# 4 6 Related Rates by 6zt86X

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Applications of Derivatives

Related Rates
Related Rates
Always involves calculating the rate at which one variable
changes from the rate at which another variable is known to
change.

Strategies for solving related rates problems

• Sketch a diagram and name the variables and constants.
• Any rate such as mph, rpm, radians/min … these are all
derivatives.
• Be careful to take note of all numerical data, you will use all of
it.
• Know what you are solving for: usually a derivative of an
equation that connects all the given variables.
• Take derivatives, substitute, and solve.

The Bubble Problem:
How fast does the radius of a spherical soap bubble change when air is
3
blown into it at a rate of 10 cm     when radius is 3cm?
sec

So, what’s this?
The rate of change of volume!

dV      cm3
 10
dt          sec

… OK, so what are we looking for? How would you describe it
mathematically?
How fast the radius is changing …
dr
dt
Which equation connects the volume of a sphere to its radius?

4
V   r3
3
LET’S TAKE DERIVATIVES
4                   dV
If   V   r 3 then find
3
dt

dV     4 2 dr
 3  r
dt     3    dt
Let’s substitute the given information…

dr
10  4 (3)  2

dt

dr
0.0884 cm         
sec       dt

DONE
One more example …

A ladder 26 ft long rests on horizontal ground and leans against a vertical wall.
The foot of the ladder is pulled away from the wall at the rate of 4 ft/sec.
How fast is the top sliding down the wall when the foot is 10 ft from the wall?

What facts do you know?

dx
 4 ft / sec
dt

When x = 10, you can find y by
Pythagorean Formula … Y = 24

The equation that connects these variables is:        y 2  262  x 2

NOTE: we need to solve for dy/dt … and you should expect the answer to
be negative!!! Why?
dy          dx
Taking derivatives produces …     2y       0  2x
dt          dt

From the problem we know that y = 24, x = 10, dx/dt = 4 … substitute

dy
2  24       2  10  4
dt
dy    5
  ft
dt    3 sec

We’re done!
Related Rates
Suppose a painter is standing on a 13
foot ladder and Joe ties a rope to the
bottom of the ladder and walks away at
the rate of 2 feet per second.
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Suppose a painter is standing on a
13 foot ladder and Joe ties a rope to
the bottom of the ladder and walks
away at the rate of 2 feet per
second.

How fast is the painter
falling when x = 5 feet?
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•Write an equation
x  y  13
2     2       2

• Differentiate the equation implicitly.
2x x’ + 2y y’ = 0 or
xx’ + yy’ = 0
• If Joe pulls at 2 ft./sec., find the speed of the painter
when x = 5.

•This means we need to solve for y’, and we need to
know length y.
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Use algebra to find y when x is 5.
x2 + y2 = 169
52 + y2 = 169
y2 = 169 – 25 = 144
y = 12
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Back to the calculus with y = 12,
x = 5, and x’ = 2 ft/sec
xx’ + yy’ =0
5(2) + 12(y’) = 0
y’ = -10/12 = -5/6 ft./sec.
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Summary
Even though Joe is walking 2 ft/sec,
the painter is only falling -5/6 ft/sec.
If the x and y values were reversed,
12(2) + 5(y’) = 0 or y’ = -24/5.
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Suppose a 6 ft tall person walks away
from a 13 ft lamp post at a speed of 5 ft
per sec. How fast is the tip of his
shadow moving when 12 ft from the
post?

6

s
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Suppose a 6 ft tall person walks away from a 13 ft
lamp post at a speed of 5 ft per sec. How fast is the
tip of his shadow moving when 12 ft from the post?

6 13

s
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Suppose a 6 ft tall person walks away from a 13 ft
lamp post at a speed of 5 ft per sec. How fast is the
tip of his shadow moving when 12 ft from the post?

6   13

s sx
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The tip of the shadow has a speed of
(s+x)’, not s’. What is s’?
s’ is the growth of the shadow and includes getting
shorter on the right.

6   13

s sx
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Cross multiplying

6( s  x)  13s        s ' x ' 
13s '
6
7s '
5
6
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Thus s’ is

30 and                  30      65
s'               s ' x '   5 
7 that the tip is moving7
Note                     almost
7
twice as fast as the walker, and more than twice as fast as
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Suppose a radar gun on first base
catches a baseball 30 feet away from
the pitcher and registers 50 feet per
second. How fast is the ball really
traveling?
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The calculus.
2  45   x  y
2       2          2

2 xx '  2 yy '
X = 30 y’ = 50 y = ?
The algebra.
2  45  30  y
2       2       2

4950  y
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X = 30 y’ = 50 y = ?4950  y
Back to the calculus.
2 xx '  2 yy '
2(30) x ' 
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Back to the calculus.
X = 30 y’ = 50 y = ?4950  y
2 xx '  2 yy '
2(30) x '  2 4950 y '
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Back to the calculus.
X = 30 y’ = 50 y = ?4950  y
2 xx '  2 yy '
2(30) x '  2 4950(50)
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x’ = 117.260 feet/sec.
2 xx '  2 yy ' 4950  y
X = 30   y’ = 50 y = ?

2(30) x '  2 4950(50)
4950 (50)
x' 
30
And now it’s time for you mathlings to learn to fly 

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