VIEWS: 17 PAGES: 25 POSTED ON: 6/27/2012 Public Domain
Applications of Derivatives Related Rates Related Rates Always involves calculating the rate at which one variable changes from the rate at which another variable is known to change. Strategies for solving related rates problems • Sketch a diagram and name the variables and constants. • Any rate such as mph, rpm, radians/min … these are all derivatives. • Be careful to take note of all numerical data, you will use all of it. • Know what you are solving for: usually a derivative of an equation that connects all the given variables. • Take derivatives, substitute, and solve. Some examples to follow … The Bubble Problem: How fast does the radius of a spherical soap bubble change when air is 3 blown into it at a rate of 10 cm when radius is 3cm? sec So, what’s this? The rate of change of volume! dV cm3 10 dt sec … OK, so what are we looking for? How would you describe it mathematically? How fast the radius is changing … dr dt Which equation connects the volume of a sphere to its radius? 4 V r3 3 LET’S TAKE DERIVATIVES 4 dV If V r 3 then find 3 dt dV 4 2 dr 3 r dt 3 dt Let’s substitute the given information… dr 10 4 (3) 2 dt dr 0.0884 cm sec dt DONE One more example … A ladder 26 ft long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 4 ft/sec. How fast is the top sliding down the wall when the foot is 10 ft from the wall? What facts do you know? dx 4 ft / sec dt When x = 10, you can find y by Pythagorean Formula … Y = 24 The equation that connects these variables is: y 2 262 x 2 NOTE: we need to solve for dy/dt … and you should expect the answer to be negative!!! Why? dy dx Taking derivatives produces … 2y 0 2x dt dt From the problem we know that y = 24, x = 10, dx/dt = 4 … substitute dy 2 24 2 10 4 dt dy 5 ft dt 3 sec We’re done! Related Rates Suppose a painter is standing on a 13 foot ladder and Joe ties a rope to the bottom of the ladder and walks away at the rate of 2 feet per second. Related Rates Suppose a painter is standing on a 13 foot ladder and Joe ties a rope to the bottom of the ladder and walks away at the rate of 2 feet per second. How fast is the painter falling when x = 5 feet? Related Rates •Write an equation x y 13 2 2 2 • Differentiate the equation implicitly. 2x x’ + 2y y’ = 0 or xx’ + yy’ = 0 • If Joe pulls at 2 ft./sec., find the speed of the painter when x = 5. •This means we need to solve for y’, and we need to know length y. Related Rates Use algebra to find y when x is 5. x2 + y2 = 169 52 + y2 = 169 y2 = 169 – 25 = 144 y = 12 Related Rates Back to the calculus with y = 12, x = 5, and x’ = 2 ft/sec xx’ + yy’ =0 5(2) + 12(y’) = 0 y’ = -10/12 = -5/6 ft./sec. Related Rates Summary Even though Joe is walking 2 ft/sec, the painter is only falling -5/6 ft/sec. If the x and y values were reversed, 12(2) + 5(y’) = 0 or y’ = -24/5. Related Rates Suppose a 6 ft tall person walks away from a 13 ft lamp post at a speed of 5 ft per sec. How fast is the tip of his shadow moving when 12 ft from the post? 6 s Related Rates Suppose a 6 ft tall person walks away from a 13 ft lamp post at a speed of 5 ft per sec. How fast is the tip of his shadow moving when 12 ft from the post? 6 13 s Related Rates Suppose a 6 ft tall person walks away from a 13 ft lamp post at a speed of 5 ft per sec. How fast is the tip of his shadow moving when 12 ft from the post? 6 13 s sx Related Rates The tip of the shadow has a speed of (s+x)’, not s’. What is s’? s’ is the growth of the shadow and includes getting shorter on the right. 6 13 s sx Related Rates Cross multiplying 6( s x) 13s s ' x ' 13s ' 6 7s ' 5 6 Related Rates Thus s’ is 30 and 30 65 s' s ' x ' 5 7 that the tip is moving7 Note almost 7 twice as fast as the walker, and more than twice as fast as the shadow regardless of x. Related Rates Suppose a radar gun on first base catches a baseball 30 feet away from the pitcher and registers 50 feet per second. How fast is the ball really traveling? Related Rates The calculus. 2 45 x y 2 2 2 2 xx ' 2 yy ' X = 30 y’ = 50 y = ? The algebra. 2 45 30 y 2 2 2 4950 y Related Rates X = 30 y’ = 50 y = ?4950 y Back to the calculus. 2 xx ' 2 yy ' 2(30) x ' Related Rates Back to the calculus. X = 30 y’ = 50 y = ?4950 y 2 xx ' 2 yy ' 2(30) x ' 2 4950 y ' Related Rates Back to the calculus. X = 30 y’ = 50 y = ?4950 y 2 xx ' 2 yy ' 2(30) x ' 2 4950(50) Related Rates x’ = 117.260 feet/sec. 2 xx ' 2 yy ' 4950 y X = 30 y’ = 50 y = ? 2(30) x ' 2 4950(50) 4950 (50) x' 30 And now it’s time for you mathlings to learn to fly