dx = 2r and so dx = r x From the information given in the question, we know = 12 ft/min (the negative is because the rope is being pulled in). We are asked to find dx when r = 107 feet. With this information, from the pythagorean equation above, we know and therefore, x = dx = 107 ( 12) ft/min ft/min The negative sign indicates that the boat is approaching the dock. Thus, the boat is approaching the dock at a rate of approximately feet per minute. Example 8. The altitude (i.e., height) of a triangle is increasing at a rate of 4 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 7.2 centimeters and the area is 67 square centimeters. Express answer to four decimal places Solution: The area of a triangle is given by A = 1 bh where b is the base and h is the 2 height. Differentiating both sides of this equation with respect to t yields = 1 2 ( h db + bdh We wish to find db, and so solving the above equation for it, we have db = 1 ( 2 ) h bdh )

6 Using the quantities dh obtain = 4, = 4.5, h = 7.2 and b = 2A/h = db = 1 ((2)(4.5) ( )(4)) cm/min , we Example 9. A baseball diamond has the shape of a square with sides 90 feet long (see figure). A player runs from second base to third base at a speed of 25 feet per second is 20 feet from third base. At what rate is the player s distance from home plate changing? Solution: Let s represent the distance of the runner to home plate and let x represent the distance of the runner to 3rd base. Then x = s 2. We differentiate implicitly to find 2x dx = 2s ds and so ds = x s dx. We know that x = 20 and so s = = = 10 85, and we know dx/ = 25. Therefore, ds = ( 25) = = The distance between the runner and home plate is decreasing at approximately 5.4 feet per second. Example 10. An airplane flies at an altitude of 5 miles toward a point directly over an observer (see the figure below). The ground speed of the plane is 460 miles per hour. Find the rate at which the distance between the observer and plane is changing when the angle of elevation θ = 70. Round answer to the nearest tenth of a mile per hour.

7 Solution: With reference to the diagram above, the distance between the observer and the plane is the hypotenuse of the triangle, we will lable that side h, and we will label the adjacent side x, we know dx = 460 miles per hour and by the Pythagorean theorem, x = h 2. So we differentiate this equation implicitly to find d dx [x2 + 25] = d dx [h2 ] 2x dx = 2hdh and so dh = x dx h. When θ = 70, we have x = 5 cot(70 ) and h = 5 csc(70 ) therefore dh = 5 cot(70 ) 5 csc(70 ) ( 460) = 460 cos(70 ) miles per hour This means the distance between the jet and observer is decreasing at a rate of miles per hour. Example 11. Suppose the edges of a cube are increasing at a rate of 5 centimeters per minute. (a) How fast is the volume changing when each edge is 8 centimeters? (b) How fast is the volume changing when each edge is 11 centimeters? Solution: First, the volume of a cube is given by V = x 3 where each side has length x units. Differentiating this implicitly with respect to time yields = 3x2 dx (a) Given x = 8 and dx = 5, we have = 3(8)2 (5) = 960 cm3 min (b) Given x = 11 and dx = 5, we have = 3(11)2 (5) = 1815 cm3 min

8 Example 12. An airplane fflying at an altitude of 6 miles passes directly over a radar station. When the airplane is 13 miles from the radar station (s = 13), the radar detects that the distance s is changing at a rate of 250 miles per hour. What is the speed of the airplane. (Round answer to nearest mile per hour). Solution: Using the Pythagorean theorem we note x = s 2, we are given s and ds and we look to find dx. Then differentiating this implicitly with respect to t, we find and so dx = s ds x. d [x2 + 36] = d [s2 ] 2x dx = 2sds When s = 13, we find that x = = 133, and we were given ds = 250 miles per hour. Therefore, dx = miles per hour 133 That is the plane is travelling approximately 282 miles per hour.