121616949-math.137 - of that side of the triangle was...

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6.2 Related Rates 123 . . . . . . . . . . (0 , a ( t )) ( b ( t ) , 0) c ( t ) P . . . . Figure 6.11 Cars moving apart. so ˙ c = a ˙ a + b ˙ b c = a ˙ a + b ˙ b a 2 + b 2 . Substituting known values we get: ˙ c = 10 · 80 + 15 · 100 10 2 + 15 2 = 460 13 127 . 6km/hr at the time of interest. Notice how this problem differs from example 6.14 . In both cases we started with the Pythagorean Theorem and took derivatives on both sides. However, in example 6.14 one of the sides was a constant (the altitude of the plane), and so the derivative of the square
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Unformatted text preview: of that side of the triangle was simply zero. In this example, on the other hand, all three sides of the right triangle are variables, even though we are interested in a specific value of each side of the triangle (namely, when the sides have lengths 10 and 15). Make sure that you understand at the start of the problem what are the variables and what are the constants....
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