RELATED RATE PROBLEMS

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QuickTime 7 free download.

1a.  An 8 foot long ladder is leaning against a wall.  The top of the ladder is sliding down the wall at the rate of 2 feet per second.  How fast is the bottom of the ladder moving along the ground at the point in time when the bottom of the ladder is 4 feet from the wall.  Click here for an animation.  

Solution:  y = distance from the top of the ladder to the ground

               x = distance from the bottom of the ladder to the wall

dy/dt = -2      Find dx/dt when x = 4 (and y = 4(3^1/2) by the Pythagorean Theorem)

x² + y² = 64

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

2x dx/dt - 4y = 0

dx/dt = 4y/(2x) = 2(3^1/2) ft/sec when x = 4 ft

1b.  A ladder 10 feet long is standing straight up against the side of a  house.  The base of the ladder is pulled away from the side of the house at the rate of 2 feet per second.  How high up the side of the house will the top of the ladder be 1 second after the base begins being pulled away from the house?  How high up the side of the house will the top of the ladder be after 2 seconds, 3 seconds, 4 seconds, 5 seconds?  Here is an animation of the ladder being pulled away from the side of the house.

Can you see what is happening to the speed at which the top of the ladder is coming down when the top of the ladder gets close to the ground?  Clicking on the picture produces an animation.

Do the problem above as a related rate problem, i.e., determine dy/dt, the rate at which the top of the ladder is coming down the wall.  Does this help you to see what is happening to the speed at which the top of the ladder is coming down when the top of the ladder gets close to the ground?  

Here is a Quicktime Version of the animation.  Try computing the average speed of the top of the ladder from time t = 4.99999998 seconds to time t = 4.99999999 seconds.  This would be found by computing

Winplot Demonstration (LadderSlide--use T to slide the ladder)    LadderSlide2 (This one includes a display of the approximate speed of the top of the ladder using a difference quotient like the one shown above.)  LadderSlide3 (This one displays the approximate speed of the top of the ladder computed from a difference quotient and compares this result to a decimal approximation of the speed of the top of the ladder using the derivative.)

Click here to read about the paradox involved in pulling the bottom of a ladder out away from a wall at a constant rate.

2.  A boat is being pulled toward a dock by means of a rope attached to the front tip of the bow.  Initially there are 30 feet of rope out and the rope is taught and being reeled in by a circular device the top of which is 10 feet higher than the point where the rope is attached to the boat.  This circular device has a radius of 1 foot and turns at the rate of one revolution every pi seconds.  How fast is the boat moving along the water when there are 15 feet of rope out?  Click here for an animation.  Quicktime animation

Solution:  y = amount of rope out

               x = distance the boat is from the base of the dock

The circumference of the circular device is 2pi ft. so 2pi feet of rope are pulled in every pi seconds.  Thus y is decreasing by 2pi/pi = 2 ft/sec. i.e., dy/dt = -2.

We want to find dx/dt when y = 15 (and so x = sqrt(15² - 10²) by the Pythagorean Theorem).

x² + 10² = y²

2x dx/dt = 2y dy/dt

dx/dt = -4y/(2x) = -6sqrt(5)/5 ft/sec when y = 15 ft (negative because x is decreasing)

What happens when the boat gets very close to the base of the dock?

3.  Water is being poured into a conical reservoir at the rate of pi cubic feet per second.  The reservoir has a radius of 6 feet across the top and a height of 12 feet.  At what rate is the depth of the water increasing when the depth is 6 feet?   Solution:  h = depth of the water in the reservoir                     r = radius of the water in the reservoir   The volume of the water in the reservoir (V) is given by    V = (1/3)pi(r2h)    and    r = (1/2)h using similar triangles so V = (1/12)pi(h3) pi = dV/dt = (1/4)pi(h2)dh/dt dh/dt = 4/h2 dh/dt when h = 6 is 1/9 ft/sec   Click here or on the top picture at the right to see an animation.  Can you determine the function I used in the animation to represent the changing depth (h) as a function of time, assuming h = 0 when t = 0?  DPGraph Animation  DPGraph Animation2

4.  A football is dropped from a height of 64 feet and at a horizontal distance of 16 feet from a light that is 64 feet above the ground at the top of a light pole.  How fast is the shadow of the ball moving along the ground one second after the ball is dropped?  Neglect air resistance so that the distance the football will have dropped as a function of time will be s = 16t² with the football dropped at t = 0.  Can you draw in the right triangle with sides of lengths 16 and 16t² missing from the top of the picture below? 

