15.4Ex38 - S E C T I O N 15.4 Differentiability, Linear...

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SECTION 15.4 Differentiability, Linear Approximation, and Tangent Planes (ET Section 14.4) 673 We now divide by I = x a y b to obtain 1 I I ax a 1 y b 1 x I + bx a y b 1 1 y I = a 1 y b 1 x x a y b + a y b 1 1 y x a y b = a 1 x x + b 1 y y That is, 1 I I a 1 x x + b 1 y y . 37. The monthly payment for a home loan is given by a function f ( P , r , N ) ,where P is the principal (the initial size of the loan), r the interest rate, and N the length of the loan in months. Interest rates are expressed as a decimal: A 6% interest rate is denoted by r = 0 . 06. If P = $100 , 000, r = 0 . 06, and N = 240 (a 20-year loan), then the monthly payment is f ( 100 , 000 , 0 . 06 , 240 ) = 716 . 43. Furthermore, with these values, we have f P = 0 . 0071 , f r = 5 , 769 , f N =− 1 . 5467 Estimate: (a) The change in monthly payment per $1,000 increase in loan principal. (b) The change in monthly payment if the interest rate increases to r = 6 . 5% and r = 7%. (c) The change in monthly payment if the length of the loan increases to 24 years. SOLUTION (a) The linear approximation to f ( P , r , N ) is 1 f f P 1 P + f r 1 r + f N 1 N We are given that f P = 0 . 0071, f r = 5769, f N 1 . 5467, and 1 P = 1000. Assuming that 1 r = 0and 1 N = 0, we get 1 f 0 . 0071 · 1000 = 7 . 1 The change in monthly payment per thousand dollar increase in loan principal is $7.1.
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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15.4Ex38 - S E C T I O N 15.4 Differentiability, Linear...

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