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anonymous
 5 years ago
Let f (x,y, z) = yz/x. Use a tangent plane (or differentials) to approximate f (1.01,2.04,2.97).
anonymous
 5 years ago
Let f (x,y, z) = yz/x. Use a tangent plane (or differentials) to approximate f (1.01,2.04,2.97).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(1.01, 2.04, 2.97) is approximately f(1, 2, 3) = 6. However, we can make it a closer approximation using a linear approximation, similar to what we've done in calc 1. In calc 1, we used lines to approximate functions. Here in 3d though, we use planes to approximate. Let f_x, f_y, and f_z be the partials of f. Then we know that the normal vector to the tangent plane at (1, 2, 3) is <f_x (1, 2, 3), f_y (1, 2, 3), f_z (1, 2, 3)>. In other words, we find the partial derivatives and evaluate them at (1, 2, 3). Just as we did in calc 1, we found the derivative and evaluated at the point we're approximating at in order to get the slope of the line. Then a better approximation to f(1.01, 2.04, 2.97) is f(1.01, 2.04, 2.97) = f(1, 2, 3) + f_x (.01) + f_y * (.04) + f_z * (0.03), where the partials are evaluated at (1, 2, 3).
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