Sample Problems. Lecture Notes Related Rates page 1

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Lecture Note Related Rate page 1 Sample Problem 1. A city i of a circular hape. The area of the city i growing at a contant rate of mi y year). How fat i the radiu growing when it i exactly 15 mi? (quare mile per. A phere i growing in uch a manner that it radiu increae at 0: m volume increaing when it radiu i m long? (meter per econd). How fat i it 3. A phere i growing in uch a manner that it volume increae at 0: m3 fat i it radiu increaing when it i 7 m long? (cubic meter per econd). How. A cube i decreaing in ize o that it urface i changing at a contant rate of 0:5 m. volume of the cube changing when it i 7 m 3? How fat i the 5. A ladder 0 ft long lean againt a vertical building. If the top of the ladder lide down at a rate of p 3 ft, how fat i the bottom of the ladder liding away from the building when the top of the ladder i 10 ft above the ground? 6. A tank, haped like a cone hown on the picture, i being lled up with water. The top of the tank i a circle with radiu 5 ft, it height i 15 ft. Water i added to the tank at the rate of V 0 (t) ft3. How fat i the water level riing when the water level i 6 ft high? (The volume of a cone with height h and bae radiu r i V r h 3.) 7. A rotating light i located 18 feet from a wall. The light complete one rotation every 5 econd. Find the rate at which the light projected onto the wall i moving along the wall when the light angle i 5 degree from perpendicular to the wall. 8. The altitude of a triangle i increaing at a rate of : centimeter/ute while the area of the triangle i increaing at a rate of 1:5 quare centimeter/ute. At what rate i the bae of the triangle changing when the altitude i 11 centimeter and the area i 87 quare centimeter? 9. The area of a rectangle i kept xed at 100 quare meter while the legth of the ide vary. Expre the rate of change of the length of the vertical ide in term of the rate of change in the length of the other ide when a) the horizontal ide i 18 meter long b) the rectangle i a quare c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 10. Two quantitie p and q depending on t are ubject to the relation 1 p + 1 q 1: a) Expre p 0 (t) in term of q 0 (t). b) At a certain moment, p (t 0 ) 3 and p0 (t 0 ) : Find q (t 0 ) and q 0 (t 0 ) : 11. The bae radiu and height of a cylinder are contantly changing but the volume of the cylinder i kept at a contant 600 in 3. a) At a time t 1 the bae radiu i r (t 1 ) 10 in and it rate of change i r 0 (t 1 ) 0: in : Compute the rate of change of the height of the cylinder h (t) at time t 1. b) At a time t the height i h (t ) 1 in and it rate of change i r 0 (t ) 0:5 in : Compute the rate of change of the radiu of the cylinder r (t) at time t. 1. An object, dropped from a height of h ha a location of y (t) 16t + h feet after t econd. We dropped a mall object from a height of 60 feet. a) Where i the object and what i it velocity after 1:5 econd? b) Suppoe there i a 30 feet tall treet light 10 feet away from the point where the object will land. How far i the hadow of the object from the bae of the treet light at t 1:5? c) How fat i the obejct hadow moving at t 1:5? c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 3 Sample Problem - Anwer 1.) 1 15 mi y.) 1: 8 m3 0: 1 385 965 99 m3 3.) 1 980 m 3: 8 06 10 m.) 0:375 m3 5.) 1 ft 6.) 1 ft 7.) 