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print barcode image c# Solution: Diagram the problem in Visual Studio .NET
Solution: Diagram the problem EAN13 Supplement 5 Encoder In .NET Using Barcode generation for Visual Studio .NET Control to generate, create European Article Number 13 image in .NET framework applications. EAN / UCC  13 Recognizer In Visual Studio .NET Using Barcode decoder for VS .NET Control to read, scan read, scan image in .NET applications. on an xy coordinate system with the first ship going in the y direction and the second ship going in the xdirection Figure 63 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 Encode Barcode In Visual Studio .NET Using Barcode printer for VS .NET Control to generate, create bar code image in Visual Studio .NET applications. Barcode Decoder In VS .NET Using Barcode decoder for .NET Control to read, scan read, scan image in Visual Studio .NET applications. + , Print European Article Number 13 In C#.NET Using Barcode encoder for .NET framework Control to generate, create EAN / UCC  13 image in .NET framework applications. GTIN  13 Printer In Visual Studio .NET Using Barcode generator for ASP.NET Control to generate, create European Article Number 13 image in ASP.NET applications. = 6  Draw EAN13 Supplement 5 In Visual Basic .NET Using Barcode maker for Visual Studio .NET Control to generate, create EAN13 image in .NET applications. Matrix 2D Barcode Printer In .NET Using Barcode drawer for .NET framework Control to generate, create 2D Barcode image in Visual Studio .NET applications. m 1000m
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The separation of the ships is a straight Pythagorean theorem problem
s = J24OO2 I28002= 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 64 This rate of separation is to be evaluated at t = 300s (Fig 64) [(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 conicalshaped 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 12fi 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 64A 3A 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 65 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 65 and 66) Notice that the triangle with sides zx 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 zx 3 2x
z 12 Fig 66 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

