calc12 unit4 notes part1€¦ · title: microsoft powerpoint - calc12_unit4_notes_part1 author:...

3. Related Rate Problems› The chain rule is a very powerful technique of finding the derivative of a composite function. The idea behind is to develop the connection between the variables and through a “chain” .

› This method has been applied to finding the derivatives of implicit functions.

› Another application of the chain rule is to solve the problems of related rates, in which the rates of change of two or more related variables in terms of time are determined.

CHAPTER 4 - APPLICATIONS OF DIFFERENTIATION (PART I) 38

› Consider a conical tank of radius , volume , and height . When water is drained out of a conical tank from the

bottom, what is the rate of change of volume related to the rates of change of radius and the height?


› First, we have to find out how , and are related. According to geometry, the volume of a cone is given by

› Then, we differentiate both sides with respect to :

› Since both and vary with time,


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ଶ

ଶ ଶ

› This gives

› Hence, to find the rate of change of volume at any given time , we only need to substitute the values of , ,

, and at into the expression above.


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› Example: Assume that there is a car whose position, , from the starting point at a given time , is given by

. The gas used by the car, liters for a given distance , is given by . How fast is the gas consumed by the car?

› Solution: Recall that

› Hence,

and


› The general procedures of solving a related rate problem– Step 1: Identify all the quantities involved in the question, and which quantity is to be determined.

– Step 2: Develop the relationship, i.e., the equation, for the quantities involved.

– Step 3: Differentiate the equation appropriately to find out how the rates of change are connected.

– Step 4: Perform substitution to find the required answer.

– Step 5: Check the final answer (i.e., Is the unit correct? Does the answer make sense?)


› Example: A raindrop falls in a puddle and the ripples spread in circles. The radii of the circles increase at the rate of 2 inches/second. Find the rate at which the area of a circle will be growing when its radius is 6 inches.

› Solution: There are three quantities in this example; namely, the area , the radius , and the time . These quantities are related by

› Differentiating with respect to gives


ଶ

› Note that

› When ,

› The unit of the answer is ଶ .


› Example: A spherical balloon is inflated so that its radius is increasing at one inch/minute. Find the rate at which the volume increases when:(a) the diameter is 2 feet.(b) the surface area is 324 in2.


› Example: A conical flower vase is 30 inches high with a radius of 6 inches at the top. If it is being filled with water at a rate of 10 cubic inches per second, find the rate at which the water level is rising when the depth is 20 inches.

› Solution: The quantities involved are: the volume of the vase: the depth of the water: the radius of water surface: time in second


› From geometry, we know

› The volume depends on both and , but we want to find only yet is unknown. Hence, we need to perform a transformation so that is given solely by .

› Using similar triangles:

› Thus .


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6 in

30 in

› So the volume formula becomes

› Differentiating on both sides gives

› Rearrange it to yield


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ଷ

ଶଶ

ଶ

› Given and

› Substituting these to the equation yields

› So, the water level is rising at a rate of about 0.20 in/s.


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› Example: A water tank is built in the shape of a cone with height 5 m and diameter of 6 m. Water is pumped into the tank at a rate of 1.6 m3/min. Find the rate at which the water level is rising when water is 2 m deep.


› Example: A 26 foot ladder is resting against a vertical wall. The foot of the ladder starts to slip horizontally away from the wall at a rate of 6 in/sec. Find the rate at which the top of the ladder descends when it is 24 feet above the ground.

› Solution: Consider the situation


› From the diagram we can deduce the relationship between the quantities and :

› Differentiating this expression implicitly yields

› Rearranging this we obtain the following


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› To find , we need , , and . The second and third quantities are already given in the question; namely

› The value of is not provided, but it can be calculated using the Pythagorean theorem:

› Hence,


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› Example: A man starts walking north at a speed of 1.5 m/s. At the same time and starting from the same point, a woman begins walking west at a speed of 2 m/s. At what rate is the distance between the man and the woman increasing one minute after they start walking?


› Example: A spotlight on the ground shines on a wall 10 m away. A 2-m tall man walks from the spotlight toward the wall at a speed of 1.2 m/s. How fast is height of the man’s shadow on the wall decreasing when he is 3 m from the wall?

› Solution: The story can be represented by the following diagram.


› Note that the quantities and are not related by the Pythagorean theorem! Instead, they are related by similar triangles:

› This gives

› Differentiating both sides with respect to yields


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› It is given

› Note that when the man is 3 m from the wall, is not 3 but . Hence,

› The shadow is decreasing at a rate of about 0.49 m/s.› Think carefully: Can we solve this question using Pythagorean theorem?


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› Example: A 6-ft figure skater is directly beneath a spotlight 30 feet above the ice. She skates from the light at a rate of 16 ft/s and the spotlight follows her.(a) How fast is the skater’s shadow lengthening when she is 25 ft from her starting position?(b) How fast is the top of her shadow’s head moving when she is 25 ft from her starting position?


› Example: A camera is mounted at a point 3000 feet from the base of a rocket. The rocket blasts off and is rising vertically at 880 feet per second. When the rocket is 4000 feet above the launching pad, how fast must the camera’s angle of elevation change so that it is still aimed at the rocket?

› Solution: We first sketch a diagram describing the motion of the rocket.


› Note that this question requires us to find . The angle and the height of the rocket, , are related by

› Differentiating this equation with respect to gives

› Thus

› To find , we need and when .


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ଶ

ଶ

› When ,

› This gives

› Since is constantly 880 ft/s, we have

› Question: What is the unit of the answer?CHAPTER 4 - APPLICATIONS OF DIFFERENTIATION

(PART I) 62

ଶ

› Example: A helicopter is holding its position 50 m from a hot air balloon and at the same altitude. An observer in the helicopter watches a parachutist jump from the balloon. The parachutist immediately opens his chute and falls at 5 m/s. How quickly do the observer’s eyes open down to keep the parachutist in sight at the following moments?(a) When the parachutist jumps.(b) When the parachutist has fallen 50 m.(c) When the parachutist has fallen 100 m.


calc12 unit4 notes part1€¦ · title: microsoft powerpoint - calc12_unit4_notes_part1 author:...

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