Almost all of the questions involving Related Rates will require one of the following relationships to link together the various changing rates:

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Naomi Hunt
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1 Related Rates All quantities that we meet in every-day life change with time, this is especially true in scientific investigations. Related Rate problems are those in which an equation epresses some relationship between two or more variables, and you are asked to find the rate of change of one variable when the rate of change of another variable in given. In these types of word problems you will typically find that the values of variables (at some instant) are given along with their respective time rate of change. The problem usually asks you to find the time rate of change of those variables at some other instant. In a Related Rate problem, the idea is to compute the rate of change (derivative) of one quantity in terms of the rate of change (derivative) of another (more easily measured) quantity. The procedure is to find an equation which relates these two quantities and then use the Chain Rule to differentiate both sides with respect to time. It may be difficult to find a relation between the two quantities, but you may find it easier to find relations between them and other quantities. There is no general method for finding relationships between two quantities, but undoubtedly the first step will be to draw a picture. The missing relationship may be discovered by the consequence of simple geometry. Almost all of the questions involving Related Rates will require one of the following relationships to link together the various changing rates: Pythagorean Theorem Similar triangles Look for a single right angle triangle. Look for two triangles (most likely both will be right angle triangles). How to Solve Related Rate Problems When you are starting these types of questions, it is best to follow the following heading guidelines until you are very comfortable with the solution techniques and can efficiently produce the solution. 1] Under the Heading LABEL. - If appropriate draw a picture, and label any quantity that seems important to the problem (choose your label so that it correspond to a word that describes the quantity.). i.e. h for height, r for radius. - Label any other quantity relevant to the problem. i.e. V=volume, A=area. ] Under the Heading GIVEN. - List all the values which are given in the problem. i.e. V=10, r=15

2 - DO NOT PUT IN SPECIFIC VALUES OF VARYING QUANTITIES ] Under the Heading REQUIRED. - Identify the Rate that is being asked for, and at what specific value. i.e. dv when r=5. 4] Under the Heading RELATIONSHIPS. - Write down any relationships that state the required variable in terms of their given or implied quantities. Note, this can include the Pythagorean Theorem or Similar Triangle relationships which may be hidden in the problem. i.e. A= r - Remember related Rates us the Chain Rule... so look for Chain Rule components. 5] Under the Heading WORK. - Using the relationships in 4], epress the required Rate in terms of the given Rate value (remember to differentiate with respect to time to get a Rate of Change). Then substitute known values to obtain an answer. Note: if you have independent variables, then you will require relationships. A 10m ladder rests against a brick wall. If the base of the ladder is pulled horizontally from the wall at the rate of m/s, how fast is the top of the ladder sliding down the wall when the base is 6m from the wall? Label:

3 Given: 6,, y Required: dy Relationship: Note that over the course of this activity, the ladder, the wall, and the ground form a right angle triangle governed by the Pythagorean Theorem. y 10 Work: Now differentiating the above with respect to time, t, yields a relationship involving the rates of change of the length of the sides. dy y 0 dy y 0 From the question, the following information was given: Now substituting in these values and solving for the required dy dy dy 9 4 Conclusion: We draw the conclusion that the top of the ladder is sliding down the wall at the rate of.5m/s when the bottom is 6m from the wall.

4 A tank is in the form of an inverted cone, having an altitude of 16 m and a base radius of 4 m. Water is flowing into the tank at a rate of m /min. How fast is the water level rising when the water is 5 m deep? The question asks for how fast the height of the water is rising. Mathematically, this translates to a requirement for dh. We are bound to the properties of the cone, in particular volume. At any given time, the water is conical in shape and hence governed by the relationship, 1 V r h Prior to differentiating, we could reduce it to a function of a single variable since r and h are not independent variables, there are bound by the given dimensions of the tank, r 4 h 16 h r 4 Substituting h r 4 into V 1 r h, we obtain,

5 V h 1 h h h 4 48 (a function of one variable) Differentiating both sides of with respect to time, (even though the function is in terms of h), remembering to apply the chain rule. dv dv dh dh Since water is flowing into the tank at a rate of m /min, we have, so h dh 16 dh h dv At h=5 m, we have dh m/ min 5. 5 We will repeat the previous, showing that it is not necessary to reduce the function to a single variable prior to differentiating. 1 V r h dv dr dh r h r Do the derivative with respect to t implicitly. Since h r, then when h=5, we have 4 5 r 4 And dr 1 dh 4 Therefore dv 5 1 dh 5 dh dh 5 dh dh dh 5

Either approach is acceptable but this presents a more general technique since it shows that the differentiation can be undertaken without knowing the behaviour of the dependent variables beforehand.

