Related Rates - Introduction

Transcription

1 Related Rates - Introduction Related rates problems involve finding the rate of change of one quantity, based on the rate of change of a related quantity.

2 Related Rates - Introduction Related rates problems involve finding the rate of change of one quantity, based on the rate of change of a related quantity. Example: RelatedRates 1 Suppose P and Q are quantities that are changing over time, t. Suppose they are related by the equation 3P 2 = 2Q 2 + Q + 3. If dp = 5 when P(t) = 1 and Q(t) = 0, then what is dq at that time?

3 Related Rates - Introduction Related rates problems involve finding the rate of change of one quantity, based on the rate of change of a related quantity. Example: RelatedRates 1 Suppose P and Q are quantities that are changing over time, t. Suppose they are related by the equation 3P 2 = 2Q 2 + Q + 3. If dp = 5 when P(t) = 1 and Q(t) = 0, then what is dq 6P dp dq = 4Q + dq 6(1)(5) = 4(0) dq + dq 30 = dq at that time?

4 Related Rates Example: RelatedRates 2 A garden hose can pump out a cubic meter of water in about 20 minutes. Suppose you re filling up a rectangular backyard pool, 3 meters wide and 6 meters long. How fast is the water rising? https: //s.iha.com/ /holiday-lettings-vila-nova-de-oliveirinha-quinta-da-mamadeira_4.jpeg

5 Related Rates Example: RelatedRates 2 A garden hose can pump out a cubic meter of water in about 20 minutes. Suppose you re filling up a rectangular backyard pool, 3 meters wide and 6 meters long. How fast is the water rising? We know the rate of change over time of the volume of water, and we want to know the rate of change over time of the height of the water. Let V be the volume of water in the cell, and let h be the height of water. Then we relate V and h: V = 3 6 h = 18h where V is measured in cubic meters, and h is measured in meters. Then we differentiate both sides with respect to t: dv = 18 dh 1 dh Now, 20 = 18, so the water level is rising at about = meters per minute, or something less than a third of a centimeter per minute. It s going to be a lot of work to fill up the pool.

6 Solving Related Rates 1. Draw a Picture 2. Write what you know, and what you want to know. Note units. 3. Relate all your relevant variables in one equation. 4. Differentiate both sides (with respect to the appropriate variable!) 5. Solve for what you want.

7 Example: RelatedRates 3 A cannonball is attached to a rope, which is attached to a pulley on a boat, at water level. The cannonball is taken 8 (horizontal) metres from its attachment point on the boat, then dropped in the water. The cannon ball sinks straight down. The rope is being let out at a constant rate of one metre per second, and two seconds have passed. How fast is the ball descending? 8

8 Example: RelatedRates 3 A cannonball is attached to a rope, which is attached to a pulley on a boat, at water level. The cannonball is taken 8 (horizontal) metres from its attachment point on the boat, then dropped in the water. The cannon ball sinks straight down. The rope is being let out at a constant rate of one metre per second, and two seconds have passed. How fast is the ball descending? Let R(t) be the length of the rope at time from the ball to the pulley after t seconds; let D(t) be the (vertical) depth of the ball. Then and so D 2 = R 2 2D dd = 2R dr After two seconds, R = = 10, so D = 6. It is given that dr 2(6) dd = 2(10)(1) So, dd = 5 3 : the ball is sinking at 5 metres per second. 3 = 1. Now:

9 Example: RelatedRates 4 You are pouring water into a large jar, through a funnel with an extremely small hole. The funnel lets water into the jar at 100mL per second, and you are pouring water into the funnel at 300mL per second. The funnel is shaped like a cone with height 20 cm and diameter at the top also 20 cm. (Ignore the hole in the bottom.) How fast is the height of the water in the funnel rising when it is 10cm high? Funnel (PSF) by Pearson Scott Foresman - Archives of Pearson Scott Foresman, donated to the Wikimedia Foundation. Licensed under Public Domain via Commons -

10 Example: RelatedRates 4 You are pouring water into a large jar, through a funnel with an extremely small hole. The funnel lets water into the jar at 100mL per second, and you are pouring water into the funnel at 300mL per second. The funnel is shaped like a cone with height 20 cm and diameter at the top also 20 cm. (Ignore the hole in the bottom.) How fast is the height of the water in the funnel rising when it is 10cm high? The volume of a cone with radius r and height h is π 3 r 2 h.