Solution:  Let x = the distance the football's shadow is from the base of the light pole.                Using similar triangles we have                   x / 64 = 16 / 16t2                 x = 64t -2                 dx/dt = -128t -3                 dx/dt (at t = 1) = -128 ft/sec   The negative answer indicates the distance between the shadow and the base of the light pole is decreasing.  Click here or on the picture at the right to see an animation and click here to see an animation with scales.  Quicktime version

5.  A man 6 feet tall is walking toward a lamppost 20 feet high at a rate of 5 feet per second.  The light at the top of the lamppost (20 feet above the ground) is casting a shadow of the man.  At what rate is the tip of his shadow moving and at what rate is the length of his shadow changing when he is 10 feet from the base of the lamppost?  Click here to see an animation.

6.  Section 2.6:  44     

A fish is reeled in at a rate of 1 foot per second from a point 10 feet above the water.  At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out?   Animation Scales Animation (Fish Has An Eye)    Quicktime Version Animation No Scales

7.  A circle starts out with a radius of 1 cm. (at time t = 0) and begins growing.  The area of the circle is increasing at the rate of 2 cm² per sec.  Find the rate of change of the radius of the circle when the radius is 5 cm.  In the animation time (t) is going from 0 to almost 40pi.  What is the radius when t = 40pi?

8.  Page 160 Review #112

See the picture in the textbook and below  and click here to see an animation.

9.  A large red balloon is rising at the rate of 20 ft/sec.  The balloon is 10 ft above the ground at the point in time that the back end of a green car is directly below the bottom of the balloon.  The car is traveling at 40 ft/sec.  Animation of the car and the balloon.  What is the rate of change of the distance between the bottom of the balloon and the point on the ground directly below the back of the car one second after the back of the car is directly below the balloon?  Click here or on the picture below for an animation.  Click here for an animated picture with no labeling scales.  Quicktime version  Quicktime extended version

Notice that if the balloon had been starting from the ground instead of 10 feet above the ground , i.e.,

10.  Section 2.6 #37

Here is a statement of the problem for Section 2.6 #37 in Calculus 8th Edition, Larson, Hostetler, Edwards Machine Design   The endpoints of a movable rod of length 1 meter have coordinates (x,0) and (0,y).  The position of the end of the rod on the x-axis is where t is the time in seconds. (a)  Find the time of one complete cycle of the rod. (b)  What is the lowest point reached by the end of the rod on the y-axis? (c)  Find the speed of the y-axis endpoint when the x-axis endpoint is (1/4,0). Click here or on the picture above to see an animation.  Click here to see an animation without scales.

11.  Section 2.6 #49

Security Camera Animation

12.  Section 2.6 #26  A Water Trough

Animation without scales Animation with scales DPGraph Animation DPGraph Animation2

13.  Section 2.6 #29

In the animation the building is in green and the red circular path is that of the end of the pipe.  The blue point moving along the red circular path is indicating the speed the end point of the pipe would move at if it was moving at a constant speed and taking the required length of time to reach the top of the building.  The blue point on the small purple circle on top of the building is indicating the speed with which the winch is rotating to pull in the purple rope at -0.2m/sec.  The pipe is in blue.  Click here or on the top picture at the right to see the animation.  Quicktime version  Quicktime Extended version

14.  The end of a pulley (at point C in the picture) is being pulled down at the rate of 1 ft/sec.  The distance from point A to point D is 2 feet.  The distance from point A to point B is y feet.  A weight pictured by the green rectangle is being lifted by the pulley.  Find the rate (dy/dt) at which the weight is rising and find dy/dt when y = 3 feet.  Also find y as a function of time (t) given that when t = 0 sec., y = 6 feet.  Click here or on the picture to see an animation of the motion.  Quicktime Animation

15.  Filling a paraboloid The top picture at the right represents a cross section of the paraboloid.  The paraboloid was formed by revolving the graph of y = x2, x going from -2 to 2, about the y-axis.  Click on the picture to see an animation. DPGraph 3D Animation DPGraph 3D Animation2 DPGraph 3D Animation3 Filling a Paraboloid Using the scrollbar to increase the transparency in DP animation3 makes for a nice effect.

17.  Find the rate of change of the volume of a cylinder when its radius is 6 feet if its height is always (3/2) times its radius and its radius is increasing at the rate of 2 feet per minute.  Here is a DPGraph Picture of the expanding cylinder.

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        Lane Vosbury, Mathematics, Seminole State College   email:  vosburyl@seminolestate.edu

        This page was last updated on 08/21/14          Copyright 2002          webstats