MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS 0 Page 1 of 12 EXAMPLE 1: A man 6 ft. tall walks at a rate of 2 ft./sec away from a lamp post that is 13 ft. high. At what rate is the length of his shadow changing when he is 25 ft. away from the lamp post. x represents his distance away from the lamp post. y represents the length of his shadow. ππ π + π = π π 13 π π π π π = π π π π π πππ = ππ + ππ x y π= π π π π π π ππ = (π) = ππ/πππ π π π π x+y The 25 did not matter. = 20 x β1 EXAMPLE 2: A spotlight on the ground shines on a wall 10 m away. A man 2 m tall walks from the spotlight toward the wall at a speed of 1.2m/s. How fast is his shadow on the wall decreasing when he is 3 m from the spotlight? x represents his distance away from the spotlight ym y represents the length of his shadow on the wall xm 10 m Use Similar Triangles Differentiate with respect to t π π = π ππ π π π π = βπππβπ π π π π ππ = ππ π π = βππ(π)βπ (π. π) π π π= ππ = πππβπ π π π π = β π/π π π π U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 2 of 12 EXAMPLE 3: A rocket rising vertically is being tracked by a radar command post on the ground 10 km from the launch pad. How fast is the rocket rising when it is 8 km high and its distance from the radar station is increasing at a rate of 1000 km/h? At any given time z km x km ππ + πππ = ππ ππ π π π π π + π = ππ π π π π π π π π =π π π π π 10 km When it is 8 km high ππ + πππ = ππ π = βππ + πππ = βπππ = πβππ = ππ. π π π π π π =π π π π π (π) π π = (πβππ)(ππππ) π π π π (πβππ)(ππππ) = = πππβππ π π π π π = ππππ π π The rocket is rising at 1600 km/h when it is 8 km high. U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 3 of 12 EXAMPLE 4: A passenger car approaches a railway crossing from the east at 50 km/h, while a locomotive approaches the crossing from the north at 110 km/h. How fast is the distance between them changing at the moment when the car is 30 m and the train is 40 m from the intersection? π π + ππ = π π Draw a diagram Differentiate with respect to t ππ π π = βπππππ/π π π z x π π π π π π π + ππ = ππ π π π π π π π π π π π π +π =π π π π π π π y π π = βππππ/π π π Find the distance between them when the car is 30 m and the train is 40 m from the crossing. πππ + ππ = ππ π = βπππ + πππ = ππ ππ Substitute into the derivative (ππ)(βπππ) + (ππ)(βππ) = (ππ) βππππ = (ππ) π π π π π π π π π π = βπππ ππ/π π π The distance between them is decreasing at 118 km/h U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 4 of 12 EXAMPLE 5: A 5 m ladder which is leaning against a wall begins to slip. The top slips down the wall at a rate of 60 cm/s. a) How fast is the bottom moving when it is 3 m from the wall? π π + ππ = ππ 5m Differentiate implicitly xm ο±ο π (ππ ) π (ππ ) π (ππ ) + = π π π π π π ym π π = βππ π/π π π π π =? ? ? ? π/π π π ππ π π π π + ππ =π π π π π Find the height of the ladder when the ladder is 3m from the π wall. π π π +π =π π = βππ β ππ = π Substitute into the derivative ππ π π π π + ππ =π π π π π π(π)(βππ) + π(π) βπππ + π π π =π π π π π =π π π π π = ππ ππ/π π π U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 5 of 12 EXAMPLE 5 continued b) At what rate is the angle formed by the ladder and the floor decreasing when the base of the ladder is 3m from the wall? (Round to 2 decimals) Using Trigonometric Ratios 5m xm π¬π’π§ π½ = ο± ym π π = βππ π/π π π ππ¨π¬ π½ π π = π π π π π½ π π π = π π π π π π When the base of the ladder is 3 m from the wall ππ¨π¬ π½ = π ππ¨π¬ π½ π π½ π π π = π π π π π π π π½ π = (βππ) π π π π π π½ π = (βππ) ( ) = βππ π π π The angle is decreasing at a rate of 20 rads/s when the base of the ladder is 3 m from the wall. U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 6 of 12 EXAMPLE 6: Two people are 50 feet apart. