20Chw4 - i Solution Think of the surface as defined...

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MATH 20C graded homework 4. (18+2 points) 14.4 # 28 (6 points): Use linear approximation to estimate 0 . 98 2 2 . 01 3 +1 . Solution: We will use the linear approximation to the function f ( x, y ) = x 2 y 3 + 1 at ( a, b ) = (1 , 2). First, f (1 , 2) = 1 9 . Also, f x ( x, y ) = 2 x y 3 + 1 , f x (1 , 2) = 2 9 , and f y ( x, y ) = - 3 x 2 y 2 ( y 3 + 1) 2 , f y (1 , 2) = - 12 81 . The linear approximation at (1 , 2) is L ( x, y ) = 2 9 ( x - 1) - 12 81 ( y - 2) + 1 9 , so 0 . 98 2 2 . 01 3 + 1 = L (0 . 98 , 2 . 01) = 2 9 ( - 0 . 02) - 12 81 (0 . 01) + 1 9 . Remark: L (0 . 98 , 2 . 01) = 71 675 . The actual value is 0 . 98 2 2 . 01 3 +1 = 0 . 1053000 to 6 decimals. 14.5 # 34 (6 points): Suppose that f P = h 2 , - 4 , 4 i . Is f increasing or decreasing at P in the direction v = h 2 , 1 , 3 i ? Solution: The unit vector in the direction of v is u = 1 14 h 2 , 1 , 3 i , so the directional derivative of f in the direction of u is D u f P = f P · u = h 2 , - 4 , 4 i · 1 14 h 2 , 1 , 3 i = 1 14 (4 - 4 + 12) = 12 14 . This is positive, so f is increasing at P in the direction v . About the grading this week: As long as there’s some indication or explanation that f is increasing because the directional derivative is positive, you get 2/2. 14.5 # 40 (6 points): Find the two points on the ellipsoid x 2 4 + y 2 9 + z 2 = 1 where the tangent plane is normal to
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Unformatted text preview: i . Solution : Think of the surface as defined implicitly by F ( x,y,z ) = x 2 4 + y 2 9 + z 2 = 1. The gradient vector is ∇ F = h 1 2 x 2 , 2 9 y 2 , 2 z i , so if ( a,b,c ) is a point on the ellipsoid, then a normal vector to the tangent plane at ( a,b,c ) is ∇ F ( a,b,c ) = ² 1 2 a, 2 9 b, 2 c ³ . For the tangent plane to be normal to v , we need to solve the system ± 1 2 a, 2 9 b, 2 c ² = λ h 1 , 1 ,-2 i , or 1 2 a = λ (1) 2 9 b = λ (2) 2 c =-2 λ. (3) These equations imply a = 2 λ , b = 9 2 λ , and c =-λ . Since ( a,b,c ) is a point on the ellipsoid, then c 2 = 1-a 2 4-b 2 9 . Substituting λ into this last equation gives λ 2 = 1-(2 λ ) 2 4-( 9 2 λ ) 2 9 or λ = ± 2 √ 17 . Therefore, the two points are ³ 4 √ 17 , 9 √ 17 ,-2 √ 17 ´ and ³-4 √ 17 ,-9 √ 17 , 2 √ 17 ´ ....
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