6 Either approach is acceptable but this presents a more general technique since it shows that the differentiation can be undertaken without knowing the behaviour of the dependent variables beforehand. A pebble is thrown into water and causes circular ripples to spread out at a rate of m/s. Find the rate of change of area in terms of when: a) seconds after the pebble falls into the water. b) the area of the ripple is 9 m / s. Let A represent the Area in m. Let r represent the radius in m. The relationship of area is A r a) We require da when t= After seconds, the radius is ()()=6 m da dr r 6 4 ;remember we are differentiating with respect to t The area is changing at the rate of 4 m / s b)we require da when A= 9 / m s

Area is 9 m / s when the radius is m da dr r 1 The area is changing at the rate of 14 m / s Water is added to a cylindrical tank of radius 5 m and a height of 10 m at a rate of 100 L/min.

7 Area is 9 m / s when the radius is m da dr r 1 The area is changing at the rate of 14 m / s Water is added to a cylindrical tank of radius 5 m and a height of 10 m at a rate of 100 L/min. Find the rate of change of the water level when the water is 6 m deep (1L=1000cm ). Let V represent the volume in cm. Let r represent the radius in cm. Let h represent the h in cm. V r h We require dh when h=600 cm

dv dr dh r h r dh 100000 500 0 600 50000 dh 100000 50000 dh 5 Therefore the height is changing at the rate of Note: since the radius does not change as the water rises, then we dr have 0 cm / min 5 A

8 dv dr dh r h r dh dh dh 5 Therefore the height is changing at the rate of Note: since the radius does not change as the water rises, then we dr have 0 cm / min 5 A sphere is growing at the rate of changing when radius is cm. 1 cm / s. Find the rate at which the radius is Let V represent the volume of the sphere in cm /s Let r represent the radius in cm. V 4 r We require dr when r dv dr 4 r dr 1 4 dr 1 16 Therefore the radius is changing at the rate of 1 cm/ s 16

9 Ship A is 5 km south of ship B and is moving due north at a rate of 5 km/h. Ship B is moving due east at the rate of 10 km/h. Find the rate of change of the distance between the two ships hours later. Let represent the distance ship B is moving east. Let y represent the distance ship A is moving north. Let s represent the distance between the two ships. s y We require ds when t=. Note: when =0 then y ds dy s y ds ds 5 Note: s y s s 5 Therefore the two ships are moving apart at the rate of 5 km/h A spot light on the ground is shining on a vertical building which is 0m from the spot light. A man m tall is walking away from the light and is walking directly towards the building at a rate of 4 m/s. Find the rate of change of the length of his shadow on the building when he is 5m from the light. Let represent the distance the man is from the light in m Let y represent the height of the shadow in m y 0 Via similar triangles 0 y 40 y

10 We require dy dy when =5 The shadow is decreasing at the rate of / 5 m s Two sides of a triangle have lengths 15m and 0m. The angle between them is increasing at rad/sec. How fast is the length of the third side changing when the 90 angle between the sides is. Label: let represent the other side Let represent the angle 15 Given: 0 One side is 15m and the other side is 0m Required: when Relationships: cosine law: ABC cos d d a b c bc A Work: cos cos 600sin d 00sin d

11 Therefore When d d 00sin 90, cos Therefore 00sin Conclusion: The third side is increasing at the rate of appro 0.50 m/s. A beacon, located a perpendicular distance of 15m from point R on a straight shoreline, revolves at 1 rev/min. how fast does its beam sweep along the shoreline at a point S on the shoreline 45m from R Label: R 15 S Given: lighthouse is 15m from shore d rad min Required: when =45m Relationships: 15 tan d d

12 Work: 15tan 15sec d d d 15sec 60 sec We need to determine the value of sec when =45. When =45 the distance from the lighthouse to point S is Now sec H 59 A Conclusion: Therefore the beam is sweeping along the shore at approimately 5580 m/min

DIFFERENTIATION RULES

3 DIFFERENTIATION RULES DIFFERENTIATION RULES If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. However,