15.4%20Ex%2030%20-%20%2038

15.4%20Ex%2030%20-%20%2038 - 670 C H A P T E R 15 D I F F E...

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670 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) The value given by a calculator is: p ( 1 . 9 )( 2 . 02 )( 4 . 05 ) 3 . 9426 30. 8 . 01 ( 1 . 99 )( 2 . 01 ) SOLUTION We use the linear approximation of the function f ( x , y , z ) = x yz at the point ( 8 , 2 , 2 ) ,whichis f ( 8 + h , 2 + k , 2 + l ) f ( 8 , 2 , 2 ) + f x ( 8 , 2 , 2 ) h + f y ( 8 , 2 , 2 ) k + f z ( 8 , 2 , 2 ) l (1) We compute the values of the function and its partial derivatives at ( 8 , 2 , 2 ) .Thisgives f ( x , y , z ) = x f ( 8 , 2 , 2 ) = 4 f x ( x , y , z ) = 1 f x ( 8 , 2 , 2 ) = 1 2 f y ( x , y , z ) = x y ( ) 1 / 2 =− 1 2 x ( ) 3 / 2 z 1 2 xy 3 / 2 z 1 / 2 f y ( 8 , 2 , 2 ) 1 f z ( x , y , z ) = x z ( ) 1 / 2 1 2 x ( ) 3 / 2 y 1 2 1 / 2 z 3 / 2 f z ( 8 , 2 , 2 ) 1 Substituting these values and h = 0 . 01, k 0 . 01, l = 0 . 01 in (1) gives the following approximation: 8 . 01 ( 1 . 99 )( 2 . 01 ) = 4 + 1 2 · 0 . 01 1 · ( 0 . 01 ) 1 · 0 . 01 = 4 . 005 The value given by a calculator is 4 . 00505. The error is 0 . 00005 and the percentage error is at most Percentage error 0 . 00005 · 100 4 . 00505 0 . 00125% 31. Estimate f ( 2 . 1 , 3 . 8 ) given that f ( 2 , 4 ) = 5 , f x ( 2 , 4 ) = 0 . 3, and f y ( 2 , 4 ) 0 . 2. We use the linear approximation of f at the point ( 2 , 4 ) f ( 2 + h , 4 + k ) f ( 2 , 4 ) + f x ( 2 , 4 ) h + f y ( 2 , 4 ) k Substituting the given values and h = 0 . 1, k 0 . 2 we obtain the following approximation: f ( 2 . 1 , 3 . 8 ) 5 + 0 . 3 · 0 . 1 + 0 . 2 · 0 . 2 = 5 . 07 . In Exercises 32–34, let I = W / H 2 denote the BMI described in Example 6. 32. A boy has weight W = 34 kg and height H = 1 . 3. Use the linear approximation to estimate the change in I if ( W , H ) changes to ( 36 , 1 . 32 ) . Let 1 I = I ( 36 , 1 . 32 ) I ( 34 , 1 . 3 ) denote the change in I . Using the linear approximation of I at the point ( 34 , 1 . 3 ) we have I ( 34 + h , 1 . 3 + k ) I ( 34 , 1 . 3 ) I W ( 34 , 1 . 3 ) h + I H ( 34 , 1 . 3 ) k For h = 2, k = 0 . 02 we obtain 1 I I W ( 34 , 1 . 3 ) · 2 + I H ( 34 , 1 . 3 ) · 0 . 02 (1) We compute the partial derivatives in (1): I W = W W H 2 = 1 H 2 I W ( 34 , 1 . 3 ) = 0 . 5917 I H = W H H 2 = W · ( 2 H 3 ) = 2 W H 3 I H ( 34 , 1 . 3 ) 30 . 9513 Substituting the partial derivatives in (1) gives the following estimation of 1 I : 1 I 0 . 5917 · 2 30 . 9513 · 0 . 02 = 0 . 5644
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SECTION 15.4 Differentiability, Linear Approximation, and Tangent Planes (ET Section 14.4) 671 33. Suppose that ( W , H ) = ( 34 , 1 . 3 ) and W increases to 35. Use the linear approximation to estimate the increase in H required to keep I constant. SOLUTION The linear approximation of I = W H 2 at the point ( 34 , 1 . 3 ) is: 1 I = I ( 34 + h , 1 . 3 + k ) I ( 34 , 1 . 3 ) I W ( 34 , 1 . 3 ) h + I H ( 34 , 1 . 3 ) k (1) In the earlier exercise, we found that I W ( 34 , 1 . 3 ) = 0 . 5917 , I H ( 34 , 1 . 3 ) =− 30 . 9513 We substitute these derivatives and h = 1 in (1), equate the resulting expression to zero and solve for k .Thisgives: 1 I 0 . 5917 · 1 30 . 9513
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.4%20Ex%2030%20-%20%2038 - 670 C H A P T E R 15 D I F F E...

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