Calculus 3 Section 14.6 Reading Assignment: Tangent Planes and Differentials

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Question 2. Read the passage that begins "To find an equation for the plane tangent to a smooth surface..." near the bottom of p. 854 (after Example 1). Explain how the equation for the tangent plane for a surface ? = ?(?, ?) on the top of p. 854 is obtained based on this passage.

Page 854

EXAMPLE 1 Find the tangent plane and normal line of the level surface The surface ctytz-9-0 faxty+z-9-0 A circular paraboloid P,(1, 2, 4) at the point P,(1, 2, 4). Solution The surface is shown in Figure 14.34. Normal line The tangent plane is the plane through P perpendicular to the gradient of f at Po. The gradient is -Tangent plane Vfly, = (2ri + 2yj + k) = 2i + 4j + k. [(.24) The tangent plane is therefore the plane 2(x - 1) + 4(y - 2) + (: - 4) = 0, or 2x + 4y + z = 14. The line normal to the surface at P, is FIGURE 14.34 The tangent plane and x - 1 + 2. y = 2 + 41. 2-4+1. normal line to this level surface at Po (Example 1). To find an equation for the plane tangent to a smooth surface z = f(x, y) at a point P(xo. You z) where zo = f(xo. yo), we first observe that the equation z = f(x, y) is equiva- lent to f(x, y) - z = 0. The surface z = f(x, y) is therefore the zero level surface of the function F(x, y, 2) = f(x, y) - z. The partial derivatives of Fare Fx = ar $(fury - D =f-0 =f. F, = = (f(x, y) - D) =f - 0 =f, F. = = (f(x. y) - 2) =0 - 1 = -1. The formula F(PX(x - x) + F(P)(y - ") + F(P.)(: - 2) = 0 for the plane tangent to the level surface at A, therefore reduces to as long as the gradient is not the zero vector at the point P