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on an x-y coordinate system with the first ship going in the y direction and the second ship going in the x-direction Figure 6-3 is for t = 0 , the time when the ships cross paths The drawing helps to visualize the problem The position of the first ship at any time t is y = (8m/s)t d y - 8 y t s dt
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The position of the second ship at any time t is x = 1000m + (6 m/s)l The separation of the ships is from the Pythagorean theorem s = d x
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The position of the first ship at 300 s is its speed (8 m/s ) times the 300 s :
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YI 300 = (8 m/s)(300 s) = 2400 m
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The position of the second ship at 300 s is the 1000m plus the 6 m/s times the 300 s :
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XI 3oo = 1000m +(6 m/s)(300 s) = (1000 +1 800)m = 2800 m
The separation of the ships is a straight Pythagorean theorem problem
s = J24OO2 -I-28002= 3688m
RELATED RATE PROBLEMS
The rate at which they are separating is the fun, er calculus, part of the problem The rate at which they are separating is, in calculus talk, dsldt, and we already have the &/dt and dy/dt Start with the separation written in Pythagorean theorem form s =(x2 + y 2 ) l J 2 and differentiate, carefully
2400m
\l/,
2800m
+
I Writing ds =-(x2 +y2)-l12d(x2 y 2 ) + 2 as the first step will help to prevent errors with 1/2's and the minus signs
1 (2xdx+ 2ydy) and finally Continuing, ds = -(x2 +y2)-l12 2
-=6- m dt s
Fig 6-4
This rate of separation is to be evaluated at t = 300s (Fig 6-4)
[(2800m)(6m/s) + (2400m)(8rn/s)]= 98m
These next two problems utilize similar triangles to write the d e f ~ equation for the g problem The first problem, concerning the rate a shadow of something is moving, is in nearly every calculus book The following problem concerning a conical-shaped container is also in nearly every calculus book in one form or another If you know how to use similar triangles to "get started" on a problem you will have mastered yet another category of related rate problems
Remember
away from a 12-fi tall penguin way light What is the length of her shadow and how fast is the tip of her shadow moving when she is 40 Et away from the light
6-4A 3-A tall penguin (Penny) is taking a leisurely stroll at O S A/S
Solution:
CHAPTER^
when you see a
triangle in a related rate problem look for similar triangles Don't start the problem looking for derivatives Concentrate on the defining equation for the problem The derivatives come later
Your first order of business in a Fig 6-5 related rate problem is to find relationshps between the variables In this problem set up the triangle, complete with known numbers, and then label some of the distances The change in length of the hypotenuse of h s triangle is not what we are looking for It is lengths along the ground: the length fkom the light to Penny and the length of her shadow Take x as the length from beneath the light to Penny, and z as the length from beneath the light to the end of her shadow The length of her shadow is z - x Draw this triangle (refer to Figs 6-5 and 6-6)
Notice that the triangle with sides z--x and 3 is similar to the triangle with sides z and 12 Similar triangles are triangles with the same angles and their sides in proportion This means that the ratios of the sides are equal z-x --3
2--x
z 12
Fig 6-6
Eliminating the fraction, 122 - 12x = 32 or 9z = 12x or 32 = 4x , produces a simple relationship between x and z The related derivative rates are dz - - _ -4 dt
3 dt
Notice that x and z don't enter into the rate relationship Penny is moving at
ft -= 05 dx
so the tip of her shadow is moving at
-= --
dz dt
dx - 4 ( o 5 t ) = o 6 7 T ft
3 dt
Since Penny is walking away from the light at 05ft/s and the tip of her shadow is growing at 067 ft/s her shadow is getting longer as she moves away from the light
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