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min. At what rate is distance between the two people changing when π = 0.5 radians? Method 1 Method 2 ππ ππ¨π¬ π½ = = πππβπ π π π¬ππ π½ = ππ π¬ππ π½ πππ§ π½ π π½ π π π = π π ππ π π β π¬π’π§ π½ π π½ π π = βπππβπ π π π π π π½ π π = π π π π β π¬π’π§ π½ π π½ βππ π π = π π π π π π πππ¬ππ π½ πππ§ π½ ππ π¬ππ π. π πππ§ π. π(π. ππ) = π π = π. ππ ππ/πππ π π π π π π π π ππ π¬π’π§ π½ π π½ = π π ππ π π When π½ = π. π πππ ππ ππ¨π¬ π. π = π ππ π= ππ¨π¬ π. π π = ππ. ππ π π (ππ. ππ)π π¬π’π§(π. π) = (π. ππ) π π ππ π π = π. ππ ππ/πππ π π U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 7 of 12 EXAMPLE 7: Optional A triangle has 2 sides of length 15 cm and 17 cm. The angle between these sides is increasing at a rate π πππ/π . At what rate is the length of the third side increasing when the opposite angle 24 π measures A πππ ? 3 of 17 cm 15 cm a Using the Law of Cosines: a2 = b2 + c2 β 2bc cos A ππ = πππ + πππ β π(ππ)(ππ) ππ¨π¬ π¨ ππ π π π π¨ = π + π β πππ π¬π’π§ π¨ π π π π π π π π¨ = βπππ π¬π’π§ π¨ π π π π π πΎπππ π¨ = πππ π ππ π ππ = πππ + πππ β π(ππ)(ππ) ππ¨π¬ ( ) π π ππ = πππ + πππ β πππ ( ) = πππ π π = βπππ π(βπππ) π π π π = βπππ π¬π’π§ ( ) ( ) π π π ππ π π = π. π π π The side is increasing at a rate of approximately 1.8 cm/s U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS ASSIGNMENT /3 Page 8 of 12 MARK______________21 + 4 = ____________% 1. A streetlight is 6 m tall. A person who is 2 m tall is walking away from the light at a rate of 1.5 m/s. How fast is the shadow cast by the person moving away from the base of the light post? x represents his distance away from the light post. y represents the length of the shadow By similar triangles π π = π+π π x ππ = ππ + ππ y ππ = ππ x+y π π π π π = π π π π π π= π π π = (π. ππ/π ) π π π π π = π. ππ π/π π π π π π The shadow is moving at a speed of 0.75 + 1.5 = 2.25 m/s away from the light post. 2. A spotlight set on the ground is shining on a wall 20 m away. A person 1.7 m tall walks away from the light toward the wall at a rate of 1.4 m/s. At what rate is the shadow cast on the wall decreasing in height when the person is 14 m from the wall? /3 y π. π π = π ππ ππ = ππ 1.7 m x 20 m dy ο½ 1.4m / s dx π ππ = ππ = βπ π πβπ πππ = βπππ π π π π π π π = βππ(ππ)βπ (π. π) π π π π = βπ. ππ π/π π π The shadow is decreasing in height at a speed of 0.24 m/s. U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 9 of 12 ππ₯ 3. A plane flies horizontally with a speed of 600 km/h ( ππ‘ ) at an altitude of 10 km and passes directly over the town of Quinton. Find the rate at which the direct ππ§ distance (ππ‘ ) is increasing when it has flown 20 km. y At any given time 102 + y2 = z2 Differentiate with respect to t. 10km z π π π π π + ππ = ππ π π π π π π π π Q π =π π π π π 0 + 2y dy = 2z dz dt dt /3 20 km 10km z=? Q dy = the z dzplane is 20 km from Find zywhen dt dt Quinton. 