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3.9 Related Rates notes

3.9 Related Rates notes - Related Rates Problems There isnt...

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Related Rates Problems There isn’t really any new theory in this section – this is an application of implicit differentiation. Example 1. Suppose x is some function of t . Suppose V = x 3 . So V is also a function of t . Now find dV dt if x = 10 and dx dt = 5. We differentiate V = x 3 w.r.t. t to obtain dV dt = 3 x 2 dx dt . So we put x = 10 and dy dt = 5 into that to obtain dV dt = 3(10) 2 (5) = 1500. Example 2. Suppose x 2 + y 2 = xy + 13. Find dx dt if x = 3, y = 4 and dy dt = 5. We implicitly differentiate: 2 x dx dt + 2 y dy dt = dx dt y + x dy dt . Now put x = 3, y = 4 and dy dt = 5 into that. We get 2(3) dx dt + 2(4)(5) = dx dt (4) + (3)(5) which we solve to obtain dx dt = - 25 / 2. Example 3. Suppose a cubical crystal is growing. At a particular point in time, it has edges 10 mm in length, and the edges are growing at a rate of 5 mm per hour. How fast is the volume of the crystal increasing at that time? To solve this problem, we know that a cube with edges x has volume V = x 3 . We are told that x = 10 and that dx dt = 5, and we are asked to find dV dt . But this is just example 1, so we know dV dt = 1500, which means the crystal is growing 1500 cubic mm per hour.
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