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Related Rates Notes Steps: 1. Draw picture.

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Related Rates Notes Steps: 1. Draw picture. 2. Write equation(s).

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Related Rates Notes Steps: 1. Draw picture. 2. Write equation(s). 3. Write all variables (include rates of change).

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Related Rates Notes Steps: 1. Draw picture. 2. Write equation(s). 3. Write all variables (include rates of change). 4. Get equation in terms of one variable (may need another equation).

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Related Rates Notes Steps: 1. Draw picture. 2. Write equation(s). 3. Write all variables (include rates of change). 4. Get equation in terms of one variable (may need another equation). 5. Take derivative with respect to t.

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Related Rates Notes Steps: 1. Draw picture. 2. Write equation(s). 3. Write all variables (include rates of change). 4. Get equation in terms of one variable (may need another equation). 5. Take derivative with respect to t. 6. Substitute and solve.

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Related Rates Notes Steps: 1. Draw picture. 2. Write equation(s). 3. Write all variables (include rates of change). 4. Get equation in terms of one variable (may need another equation). 5. Take derivative with respect to t. 6. Substitute and solve. 7. Answer the question.

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Related Rates Notes Right Δ's: a 2 + b 2 = c 2 (Ladder problems)

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Related Rates Notes Right Δ's: a 2 + b 2 = c 2 (Ladder problems) Circles: A = πr 2 C = 2 πr

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Related Rates Notes Right Δ's: a 2 + b 2 = c 2 (Ladder problems) Circles: A = πr 2 C = 2 πr Cones: V = (1/3)πr 2 h

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Related Rates Notes Right Δ's: a 2 + b 2 = c 2 (Ladder problems) Circles: A = πr 2 C = 2 πr Cones: V = (1/3)πr 2 h Cylinders: V = πr 2 h

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Related Rates Notes Right Δ's: a 2 + b 2 = c 2 (Ladder problems) Circles: A = πr 2 C = 2 πr Cones: V = (1/3)πr 2 h Cylinders: V = πr 2 h Spheres: V = (4/3)πr 3 S = 4πr 2

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Related Rates Notes Example: The famous ladder problem -- A.P.' s absolute favorite!

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Related Rates Notes Example: The famous ladder problem -- A.P.' s absolute favorite! Question: A ladder, 10 feet long is leaning against a building.

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Related Rates Notes Example: The famous ladder problem -- A.P.' s absolute favorite! Question: A ladder, 10 feet long is leaning against a building. The bottom of the ladder is moved away from the building at the constant rate of 0.5 ft/sec.

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Related Rates Notes Example: The famous ladder problem -- A.P.' s absolute favorite! Question: A ladder, 10 feet long is leaning against a building. The bottom of the ladder is moved away from the building at the constant rate of 0.5 ft/sec. Find the rate at which the ladder is falling down the building when the ladder is 6 feet from the building.

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Related Rates Notes Step #1 Draw Picture

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Related Rates Notes Step #1 Draw Picture 10 x y

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Related Rates Notes Step #2 Write equation(s). 10 x y

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Related Rates Notes Step #2 Write equation(s). x 2 + y 2 = 10 2 10 x y

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Related Rates Notes Step #2 Write equation(s). x 2 + y 2 = 10 2 10 x y Note: You may substitute a value only AFTER taking the derivative if that value is changing!

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6dx/dt = 0.5

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6dx/dt = 0.5 y 2 = 8

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6dx/dt = 0.5 y 2 = 8 (6 2 + y 2 = 10 2 )

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6dx/dt = 0.5 y 2 = 8 (6 2 + y 2 = 10 2 )dy/dt = ?

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6dx/dt = 0.5 y 2 = 8 (6 2 + y 2 = 10 2 )dy/dt = ? Step #4 Get equation in terms of one variable (may need another equation).

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Related Rates Notes Step #3 Write all variables (include rates of change). x 2 = 6dx/dt = 0.5 y 2 = 8 (6 2 + y 2 = 10 2 )dy/dt = ? Step #4 Get equation in terms of one variable (may need another equation). No second equation to substitute for this problem

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Related Rates Notes Step #5Take derivative with respect to t. x 2 + y 2 = 10 2

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Related Rates Notes Step #5Take derivative with respect to t. x 2 + y 2 = 10 2 (dx/dt)(2x) + (dy/dt)(2y) = 0

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Related Rates Notes Step #5Take derivative with respect to t. x 2 + y 2 = 10 2 (dx/dt)(2x) + (dy/dt)(2y) = 0 Recall: x = 6 and y = 8 when dx/dt = 0.5

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Related Rates Notes Step #5Take derivative with respect to t. x 2 + y 2 = 10 2 (dx/dt)(2x) + (dy/dt)(2y) = 0 Step #6(0.5)(2*6) + (dy/dt)(2*8) = 0 Recall: x = 6 and y = 8 when dx/dt = 0.5

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Related Rates Notes Step #5Take derivative with respect to t. x 2 + y 2 = 10 2 (dx/dt)(2x) + (dy/dt)(2y) = 0 Step #6(0.5)(2*6) + (dy/dt)(2*8) = 0 6 + 16(dy/dt) = 0

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Related Rates Notes Step #5Take derivative with respect to t. x 2 + y 2 = 10 2 (dx/dt)(2x) + (dy/dt)(2y) = 0 Step #6(0.5)(2*6) + (dy/dt)(2*8) = 0 6 + 16(dy/dt) = 0 dy/dt = -6/16 = -3/8

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Related Rates Notes Step #7 The ladder is falling at a rate of 3/8 feet per second.

