Related Rates MATH 151 Calculus for Management J. Robert Buchanan Department of Mathematics 2014
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.
Example Given an equation relating x and y such as x 2 y + 3x + y 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. dx = 2
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 2 + 3 dx 2(1)(1)(2) + (1 2 ) dx = 2 d [ x 2 y + 3x + y 2] = d [5] + 2y + 3(2) + 2(1) 4 + + 6 + 2 = 0 = 0 = 0 = 10 3
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.
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
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?
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
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?
Solution We may use the Pythagorean Theorem. 2x dx x 2 + y 2 = (15) 2 + 2y = 0 = x dx y
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.
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. Substituting into the rate equation above, we have = (10 2)(22) 5 = 44 2 62.23 ft/sec.
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.
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
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 = 10 10.
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 = 10 10. Substituting into the rate equation above, we have dz = (30)(15) + (10)(10) 10 10 = 55 10 14.23 nm/hr.
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?
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
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 = 7.26. By using the volume formula we have h = 3 3V π = 3 3(29) π 3.02545.
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 = 7.26. By using the volume formula we have 3V 3(29) h = 3 π = 3 3.02545. π Substituting into the rate equation above, we have dh 7.26 = 0.2525 in/sec. π(3.02545) 2
Example A company that manufactures toys calculates that its costs and revenue can be modeled by the equations C = 57, 000 + 1.15x 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.