Related Rates. MATH 151 Calculus for Management. J. Robert Buchanan. Department of Mathematics. J. Robert Buchanan Related Rates

Transcription

1 Related Rates MATH 151 Calculus for Management J. Robert Buchanan Department of Mathematics 2014

2 Related Rates Problems Another common application of the derivative involved situations in which two or more related quantities are changing with time. These are called related rates problems.

3 Example Given an equation relating x and y such as x 2 y + 3x + y 2 = 5

4 Example Given an equation relating x and y such as then if we can find x 2 y + 3x + y 2 = 5 x = 1, y = 1, and using implicit differentiation. dx = 2

5 Example Given an equation relating x and y such as then if we can find x 2 y + 3x + y 2 = 5 x = 1, y = 1, and using implicit differentiation. 2xy dx + x dx 2(1)(1)(2) + (1 2 ) dx = 2 d [ x 2 y + 3x + y 2] = d [5] + 2y + 3(2) + 2(1) = 0 = 0 = 0 = 10 3

6 Guidelines 1 If possible, draw a picture to illustrate the problem and label the pertinent quantities. 2 Set up an equation relating all of the relevant quantities. 3 Differentiate (implicitly) both sides of the equation with respect to time (t). 4 Substitute in the values for all known quantities and rates. 5 Solve for the remaining unknown rate.

7 Useful, Common Formulas Pythagorean Theorem : a 2 + b 2 = c 2 Area of triangle : A = 1 2 b h Circumference of circle : C = 2πr Area of circle : A = πr 2 Surface area of cube : S = 6x 2 Volume of cube : V = x 3 Volume of sphere : V = 4 3 πr 3 Volume of cone : V = 1 3 πr 2 h

8 Example Suppose the radius of a circle is increasing at 7 cm/s. How fast is the area increasing when the radius is 20 cm?

9 Example Suppose the radius of a circle is increasing at 7 cm/s. How fast is the area increasing when the radius is 20 cm? da A = πr 2 = 2πr dr = 2π(20)(7) = 280π cm 2 /s

10 Example A painter is painting a house using a ladder 15 feet long. A dog runs by the ladder dragging a leash that catches the bottom of the ladder and drags it directly away from the house at a rate of 22 feet per second. How fast is the top of the ladder moving down the wall when the top of the ladder is 5 feet from the ground?

11 Solution We may use the Pythagorean Theorem. 2x dx x 2 + y 2 = (15) 2 + 2y = 0 = x dx y

12 Solution We may use the Pythagorean Theorem. 2x dx x 2 + y 2 = (15) 2 + 2y = 0 = x dx y We are told dx = 22, y = 5, and by the Pythagorean Theorem x = (15) 2 (5) 2 = 200 = 10 2.

13 Solution We may use the Pythagorean Theorem. 2x dx x 2 + y 2 = (15) 2 + 2y = 0 = x dx y We are told dx = 22, y = 5, and by the Pythagorean Theorem x = (15) 2 (5) 2 = 200 = Substituting into the rate equation above, we have = (10 2)(22) 5 = ft/sec.

14 Example Two ships sail from the same port. The first ship leaves port at 1:00AM and travels east at a speed of 15 nautical miles per hour. The second ship leaves port at 2:00AM and travels north at a speed of 10 nautical miles per hour. Determine the rate at which the ships are separating at 3:00AM.

15 Solution We may use the Pythagorean Theorem. 2(x + 15) dx (x + 15) 2 + y 2 = z 2 + 2y dz = 2z dz dx (x + 15) + y = z

16 Solution We may use the Pythagorean Theorem. 2(x + 15) dx (x + 15) 2 + y 2 = z 2 + 2y dz = 2z dz dx (x + 15) + y = z We are told dx = 15, = 10, x = 15, y = 10, and by the Pythagorean Theorem z = (30) 2 + (10) 2 = 1000 =

17 Solution We may use the Pythagorean Theorem. 2(x + 15) dx (x + 15) 2 + y 2 = z 2 + 2y dz = 2z dz dx (x + 15) + y = z We are told dx = 15, = 10, x = 15, y = 10, and by the Pythagorean Theorem z = (30) 2 + (10) 2 = 1000 = Substituting into the rate equation above, we have dz = (30)(15) + (10)(10) = nm/hr.

18 Example Flour sifted onto waxed paper forms a pile in the shape of a cone with equal radius and height. The volume of flour in the pile is increasing at a rate of 7.26 in 3 /s. How fast is the height of the flour increasing when the volume is 29 in 3?

19 Solution We may use the formula for the volume of a cone. V = 1 3 πr 2 h dv dh = 1 3 πh3 = πh 2 dh = dv πh 2

20 Solution We may use the formula for the volume of a cone. V = 1 3 πr 2 h = 1 3 πh3 dv = πh 2 dh dv dh = πh 2 We are told V = 29 and dv = By using the volume formula we have h = 3 3V π = 3 3(29) π

21 Solution We may use the formula for the volume of a cone. V = 1 3 πr 2 h = 1 3 πh3 dv = πh 2 dh dv dh = πh 2 We are told V = 29 and dv = By using the volume formula we have 3V 3(29) h = 3 π = π Substituting into the rate equation above, we have dh 7.26 = in/sec. π( ) 2

22 Example A company that manufactures toys calculates that its costs and revenue can be modeled by the equations C = 57, x R = 505x x 2 50 where x is the number of toys produced in a week. If the level of production is 4500 toys and is increasing at a rate of 200 toys per week, find 1 the rate at which the cost is changing, 2 the rate at which the revenue is changing, 3 the rate at which the profit is changing.