2 2 2 20 + 10 = z z2 = 500 π = βπππ = ππβπ = ππ. πππ Substitute into the derivative. Give your answer to the nearest whole km/h. π π (ππ)(πππ) = ππβπ π π π π πππππ = = πππβπ ππ πππ ππ/π π π ππβπ U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 10 of 12 4. Joe is driving west at 60 km/h and Dave is driving south at 70 km/h. Both cars are approaching the intersection of the two roads. At what rate is the distance between the cars decreasing when Joeβs car is 0.4 km and Daveβs is 0.3 km from the intersection? π π + ππ = π π β 70 km/h π π π π π π ππ + ππ = ππ π π π π π π /3 z x π (π. π)π + (π. π) = ππ π = π. π ππ π π π π π π +π =π π π π π π π β 60 km/h y (π. π ππ)(β ππ ππ/π ) + (π. π ππ)(β ππ ππ/π ) = (π. π ππ) π π π π π π (π. π ππ)(β ππ ππ/π ) + (π. π ππ)(β ππ ππ/π ) = =β ππππ/π π π π. πππ 5. A ladder 4 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 0.30 m/s. a) How quickly is the top of the ladder sliding down the wall when the bottom of the ladder is 2 m from the wall? 4m x ππ¦ = 0.30 π/π ππ‘ y At any given time x2 + y2 = 42 Differentiate with respect to t. π π π π ππ + ππ =π π π π π π π π π π +π =π π π π π Find x when the ladder is 2 m from the wall. x2 + 22 = 42 x2 = 12 π = βππ = πβπ = π. π Substitute into the derivative. Give your answer to the nearest hundredth m/s. π π πβππ + ππ(π. πππ/π ) = π π π π π βππ(π. πππ/π ) βπ. ππ = = ππ β π. ππππ/π π π πβπ π βπ U4 L5b ANS Related Rates Distances with Cos Law 2015 MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 11 of 12 b) How fast is the angle between the ladder and the ground decreasing when the bottom of the ladder is 2 m from the wall? /3 From above π = βππ = πβπ ππ π. π when the ladder is y = 2 m from the wall. ππ₯ = β0.173 π/π ππ‘ Method 2 Method 1 π π π¬π’π§ π½ = = π π π ο 4m x ππ¨π¬ π½ π½ y=2m When x = πβπ π¬π’π§ π½ = π¦ 1 cos π = = π¦ 4 4 π π½ π π π = π π π π π β sinΞΈ π π π½ π = (β π. πππ) π π π π ππ¦ = 0.3 π/π ππ‘ π π½ π π = (βπ. πππ) × π π π π = βπ. ππππππ /πππ β ππ ππ‘ 40 x cm ο± rad y cm π¬π’π§ π½ = ππ¨π¬ π½ /3 π ππ π π½ π π π = π π ππ π π ππ π π½ π = × (βπ) ππ π π ππ π π½ π ππ π = × (βπ) × =β πππ /π π π ππ ππ π U4 L5b ANS Related Rates Distances with Cos Law 2015 = βπ π β3 ππ 1 = (0.30) 2 ππ‘ 4 1 = 4 (0.30) × β 2 β3 = β0.087rad/sec 6. The length of the hypotenuse of a right triangle is constant at 64 cm. The vertical leg of the triangle is decreasing at a rate of 5 cm/s. At what rate is the angle opposite the vertical side changing when the horizontal side is 40 cm long? When y = 40 cos π = 64 π ππ 1 ππ¦ = ππ‘ 4 ππ‘ The angle between the ladder and the ground is decreasing at a rate of β 0.0867 rad/s 64 cm πβπ MATH 31 UNIT 4 LESSON #5b RELATED RATES: DISTANCES NAME ANSWERS Page 12 of 12 Question 7 is a bonus question 7. Two sides of a triangle have fixed lengths of 20 cm and 24 cm. The third side is increasing at a rate of 3.5 cm/s. When the third side is 15 cm in length, what is the rate of change for the angle between the fixed sides? ο±ο 20 cm /4 da ο½ 3.5cm / s dt 24 cm Using the Law of Cosines:a a2 = b2 + c2 β 2bc cos A da ο½ 3.5cm / s dt cm ππ = πππ + πππ β π(ππ)(ππ) ππ¨π¬ π¨ ππ = πππ β πππ ππ¨π¬ π¨ ππ π π π π¨ = π β πππ (βπ¬π’π§ π¨) π π π π ππ π π π π¨ = πππ π¬π’π§ π¨ π π π π When a = 15 cm πππ + πππ β πππ ππ¨π¬ π¨ = π(ππ)(ππ) A= 0.672 rad π π¨ ππ π π = π π πππ π¬π’π§ π¨ π π π π¨ π(ππ)(π. π) = = π. πππππ /π π π πππ(π¬π’π§ π. πππ) U4 L5b ANS Related Rates Distances with Cos Law 2015
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