36 5 ec (5 ) ft : 796 ft 9.) a) v 0 (t) 5 81 h0 (t) b) v 0 (t) h 0 (t) 10.) a) p 0 p q q0 b) 18 11.) a) 0: in b) 5 p in 8 0:17 31 in 8.) :89091 cm 1.) a) y (1:5) ft and y 0 (1:5) 8 ft b) 50 ft c) 00 ft Sample Problem - Solution 1. A city i of a circular hape. The area of the city i growing at a contant rate of mi (quare mile per y year). How fat i the radiu growing when it i exactly 15 mi? Solution: The area of a circle with radiu r i A r : Only thi time, both A and r are function of time: A (t) r (t) : It i alo given that A 0 (t) mi y : We di erentiate both ide of A (t) r (t) with repect to t. Let t 1 be the time when r (t 1 ) 15 mi: Then A (t) r (t) A 0 (t) r (t) r 0 (t) A 0 (t 1 ) r (t 1 ) r 0 (t 1 ) mi A 0 (t 1 ) r 0 (t 1 ) ) r 0 (t 1 ) A0 (t 1 ) r (t 1 ) r (t 1 ) y 15 mi 1 15 mi y. A phere i growing in uch a manner that it radiu increae at 0: m volume increaing when it radiu i m long? (meter per econd). How fat i it V (t) 3 r3 (t) Let t 1 be the time when r (t 1 ) m: Then V 0 (t) 3 3r (t) r 0 (t) r (t) r 0 (t) V 0 (t 1 ) r (t 1 ) r 0 (t 1 ) ( m) 0: m 1:8 m3 0: 1 386 m3 c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 3. A phere i growing in uch a manner that it volume increae at 0: m3 fat i it radiu increaing when it i 7 m long? (cubic meter per econd). How V (t) 3 r3 (t) or V 3 r3 V 0 (t) 3 3r (t) r 0 (t) V 0 3 3r r 0 V 0 (t) r (t) r 0 (t) V 0 r r 0 r 0 (t) V 0 (t) r (t) Let t 1 be the time when r (t 1 ) 7 m: Then m 3 r 0 V 0 r r 0 (t 1 ) V 0 (t 1 ) 0: r (t 1 ) (7 m) 0:000381 m. A cube i decreaing in ize o that it urface i changing at a contant rate of 0:5 m. volume of the cube changing when it i 7 m 3? How fat i the Solution: Let (t) denote the length of the edge of the cube at a time t. Let A (t) denote the urface area of the cube at a time t, and V (t) denote the volume of the cube. Recall the formula A 6 and V 3. A (t) 6 (t) A 0 (t) 6 (t) 0 (t) A 0 (t) 1 (t) 0 (t) V (t) 3 (t) V 0 (t) 3 (t) 0 (t) 3 (t) A 0 (t) 1 (t) (t) A0 (t) Let t 0 be the time at which the volume i 7 m 3. At that time, the edge are 3 m long. V 0 (t 0 ) (t 0) A 0 3 m 0:5 m (t 0 ) 0:375 m3 5. A ladder 0 ft long lean againt a vertical building. If the top of the ladder lide down at a rate of p 3 ft, how fat i the bottom of the ladder liding away from the building when the top of the ladder i 10 ft above the ground? Solution: Let x (t) and y (t)denote the horizontal and vertical poition of the endpoint of the ladder. With thi notation, y 0 (t) p 3 ft : By the Pythagorean theorem, x (t) + y (t) 00. We di erentiate both ide and olve for x 0 (t) : x (t) + y (t) 00 or x + y 00 x (t) x 0 (t) + y (t) y 0 (t) 0 xx 0 + yy 0 0 x 0 (t) y (t) y 0 (t) x 0 yy0 x (t) x c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 5 Let t 1 denote the time when x (t 1 ) 10 ft: x 0 (t 1 ) y (t 1) y 0 (t 1 ) x (t 1 ) p3 ft (10 ft) q(0 ft) (10 ft) 10 p 3 ft p 300 ft 1 ft 6. A tank, haped like a cone hown on the picture, i being lled up with water. The top of the tank i a circle with radiu 5 ft, it height i 15 ft. Water i added to the tank at the rate of V 0 (t) ft3. How fat i the water level riing when the water level i 6 ft high? (The volume of a cone with height h and bae radiu r i V r h 3.) Solution: Let h (t) and r (t) denote the height and radiu of the urface. There i a imple connection between thee two: The two triangle hown on the picture above are imilar. r 5 h and o r h 15 3. V (t) 1 3 r (t) h (t) 1 3 h (t) 3 h (t) 1 7 h3 (t) V 0 (t) 1 7 3h (t) h 0 (t) 1 9 h (t) h 0 (t) V 0 (t) 1 9 h (t) h 0 (t) 9V 0 (t) h (t) h 0 (t) Let t 1 denote the time when h (t 1 ) 6 ft: h 0 (t 1 ) 9V 0 9 ft3 (t 1 ) h (t 1 ) (6 ft) 1 ft c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 6 7. A rotating light i located 18 feet from a wall. The light complete one rotation every 5 econd. Find the rate at which the light projected onto the wall i moving along the wall when the light angle i 5 degree from perpendicular to the wall. Solution: Let x denote the ditance one the wall between the location of the light and the perpendicular ditance. Let denote the angle from the perpendicular. Uing thi notation, we are given d and we are aked dx. (t) and x (t) are related to each other. that 0 rad (t) for all t. 5 We need to nd how the two quantitie, It i given We di erentiate both ide with repect to t. tan ( (t)) x (t) 18 ft In the left-hand ide, we apply the chain rule: ec ( (t)) 0 (t) x0 (t) 18 ft (18 ft) ec ( (t)) 0 (t) x 0 (t) Let t 0 be the time when (t 0 ) 5. Then x 0 (t 0 ) (18 ft) ec ( (t 0 )) 0 (t 0 ) (18 m) ec (5 ) rad 5 The exact value of the anwer i 36 5 ec (5 ) ft. The approximate value i : 796 ft 8. The altitude of a triangle i increaing at a rate of : centimeter/ute while the area of the triangle i increaing at a rate of 1:5 quare centimeter/ute. At what rate i the bae of the triangle changing when the altitude i 11 centimeter and the area i 87 quare centimeter? Solution: The formula for the area of a triangle i A 1 bh. In thi cae, thee quantitie are function of time, i.e. they are A (t), b (t), and h (t). We olve for b. A 1 bh A bh b A h or rather b (t) A (t) h (t) c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 7 We di erentiate both ide. We will ue the quotient rule. 1:5 cm b 0 (t) A0 (t) h (t) A (t) h 0 (t) h (t) (11 cm) 87 cm : cm (11 cm) :89091 cm The negative ign indicate that at the time indicated, the ide b i becog horter at a rate of :89091 centimeter per ute. 9. The area of a rectangle i kept xed at 100 quare meter while the legth of the ide vary. Expre the rate of change of the length of the vertical ide in term of the rate of change in the length of the other ide when a) the horizontal ide i 18 meter long b) the rectangle i a quare Solution: Let v and h denote the length of the horizontal and vertical ide, repectively. Then clearly vh 100. We di erentiate both ide of thi with repect of time, and then olve for v 0 in term of h 0. h (t) v (t) 100 or vh 100 h 0 (t) v (t) + h (t) v 0 (t) 0 v 0 h + vh 0 0 v 0 (t) h 0 (t) v (t) v 0 h0 v h (t) h a) When h 18, then v 100 18 50 9 v 0 (t) and o h0 (t) v (t) h (t) 50 h 0 (t) 9 18 5 81 h0 (t) b) When the rectangle i a quare, then v h 10. v 0 (t) h0 (t) v (t) h (t) h0 (t) (10) 10 h 0 (t) 10. Two quantitie p and q depending on t are ubject to the relation 1 p + 1 q 1: a) Expre p 0 in term of q 0. 1 p + 1 q 0 ) p 0 q 0 p q 0 ) p0 p q q0 b) At a certain moment, p (t 0 ) 3 and p0 (t 0 ) : Find q (t 0 ) and q 0 (t 0 ) : 1 p (t 0 ) + 1 q (t 0 ) 1 ) 1 3 + 1 q (t 0 ) 1 ) q (t 0) q 0 q p p0 18 3 c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 8 11. The bae radiu and height of a cylinder are contantly changing but the volume of the cylinder i kept at a contant 600 in 3. a) At a time t 1 the bae radiu i r (t 1 ) 10 in and it rate of change i r 0 (t 1 ) 0: in : Compute the rate of change of the height of the cylinder h (t) at time t 1. Solution: V r h h V r V r h 0 rr 0 h + r h 0 0 rr 0 h + r h 0 rr 0 h r h 0 h 0 r 0 h r V r 0 r r r0 V r 3 0: in 600 in 3 (10 in) 3 0: in