Calculus More Related Rates
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More Relatives
More Related Rates


 Suppose we have a 10 foot ladder leaning against a wall ...



 Say the bottom of the ladder is sliding away from the wall
 at the rate of 2 feet per minute.
 How fast is the top of the ladder sliding down when it is 4 feet off the ground?
 Assume that the wall and the ground meet at a right angle and the ground is flat.
 The ladder, the wall, and the ground will form a right triangle.
 That means we are going to see our old pal, The Pythagorean theorem!
 Call the direction of the wall Y and the direction of the ground X.


 We know that 10 2 = X 2 + Y 2
 Taking the derivative of this "with respect to time" gives us:
0 = 2X dX + 2Y dY

dt dt
 We know some of the values at the time we are interested in ...
Y = 4 dY

 = ?

X = ? dX

 = 2 ft/min

 We have a bit of a problem here because we have 2 unknowns.
 But Wait!

 X2 + Y2 = 100   and Y = 4

 X2 + 16 = 100

X2 = 84 

X =  

   X = 9.17

 Now we can find dY/dt.

0 = 2(2)(9.17) + 2(4)



0 = 36.68 + 8 





 = -4.585



 The minus sign is there because the ladder is moving down (towards zero height).
 Another Example:
 Someone who is 6 feet tall is walking away from a lamp post
 at a rate of 5 feet per minute.
 The lamp post is 20 feet tall.
 The person casts a shadow on the ground in front of them.
 How fast is the shadow growing when the person is 30 feet from the lamp post.


 To solve this problem we need a theorem from geometry called similar triangles
 Hey, that and old Pythagoras are about the only big deals in geometry.
 Similar triangles goes like this
 If two triangles can be drawn
 so that one triangle fits into the corner of the other triangle,
 and the third sides of the two triangles are parallel to each other,
 then the ratio of any two sides in the little triangle
 is the same as the ratio of the corresponding two sides in the larger triangle.


b = B and c = C and b = B

a A a A c C
 Maybe you remember from trig we used this idea 
to make something called the Law of Sines
 Back to the Problem!
 Now for the all important setup ...
 Call the distance from the lamp post to the person X 
 and the distance from the person to the tip of the shadow Y.


 The person is walking 5 feet per minute, so dX/dt = 5.
 We want to find out how fast the end of the shadow is moving
 so we are looking for dY/dt.
 From the similar triangle idea we get the equation:
Y = X + Y

6 20
 We can simplify this puppy. 
 First multiply by 120

 20Y = 6(X + Y)

20Y = 6X + 6Y)

 14Y = 6X

 Find the derivative of this

 with respect to time.
14 dY  = 6 dX

dt dt
 But we already know that dX/dt = 5, so we can solve for dY/dt.
14 dY  = 6(5)

dY = 30 = 15

dt 14 7

 But wait!

 The problem wanted to know how fast the end of he shadow was growing 
 when the person was 30 feet from the wall.
 We didn't use the 30 feet anywhere.
 No we didn't.
 That means that it doesn't matter how far the person is from the wall. 
 The shadow is growing at the same rate all the time.
 I wonder why????
 One Last Example:
 A triangular trough 4 feet wide at the top, 6 feet deep, and 20 feet long 
 is being filled with water at the rate of 10 cubic feet per minute.


 How fast is the water rising when it is 2 feet deep?
 The volume of the water in the trough is equal to the area of the triangle 
 on the end times the length.
V = 1 b x h x l

 As the trough fills, both the b and the h of the water change.
 That means we have two variables.
 But here's the trick ...
 We can use the idea of similar triangles here too!

 b = 4  and  h = 6

b = 4 = 2

h 6 3
b =  2


 So at any time during the fill,
 the base (width) of the water triangle will be 2/3 of the height.
 Looking at the trough from the end ...


 With this information, we can go back to the volume equation
 and substitute for either b or h.
 Since we want to find dh/dt, we will substitute for b and keep h around.
V = 1 b x h x l


b = 


h      and      L = 20 (L doesn't change)

V = 1 ( 2

h) h x 20

2 3
V = 20


 Finding the derivative of this with respect to time, we have:
dV = 20 2h dh

dt 3 dt
 In the original problem,
 we were told that there was 10 cubic feet of water per minute
 coming into the trough.
 We are interested in the time when the water in the trough was 2 feet deep.
 That is, where:

= 10   and   h = 2

 So substituting these into the equation, we have

10 =

20 2(2) dh

3 dt
 Solving this for dh/dt.
dh = 3


dt 8
 So when the water in the trough is 2 feet deep, 
 the water is rising at the rate of 3/8 ft per minute.

   copyright 2005 Bruce Kirkpatrick

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