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Related Rates Notes The famous CONE problem -- A.P.'s 2 nd favorite question!

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Related Rates Notes The famous CONE problem -- A.P.'s 2 nd favorite question! If a cone is expanding or contracting, then the radius and height remain proportional r 1 /h 1 = r 2 /h 2

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Related Rates Notes The famous CONE problem -- A.P.'s 2 nd favorite question! If a cone is expanding or contracting, then the radius and height remain proportional r 1 /h 1 = r 2 /h 2 Thus, you can substitute for r such that your V = (1/3)πr 2 h equation will only have h.

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Related Rates Notes The famous CONE problem -- A.P.'s 2 nd favorite question! If a cone is expanding or contracting, then the radius and height remain proportional r 1 /h 1 = r 2 /h 2 Thus, you can substitute for r such that your V = (1/3)πr 2 h equation will only have h. Thus, you won't have to do the product rule!

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Related Rates Notes Example: Water is withdrawn from a conical reservoir 20 ft in diameter and 50 ft deep (vertex down) at the constant rate of 12 cubic feet per minute.

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Related Rates Notes Example: Water is withdrawn from a conical reservoir 20 ft in diameter and 50 ft deep (vertex down) at the constant rate of 12 cubic feet per minute. How fast is the water level falling when the depth of water in the reservoir is 20 ft?

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Related Rates Notes Step #1 Draw Picture

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Related Rates Notes Step #1 Draw Picture 10 50

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Related Rates Notes Step #1 Draw Picture 10 50 h2h2

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Related Rates Notes Step #1 Draw Picture 10 50 h2h2

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Related Rates Notes Step #1 Draw Picture 10 50 h2h2

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Related Rates Notes Step #1 Draw Picture 10 50 h2h2

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Related Rates Notes Step #1 Draw Picture 10 50 h2h2

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Related Rates Notes Step #1 Draw Picture 10 50 h2h2 r2r2

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Related Rates Notes Step #2 V = (1/3)πr 2 h 10 50 h2h2 r2r2

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Related Rates Notes Step #2 V = (1/3)πr 2 h and (10/50) = (r 2 /h 2 ) 10 50 h2h2 r2r2

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Related Rates Notes Step #2 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2

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Related Rates Notes Step #2 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2 r 2 = (10/50) h 2

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Related Rates Notes Step #2 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2

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Related Rates Notes Step #3 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12

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Related Rates Notes Step #3 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ?

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Related Rates Notes Step #4 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ?

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Related Rates Notes Step #4 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) 10 50 h2h2 r2r2 So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ?

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Related Rates Notes Step #4 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/3)π(0.2h) 2 h

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Related Rates Notes Step #4 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/3)π(0.2h) 2 h V = (1/3)π (.04h 2 )h

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Related Rates Notes Step #4 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/3)π(0.2h) 2 h V = (1/3)π (.04h 2 )h V = (1/75)πh 3

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Related Rates Notes Step #4 V = (1/3)πr 2 h and (10/50) = (r 2 / h 2 ) So, 50r 2 = 10 h 2 r 2 = (10/50) h 2 r 2 = 0.2 h 2 h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/3)π(0.2h) 2 h V = (1/3)π (.04h 2 )h V = (1/75)πh 3 Use this equation for Volume!

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Related Rates Notes Step #5 Take derivative with respect to time h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3

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Related Rates Notes Step #5 Take derivative with respect to time h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt) -12=(1/25)π(20) 2 (dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt) -12=(1/25)π(20) 2 (dh/dt) -12=(1/25)π(400)(dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt) -12=(1/25)π(20) 2 (dh/dt) -12=(1/25)π(400)(dh/dt) -12= 16π(dh/dt)

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Related Rates Notes Step #6 Solve for dh/dt h 2 = 20 dv/dt = -12 V = r 2 = 0.2*20 = 4 dh/dt = ? V = (1/75)πh 3 (dv/dt) = (1/25)πh 2 (dh/dt) -12 = (1/25)πh 2 (dh/dt) -12=(1/25)π(20) 2 (dh/dt) -12=(1/25)π(400)(dh/dt) -12= 16π(dh/dt) dh/dt = -3/(4π)

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Related Rates Notes Step #7 The water level is falling at a rate of -3/(4π) feet per minute!

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