2. Related Rates Kirsten Maund Dahlia Sweeney

3. Background Calculus was invented to predict phenomena of change: planetary motion, objects in freefall, varying populations, etc. In many practical applications, several related rates vary together. Naturally, the rates at which they vary are also related to each other. With calculus, we can describe and calculate such related rates.

4. What are related rates?  A related rates problem involves two or more quantities that vary with time and an equation that expresses some relationship between them.  Typically, the values of these quantities at some instant are given together with all their time rates of change but one. The problem is usually to find the time rate of change this is not given, at some instant specified in the problem.

5. How to Solve Related Rates  One common method for solving such a problem is to begin with implicit differentiation of the equation that relates the given quantities.  For example, suppose that x and y are each functions of time such that: x^2 + y^2 = a^2 (a is a constant) x^2 + y^2 = a^2 (a is a constant)  Differentiate both sides of this equation with respect to time t. This produces the equation: 2x dx/dt + 2y dy/dt = 0 2x dx/dt + 2y dy/dt = 0

6.  If the values of x, y, and dx/dt at a certain instant t are known, then the last equation can be solved for the value of dy/dt at time t.  Note that it is not necessary to know x and y as functions of t.  It is typical for a related rates problem to contain insufficient information to express x and y as functions of t.

7. WARNING! The most common error to be avoided is the premature substitution of the given data, before rather than after implicit differentiation.

8. Strategy for Solving  Step 1 : Make a drawing of the situation if possible.

9.  Step 2: Use letters to represent the variables involved in the situation - say x, y.

10.  Step 3: Identify all rates of change given and those to be determined, Use the calculus notation (dx/dt, dy,dt, etc) to represent them.

11.  Step 4: Determine an equation that involves both  The variables from step two  The derivative of step three

12.  Step 5: Differentiate (by implicit differentiation) the equation of step four

13.  Step 6: Substitute all know values into the differentiated equation

14.  Step 7: Use algebraic manipulation,if necessary, to solve for the unknown rate or quantity

15. Formulas You May Need To Know

16. Example #1  A ladder 10 feet long is resting against a wall. If the bottom of the ladder is sliding away from the wall at a rate of 1 foot per second, how fast is the top of the ladder moving down when the bottom of the ladder is 8 feet from the wall?  First, draw the picture:

17.  We have dx/dt is one foot per second. We want to find dy/dt.  X and y are related by the Pythagorean Thereom  Differentiate both sides of this equation with respect to t to get  When x = 8 ft, we have  Therefore  The top of the ladder is sliding down (because of the negative sign in the result) at a rate of 4/3 feet per second.

18. Example #2  A man 6 ft tall walks with a speed of 8 ft per second away from a street light atop an 8 foot pole. How fast is the tip of his shadow moving along the ground when he is 100 feet from the light pole. 18 ft z - xx z 6 ft

19.  Let x be the man’s distance from the pole and z be the distance of the tip of his shadow from the base of the pole.  Even though x and z are functions of t, we do not attempt to obtain implicit formulas for either.  We are given that dx/dt = 8 (ft/sec), and we want to find dz/dt when x = 100 (ft).  We equate ratios of corresponding sides of the two similar triangles and find that z/18 = (z-x)/6  Thus 2z = 3x

20.  Implicit differentiation now gives 2 dz/dt = 3 dx/dt  We substitute dx/dt = 8 and find that (dz/dt = 3/2) * (dx/dt = 3/2) * (8) = 12 So the tip of the man’s shadow is moving at 12 ft per second.

21. Try Me!  A ladder 25 ft long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec, how fast is the top of the ladder sliding down the wall, when the bottom is 15 ft from the wall?

22. Solution  t = the number of seconds in time that has elapsed since the ladder started to slide down the wall.  y = the number of feet in distance from the ground to the top of the ladder at t seconds.  x = the number of feet in the distance from the bottom of the ladder to the wall at t seconds.

23.  Because the bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec, dx/dt = 3. We wish to find dy/dt when x = 15.  From the Pythagorean Thereom, we have y^2 = 625 – x^2  Because x and y are functions of t, we differentiate both sides of equation one with respect to t and obtain 2y dy/dt = -2x dx/dt giving us dy/dt = -x/y dx/dt

24.  When x = 15, it follows from equation one that y = 20.  Because dx/dt = 3, we get from equation two: dy/dt = (-15/20) * 3 = -9/4  Therefore, the top of the ladder is sliding down the wall at the rate of 2 ¼ ft/sec when the bottom is 15 ft from the wall.  The significance of the minus sign is that y is decreasing as t is increasing.

25. Was Your Answer Correct?

26. Bibliography http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217ladder.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217ladder.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217ladder.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217ladder.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217baseball.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217baseball.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217baseball.htm http://www.math.dartmouth.edu/~klbook site/2.17/217examples/217baseball.htm © Maund and Sweeney 2011