11 Example: RelatedRates 4 You are pouring water into a large jar, through a funnel with an extremely small hole. The funnel lets water into the jar at 100mL per second, and you are pouring water into the funnel at 300mL per second. The funnel is shaped like a cone with height 20 cm and diameter at the top also 20 cm. (Ignore the hole in the bottom.) How fast is the height of the water in the funnel rising when it is 10cm high? The volume of a cone with radius r and height h is π 3 r 2 h. Let V be the volume of water in the funnel, so dv = = 200 ml/sec. Let h be the height of water in the funnel: we re concerned with the instant when h = 10. Then we need to know r, the radius of the cone shape made by the water in the funnel. By using similar triangles, we see that r h = 10 20, so r = h 2. Now, V = π 3 r 2 h = π ( ) h 2 h = π h3 We differentiate both sides with t as our variable: dv So, 200 = π dh 4 (10)2, leaving = π 4 h2 dh dh = 8 π cm/sec

12 Chapter 3: Applications of Derivatives Example: RelatedRates 5 A sprinkler is 3m from a long, straight wall. The sprinkler sprays water in a circle, making three revolutions per minute. Let P be the point on the wall closest to the sprinkler. The water hits the wall at some spot, and that spot moves as the sprinkler rotates. When the spot where the water hits the wall is 1m away from P, how fast is the spot moving? JJ Harrison

13 x P 3 θ sprinkler Given the labels in the picture above, we know dθ can relate x to θ by tan(θ) = x 3 sec 2 θ dθ = 1 3 sec θ = hyp dx = 10 adj 3 = 3(2π) radians per minute, and we. Then differentiate both sides with respect to t to get:. We only need to find sec θ, which we get from our triangle: ( 10 ) 2 dx, so = 3 3 6π = 20π meters per minute.

14 Example: RelatedRates 6 A roller coaster has a track shaped in part like the folium of Descartes: x 3 + y 3 = 6xy. When it is at the position (3, 3), its horizontal position is changing at 2 units per second in the negative direction. How fast is its vertical position changing?

15 Example: RelatedRates 6 A roller coaster has a track shaped in part like the folium of Descartes: x 3 + y 3 = 6xy. When it is at the position (3, 3), its horizontal position is changing at 2 units per second in the negative direction. How fast is its vertical position changing? We know dx dy = 2 when x = y = 3, and we would like to find relationship between x and y is given. We differentiate implicitly: 3x 2 dx + 3y 2 dy = 6(x dy + y dx ) at the same time. The 3(9)( 2) + 3(9) dy dy = 6(3 + 3( 2)) dy = 2 So, the roller coaster is moving 2 units per second in the positive y direction.

16 Example: RelatedRates 7 Two dogs are tied with elastic leashes to a lamp post that is 2 metres from a straight road. At first, both dogs are on the road, at the closest part of the road to the lamp post. Then, they start running in opposite directions: one dog runs 3 metres per second, and the other runs 2 metres per second. After one second of running, how fast is the angle made by the two leashes increasing? doggy 1 doggy 2 road θ lamp post

17 Example: RelatedRates 7 Two dogs are tied with elastic leashes to a lamp post that is 2 metres from a straight road. At first, both dogs are on the road, at the closest part of the road to the lamp post. Then, they start running in opposite directions: one dog runs 3 metres per second, and the other runs 2 metres per second. After one second of running, how fast is the angle made by the two leashes increasing? doggy 1 doggy 2 road θ lamp post The angle θ made by the leashes is made up of θ 1 + θ 2, as shown on either side of the dashed line above. tan θ 1 = 2t = t, so 2 sec2 θ 1 θ 1 = 1, so θ 1 = cos 2 θ. When doggy 2 is ( ) 2 two metres away from its starting position, θ = 2 2 = 1. Similarly, 2 ( ) 2 θ 2 = 3 2 cos2 θ 2 = = 6. So, all together, 13 θ = = 25 radians per second

18 Portrait of Newton, Godfrey Kneller Chapter 3: Applications of Derivatives Copy of a potrait of Leibniz by Andreas Scheits Example: RelatedRates 8 Newton and Leibniz are attending a conference at the same university. Eager not to cross paths, Newton is one kilometer due east of the math building, heading east, and Leibniz is two kilometers due north of the math building, heading north. If Newton is walking at 5kph, and Leibniz is walking 7kph, how fast is the distance between them increasing?

19 L y D math bldg x N We relate all the variables to each other: Note D = x 2 + y 2 = = 5. Differentiate with respect to time: 2D dd D 2 = x 2 + y 2 dx dy = 2x + 2y and fill in what we know so far: 2 5 dd = 2(1)(5) + 2(2)(7) Then their distance is increasing at kph.

20 Example: RelatedRates 9 A triangle has one side that is 1cm long, and another side that is 2cm, and the third side is formed by an elastic band that can shrink and stretch. The two fixed sides are rotated so that the angle they form, θ, grows by 1.5 radians each second. Find the rate of change of the area inside the triangle when θ = π/4.

21 Example: RelatedRates 9 A triangle has one side that is 1cm long, and another side that is 2cm, and the third side is formed by an elastic band that can shrink and stretch. The two fixed sides are rotated so that the angle they form, θ, grows by 1.5 radians each second. Find the rate of change of the area inside the triangle when θ = π/4. 1 θ h 2

22 Example: RelatedRates 9 A triangle has one side that is 1cm long, and another side that is 2cm, and the third side is formed by an elastic band that can shrink and stretch. The two fixed sides are rotated so that the angle they form, θ, grows by 1.5 radians each second. Find the rate of change of the area inside the triangle when θ = π/4. 1 θ h 2 sin θ = opp = h = h, and A = 1 bh, so hyp 1 2 A = 1 (2) sin θ = sin θ 2 Now da = cos θ dθ = cos(π/4)(1.5) = 3 2 2