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Unformatted text preview: Math 23 B. Dodson Week 7 Homework: [Revised] 14.4 tangent plane, differentials 14.5 chain rule Problem 14.4.3: Find the tangent plane to the surface the z = f (x, y) = Solution: We recall that the tangent plane at (a, b, f (a, b)) is the plane with equation z - f (a, b) = fx (a, b)(x - a) + fy (a, b)(y - b).
1 We compute fx = 2 (4 - x2 - 2y 2 )- 2 (-2x), 1 and fy = 2 (4 - x2 - 2y 2 )- 2 (-4y),
1 1 4 - x2 - 2y 2 at (1, -1, 1). so fx (1, -1) = -(4 - 1 - 2)- 2 = -1, fy (1, -1) = 2(4 - 1 - 2)- 2 = 2, giving z - 1 = -(x - 1) + 2(y + 1), or -x + 2y - z + 2 = 0, the plane through (1, -1, 1) with normal < fx , fy , -1 >=< -1, 2, -1 > . As an example of differential (linear) approximation, we solve #19, 14.4.
1 1 2 Week 7 Homework: 14.5 chain rule Problem 14.5.3: Find f (t) when f (x, y) = sin x cos y, with x = x(t) = t, y = y(t) = t. Solution: We have f (t) = fx x + fy y , with fx and fy evaluated at (x, y) = (x(t), y(t)). Here fx = cos x cos y, fy = sin x(- sin y), so x = x(t) = t, y = y(t) = t gives fx (t, t) = cos(t) cos( t), fy = sin(t)(- sin t). Then x = , y = 1 t- 2 , so 2 1 f (t) = cos(t) cos( t) + 1 t- 2 sin(t)(- sin t). 2 We note that this calculation gives vector structure to what is otherwise a straight calculus 1 calculation, d (sin(t) cos( t)), dt using the product rule and the 1-variable chain-rule.
1 The vector formulation is as a dot product, of a vector < fx , fy >, with < x , y >=< x (t), y (t) >, where < fx , fy > is evaluated at (x, y) = (x(t), y(t)). As a preview, the first vector is called the gradient of f . We also solved #13 and #23 of section 14.5. ...
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This note was uploaded on 02/29/2008 for the course MATH 23 taught by Professor Yukich during the Spring '06 term at Lehigh University .
- Spring '06
- Chain Rule