Related Rates Notes Steps: 1. Draw picture.. Related Rates Notes Steps: 1. Draw picture. 2. Write...

Related Rates Notes

Steps:1. Draw picture.

Related Rates Notes

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

Related Rates Notes

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

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).

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.

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.

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.

Related Rates NotesRight Δ's: a2 + b2 = c2 (Ladder problems)

Circles: A = πr2 C = 2 πr

Cones: V = (1/3)πr2h

Cylinders: V = πr2h

Spheres: V = (4/3)πr3 S = 4πr2

Related Rates Notes

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

Related Rates Notes

Question: A ladder, 10 feet long is leaning against a building.

Related Rates Notes

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.

Related Rates Notes

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.

Related Rates NotesStep #1 Draw Picture

Related Rates NotesStep #2 Write equation(s).

Related Rates NotesStep #2 Write equation(s). x2 + y2 = 102

Note: You may substitute a value only AFTER taking the derivative if that value is changing!

Related Rates Notes

Step #3 Write all variables (include rates of change).

x2 = 6

Related Rates Notes

x2 = 6 dx/dt = 0.5

Related Rates Notes

x2 = 6 dx/dt = 0.5

y2 = 8

Related Rates Notes

x2 = 6 dx/dt = 0.5

y2 = 8 (62 + y2 = 102)

Related Rates Notes

x2 = 6 dx/dt = 0.5

y2 = 8 (62 + y2 = 102) dy/dt = ?

Related Rates Notes

x2 = 6 dx/dt = 0.5

y2 = 8 (62 + y2 = 102) dy/dt = ?

Step #4

Get equation in terms of one variable (may need another equation).

Related Rates Notes

x2 = 6 dx/dt = 0.5

y2 = 8 (62 + y2 = 102) dy/dt = ?

Step #4

Get equation in terms of one variable (may need another equation).

No second equation to substitute for this problem

Related Rates Notes

Step #5 Take derivative with respect to t.

x2 + y2 = 102

Related Rates Notes

x2 + y2 = 102

(dx/dt)(2x) + (dy/dt)(2y) = 0

Related Rates Notes

x2 + y2 = 102

(dx/dt)(2x) + (dy/dt)(2y) = 0

Recall: x = 6 and y = 8 when dx/dt = 0.5

Related Rates Notes

x2 + y2 = 102

(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

Related Rates Notes

x2 + y2 = 102

(dx/dt)(2x) + (dy/dt)(2y) = 0

Step #6 (0.5)(2*6) + (dy/dt)(2*8) = 0

6 + 16(dy/dt) = 0

Related Rates Notes

x2 + y2 = 102

(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

Related Rates Notes

Step #7

The ladder is falling at a rate of 3/8 feet per second.

Related Rates Notes

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

Related Rates Notes

If a cone is expanding or contracting, then the radius and height remain proportional r1/h1 = r2/h2

Related Rates Notes

Thus, you can substitute for r such that your V = (1/3)πr2h equation will only have h.

Related Rates Notes

Thus, you can substitute for r such that your V = (1/3)πr2h equation will only have h.

Thus, you won't have to do the product rule!

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.

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?

Related Rates NotesStep #1 Draw Picture

Related Rates NotesStep #2 V = (1/3)πr2h

Related Rates NotesStep #2 V = (1/3)πr2h and (10/50) = (r2/h2)

Related Rates NotesStep #2 V = (1/3)πr2h and (10/50) = (r2/ h2)

So, 50r2 = 10 h2

r2 = (10/50) h2

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/3)π(0.2h)2h

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/3)π(0.2h)2hV = (1/3)π (.04h 2 )h

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/3)π(0.2h)2hV = (1/3)π (.04h 2 )h

V = (1/75)πh 3

So, 50r2 = 10 h2

r2 = (10/50) h2

r2 = 0.2 h2

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/3)π(0.2h)2hV = (1/3)π (.04h 2 )h

V = (1/75)πh 3

Use this equation for Volume!

Related Rates NotesStep #5 Take derivative with respect to time

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/75)πh 3

Related Rates NotesStep #5 Take derivative with respect to time

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/75)πh 3 (dv/dt) = (1/25)πh 2(dh/dt)

Related Rates NotesStep #6 Solve for dh/dt

h2 = 20

dv/dt = -12

r2 = 0.2*20 = 4

dh/dt = ?

V = (1/75)πh 3 (dv/dt) = (1/25)πh 2(dh/dt)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

h2 = 20

dv/dt = -12

r2 = 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)

Related Rates Notes

Step #7

The water level is falling at a rate of -3/(4π) feet per minute!

Related Rates Notes Steps: 1. Draw picture.. Related Rates Notes Steps: 1. Draw picture. 2. Write...

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