MATH 1B SOLUTION SET FOR CHAPTERS 11 12 9 1 9 2 Problem 11 12 25 Use Taylor s Inequality to determine the number of terms of the Maclaurin series for ex that should be used to estimate e0 1 to within 10 5 Solution First of all every derivative of ex is ex Since ex is increasing the maximum of every derivative of ex on 0 0 1 is e0 1 itself Now if we actually knew e0 1 we wouldn t need to estimate it Still we know that it is well less than 2 the square root of 3 is and this is the 10th root of something less than 3 Our error bound for truncating at the nth term is thus at most 2 10 n 1 Rn n 1 We wish to ensure that this error is less than 10 5 This will clearly be met for n 4 and with a bit more work for n 3 At n 2 it won t be met Thus we have 1 1 e0 1 1 0 1 0 1 2 0 1 3 2 6 1 10517 Problem 11 12 26 How many terms of the Maclaurin series for ln 1 x do you need to use to estimate ln 1 4 to within 0 001 n 1 n P Solution First ln 1 x n 1 1 n x We first must find the nth derivative d 1 of ln 1 x Well dx ln 1 x 1 x It s easy to show that the nth derivative of 1 1 x is 1 n n 1 x n 1 it s true for the 0th derivative and if it s true for the nth then n n 1 1 n 1 d These it s true for the next derivative as well Thus dx n ln 1 x 1 x n functions have strictly decreasing absolute value on 0 0 4 so we may take M 1 Thus 0 4n 1 Rn 0 4 n 1 4n 1 10 n 1 Obviously n 1 3 we have 64 10 which is 6 For simplicity we can write this as we needn t consider anything below n 2 At n 2 clearly greater than 4 3 10 3 At n 3 we have 64 10 which is still greater than 10 AT n 4 we 6 5 128 10 3 which is less than 10 We must therefore keep everything out to have 3 the n 4 term so we must keep 5 terms Problem 11 12 31 An electric dipole consists of two electric charges of equal magnitude and opposite signs If the charges are q and q and are located at a distance d from each other then the electric field E at the point P in the figure is E q q D2 D d 2 Typeset by AMS TEX 1 2 MATH 1B SOLUTION SET FOR CHAPTERS 11 12 9 1 9 2 d show that E is By expanding this expression for E as a series of powers in D 1 approximately proportional to D3 when P is far from the dipole Solution First let s write the expression in terms of q q E 2 2 d 2 D D 1 D d D Now expand the second term on the right as a binomial series n q q X 2 d E 2 2 D D n 0 n D At this point we argue that since d D we can reasonably approximate E by truncating the series at two terms leaving d q E 2 1 1 2 D D 2qd 3 D Problem 9 1 1 Show that y x x 1 is a solution of the differential equation xy 0 y 2x Solution Here we just plug in xy 0 y x 1 x 2 x x 1 x x 1 x x 1 2x as desired Problem 9 1 2 Verify that y sin x cos x cos x is a solution of the initial value problem y 0 tan x y cos2 x y 0 1 on the interval 2 x 2 Solution First y 0 sin 0 cos 0 cos 0 1 satisfying the initial condition Next plugging in we have cos2 x sin2 x sin x sin2 x sin x cos2 x as desired so the solution is good as long as tan x is defined which imposes the range Problem 9 1 3 a For what nonzero values of k does the function y sin kt satisfy the differential equation y 00 9y 0 b For those values of k verify that every member of the family of functions y A sin kt B cos kt is also a solution Proof a Pluggin in the putative solution we get k 2 sin kt 9 sin kt 0 whence k 2 9 or k 3 b If y A sin 3t B cos 3t we have y 0 3A sin 3t 3B cos 3t and 00 y 9A sin 3t 9B cos 3t Thus certainly y 00 9y 0 and these functions also satisfy the differential equation MATH 1B SOLUTION SET FOR CHAPTERS 11 12 9 1 9 2 3 Problem 9 1 10 A function y t satisfies the differential equation dy y 4 6y 3 5y 2 dt a What are the constant solutions of the equation b For what values of y is y increasing c For what values of y is y decreasing Solution a For the solution to be constant we must have dy dt 0 This requires y 4 6y 3 5y 2 0 y 2 y 2 6y 5 0 y 2 y 5 y 1 0 Thus the constant solutions are y 0 y 1 and y 5 dy b c The function y is increasing where dy dt 0 and decreasing where dt 0 2 The factor y is greater than zero everywhere except at 0 where it is zero The factor y 1 is negative on y 1 positive on y 1 and zero at 1 The factor y 5 is negative on y 5 positive on y 5 and zero at 5 Just keeping track of the signs we see that y is increasing on 0 0 1 5 decreasing on 1 5 and constant on 0 1 5 As an aside we note that this means that the equilibrium solution at 1 is stable the equilibrium solution at 5 is unstable and the equilibrium solution at 0 is unstable under positive perturbations but stable under negative perturbations for most practical purposes this simply means unstable as you usually can t count on all perturbations being in a favorable direction Problem 9 1 11 Explain why the function with the given graphs not reproduced here see p 592 of Stewart can t be solutions of the differential equation dy et y 1 2 dt Solution a can t work because it has negative slope on portions of its solutions The slope et y 1 2 is a nonnegative function b can t work because it has positive slope at y …
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