b) At a time t the height i h (t ) 1 in and it rate of change i r 0 (t ) 0:5 in : Compute the rate of change of the radiu of the cylinder r (t) at time t. r V Solution: V r h r h V r h 0 rr 0 h + r h 0 0 rr 0 h + r h 0 r 0 r h 0 rh rh0 h r! V h 0 h h! 600 in 3 0:5 in (1 in) (1 in) p 50 in 0:5 in in 5 p in 0:5 in in 5 8 p in 0:17 31 in 1. An object, dropped from a height of h ha a location of y (t) 16t + h feet after t econd. We dropped a mall object from a height of 60 feet. a) Where i the object and what i it velocity after 1:5 econd? Solution: we ubtitute t 1:5 into the formula y (t) 16t + 60 and get y (1:5 ) ft For the velocity, we di erentiate y (t) with reepct to t and evaluate the derivative at t 1:5. y 0 (t) 3t y 0 (1:5) 3 (1:5) 8 So y 0 (1:5 ) 8 ft c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 9 b) Suppoe there i a 30 feet tall treet light 10 feet away from the point where the object will land. How far i the hadow of the object from the bae of the treet light at t 1:5? Solution: hadow. Let u denote by y (t) the vertical poition of the object and by x (t) the horizontal poition of it By imilar triangle, we have that We ue thi equation rt to olve for x (t) In particular, x (1:5) y (t) 30 clear denoator x (t) 10 x (t) x (t) y (t) 30 (x (t) 10) x (t) y (t) 30x (t) 300 300 30x (t) x (t) y (t) 300 x (t) (30 y (t)) 300 30 y (t) x (t) 300 30 y (1:5) 300 30 50. c) How fat i the obejct hadow moving at t 1:5? Solution: To compute x 0 (1:5), we di erentiate x (t) y (t) 30 (x (t) 10) So x 0 (1:5) i y (t) x (t) 30 (x (t) 10) x 0 (t) y (t) + x (t) y 0 (t) 30x 0 (t) Solve for x 0 (t) x (t) y 0 (t) 30x 0 (t) x 0 (t) y (t) x (t) y 0 (t) x 0 (t) (30 y (t)) x (t) y 0 (t) 30 y (t) x 0 (t) (50 ft) 8 ft x 0 (1:5) 00 ft 30 ft ft The negative ign indicate that the hadow i traveling toward the left. c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 10 13. Two boat leave a port at the ame time, one traveling wet at 0 mi/hr and the other traveling outhwet at 15 mi/hr. At what rate i the ditance between them changing 30 after they leave the port? Solution: Let (x 1 ; y 1 ) denote the location of the car moving to the wet. Let (x ; y ) denote the location of the car moving outhwet. Then x 1 (t) 0t, y 1 (t) 0 and x (t) p 15 t and y (t) p 15 t. And then dx 1 dx p 15 15p and dy 15 Firt we will expre the ditance between the car. p 15p. 0, dy 1 For that, we will ue the Pythagorean theorem. 0, (x x 1 ) + ( y ) We di erentiate both ide with repect to t and olve for d d dx (x x 1 ) d d (x x 1 ) dx dx (x x 1 ) dx 1 dx 1 dx 1 + ( y ) dy + y dy + y q (x x 1 ) + ( y ) dy After 30 ute, x 1 10 y 1 0 and x p 15 1 15 p and y (t) 15 1 p time, d dx (x x 1 ) dx 1 dy + y q (x x 1 ) + ( y ) 15p. At that c Hidegkuti, Powell, 009 Lat revied: November, 015

Lecture Note Related Rate page 11 d d dx (x x 1 ) dx 1 dy + y q (x x 1 ) + ( y ) 15 p 15 p ( 10) v u t 15 ( 0)! + p ( 10) + 15p 15 p! 15 p! 3:75 p + 10 7:5 p + 0 + 5 1: 168 131 3:75 p + 10 + 15 Or: Let a be the diance from the origin of one car and b the ditance of the origin from the other car. Then a + b ab co 5 we di erentiate both ide: p! 0 aa 0 + bb 0 a 0 b + ab 0 p! aa 0 + bb 0 (a 0 b + ab 0 ) 0 p! 10 (0) + 7:5 (15) (0 (7:5) + 10 (15)) p 1: 168 131 10 + 7:5 (10) (7:5) co 5 For more document like thi, viit our page at http://www.teaching.martahidegkuti.com and click on Lecture Note. E-mail quetion or comment to mhidegkuti@ccc.edu. c Hidegkuti, Powell, 009 Lat revied: November, 015