



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. 





OK, 


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. 


So: 





We know
that 10 ^{2} = X ^{2} + Y ^{2} 





Taking
the derivative of this "with respect to time" gives us: 











We know
some of the values at the time we are interested in ... 


Y
= 4 
dY 
= ? 

dt 



X
= ? 
dX 
= 2
ft/min 

dt 






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





But
Wait! 


X^{2}
+ Y^{2} = 100 and Y = 4



X^{2}
+ 16 = 100



X^{2} =
84



X
= 




X = 9.17






Now we
can find ^{dY}/dt. 


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

dt 












dY 
= 
36.68 
=
4.585 


dt 
8 






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: 











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. 








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

















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. 








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


















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. 











b = 
2 
h
and L = 20 (L doesn't change) 

3 


















Finding
the derivative of this with respect to time, we have: 











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: 








So
substituting these into the equation, we have 








Solving
this for ^{dh}/dt. 








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 
