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Suppose
we have a 10 foot ladder leaning against a wall ...
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Say the
bottom of the ladder is sliding away from the wall |
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at the rate of 2
feet per minute. |
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How fast
is the top of the ladder sliding down when it is 4 feet off the
ground? |
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Assume
that the wall and the ground meet at a right angle and the ground is
flat. |
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OK, |
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The
ladder, the wall, and the ground will form a right triangle. |
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That
means we are going to see our old pal, The Pythagorean theorem! |
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Call the
direction of the wall Y and the direction of the ground X. |
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So: |
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We know
that 10 2 = X 2 + Y 2 |
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Taking
the derivative of this "with respect to time" gives us: |
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We know
some of the values at the time we are interested in ... |
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| Y
= 4 |
dY |
= ? |
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| X
= ? |
dX |
= 2
ft/min |
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| dt |
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We have a
bit of a problem here because we have 2 unknowns. |
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But
Wait! |
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X2
+ Y2 = 100 and Y = 4
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X2
+ 16 = 100
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X2 =
84
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X
= |
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X = 9.17
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Now we
can find dY/dt. |
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0 = 2(2)(9.17)
+ 2(4) |
dY |
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| dY |
= |
-36.68 |
=
-4.585 |
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| dt |
8 |
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The minus
sign is there because the ladder is moving down (towards zero
height). |
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Another
Example: |
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Someone
who is 6 feet tall is walking away from a lamp post |
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at a rate of 5
feet per minute. |
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The lamp
post is 20 feet tall. |
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The
person casts a shadow on the ground in front of them. |
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How fast
is the shadow growing when the person is 30 feet from the lamp post. |
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To solve
this problem we need a theorem from geometry called similar
triangles |
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Hey, that
and old Pythagoras are about the only big deals in geometry. |
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Similar
triangles goes like this |
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If two
triangles can be drawn |
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so that one triangle fits into the corner of
the other triangle, |
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and the
third sides of the two triangles are parallel to each other, |
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then the
ratio of any two sides in the little triangle |
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is the
same as the ratio of the corresponding two sides in the larger
triangle. |
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| b |
= |
B |
and |
c |
= |
C |
and |
b |
= |
B |
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| a |
A |
a |
A |
c |
C |
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Maybe you
remember from trig we used this idea |
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to make something called the
Law of Sines |
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Back to
the Problem! |
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Now for
the all important setup ... |
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Call the
distance from the lamp post to the person X |
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and the
distance from the person to the tip of the shadow Y. |
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The
person is walking 5 feet per minute, so dX/dt = 5. |
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We want
to find out how fast the end of the shadow is moving |
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so we are
looking for dY/dt. |
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From the
similar triangle idea we get the equation: |
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We can
simplify this puppy. |
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First
multiply by 120 |
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20Y
= 6(X + Y)
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20Y = 6X + 6Y) |
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14Y
= 6X
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Find
the derivative of this
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with
respect to time. |
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But we
already know that dX/dt = 5, so we can solve for dY/dt. |
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But
wait!
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The
problem wanted to know how fast the end of he shadow was
growing |
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when the
person was 30 feet from the wall. |
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We didn't
use the 30 feet anywhere. |
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No we
didn't. |
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That
means that it doesn't matter how far the person is from the
wall. |
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The
shadow is growing at the same rate all the time. |
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I wonder
why???? |
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One
Last Example: |
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A
triangular trough 4 feet wide at the top, 6 feet deep, and 20 feet
long |
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is being
filled with water at the rate of 10 cubic feet per minute. |
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How fast
is the water rising when it is 2 feet deep? |
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The
volume of the water in the trough is equal to the area of the
triangle |
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on the end
times the
length. |
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As the
trough fills, both the b and the h of the water change. |
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That
means we have two variables. |
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But
here's the trick ... |
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We can
use the idea of similar triangles here too! |
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b
= 4 and h = 6
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So at any
time during the fill, |
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the base (width) of the water triangle will be
2/3 of the height. |
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Looking
at the trough from the end ... |
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With this
information, we can go back to the volume equation |
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and
substitute for either b or h. |
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Since we
want to find dh/dt, we will substitute for b and keep h around. |
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b = |
2 |
h
and L = 20 (L doesn't change) |
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Finding
the derivative of this with respect to time, we have: |
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In the
original problem, |
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we were
told that there was 10 cubic feet of water per minute |
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coming into
the trough. |
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We are interested in the time when the water in the trough was 2 feet
deep. |
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That is,
where: |
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So
substituting these into the equation, we have |
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Solving
this for dh/dt. |
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So when
the water in the trough is 2 feet deep, |
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the water
is rising at the rate of 3/8 ft per minute. |
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copyright 2005 Bruce Kirkpatrick |
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