= =5 (0:4) 4 10 = = = = = 2:005 32:4 2: :

Transcription

1 MATH LEC SECOND EXAM THU NOV 0 PROBLEM Part (a) ( 5 oints ) Aroximate 5 :4 using a suitable dierential. Show your work carrying at least 6 decimal digits. A mere calculator answer will receive zero credit. 5 :4 5 + d 5 x x (0:4) : (0:4) To contrast this aroximation with the exact answer we rovide a calculator aroximation 5 :4 : : 5 :4 :005 Part (b) ( 5 oints ) Use a suitable dierential to aroximate tan Show your work carrying at least 6 decimal digits. A mere calculator answer will receive zero credit. tan 6:05 tan + 6 d tan(x) + sec 6 + sec 6 + cos : x 6 6:05 6 6:05 6:05 6 0:05 6: :05 : To contrast this aroximation with the exact answer we rovide a calculator aroximation tan 0: : 6:05 tan 6:05 0:

2 MATH LEC SECOND EXAM THU NOV 0 PROBLEM ( 0 oints) A farmer has 750 f t of fencing material available. She wants to fence in a rectangular area and also subdivide the area into four ens by using fencing arallel to one side of the rectangle. Find the lengths of the sides of a rectangle of maximal enclosed area which uses all available fencing material. Let x and y denote the dimensions of the rectangle. Assume arts of the fencing searating the ens are of length x. Then for maximal area all the fencing material is used and 750 5x + y so that y 750 5x : Area (x) xy 750 x 75x 5x 5 x Area 0 (x) 75 5x : 0 x critical y critical 750 5x critical :5 x 75 f t y 87:5 f t

3 MATH LEC SECOND EXAM THU NOV 0 PROBLEM Let f(x) x x ; x 6 0 : Part(a) ( 4 oints ) Where is f(x) increasing and decreasing? What are the local maximum and minimum values? f(x) x x f 0 (x) (x) x + x ( )x 4 x (x ) 9 x x 4 x 4 f 0 (x) > 0, 9 x > 0, x < 9, 0 < jxj < f 0 (x) < 0, 9 x < 0, x > 9, jxj > f ": x [ ; 0) [ (0; ] f #: x ( ; ] [ [; ) minimum 9 at x and maximum 9 at x : Part (b) ( oints ) inection. Where is f(x) concave u and concave down? Find all oints of f 0 (x) 9 x f 00 (x) x 4 f 00 (x) > 0, x f 00 (x) < 0, x x 4 (9 x ) (x 8) x x 5 x 5 ; 0 [ ; ; [ 0; x + x 5 S f : x ; 0 [ T ; f : x ; [ 0; Inections oints at ; 5 6 Part (c) ( oints ) Sketch the grah of y f(x). Find all vertical and horizontal asymtotes. Vertical asymtote: x 0 : Horizontal asymtote: y 0 :

4 MATH LEC SECOND EXAM THU NOV 0 PROBLEM 4 ( 0 oints ) Comute 6 Use y cos and dy d 0 sin cos d : sin in the Substitution Rule for Denite Integrals : 6 0 sin 6 cos d y 0 cos(6) cos 0 y sin cos d y dy 6 6

5 MATH LEC SECOND EXAM THU NOV 0 PROBLEM 5 Part (a) ( 8 oints ) Find the general solution of dy y x 5 + x : y dy x 5 + x y dy x5 + x y x6 y C y 6 + x + C e x 6 x C x 6 x y C x 6 x Part (a) ( oints ) y 7 when x 0. 7 C C Find the articular solution of the above equation which satises 7 C C y 49 x 6 x C y 49 x 6 x

6 MATH LEC SECOND EXAM THU NOV 0 PROBLEM 6 Comute the integral using the denition of a denite integral : 0 x x Part (a) ( 8 oints ) For an integer n > write down and comute R n, the Riemann sum for this integral using the regular artition of the interval and right endoint samle oints. 4x n x i i i 0 n i n n x i 0 + i 0 R n i i i n x i n i n i n x i 4x x i x i i n i i n n i n n(n + )(n + ) 6n + n n i + n i n(n + ) + n n R n + n + n + n Part (b) ( oints ) Use the exression for R n found in art (a) to comute the limit lim n! R n. lim n! R n lim + + n! n n + n lim n! R n

7 MATH LEC SECOND EXAM THU NOV 0 PROBLEM 7 Part (a) ( 5 oints ) Comute d x v dv : d x v dv d x v dv v vx x R d x v dv x Part (b) ( 5 oints ) Comute Let F (y) R y x D x t + dt : t + dt so that F 0 (y) y + : Then x D x t + dt D xf (x ) F 0 (x ) x y + x 4 + x x x 4 + yx x D x R x t dt + x x 4 +

8 MATH LEC SECOND EXAM THU NOV 0 PROBLEM 8 Part (a) ( 5 oints ) State the recise form of the Mean Value Theorem for Derivatives. If f is continuous on a closed interval [a; b] and dierentiable on the oen interval (a; b), then there exists at least one number c in (a; b) with f(b) b f(a) a f 0 (c) : () Part (b) ( 5 oints ) Determine if f(x) x satises the conditions of the Mean Value Theorem on the interval [; ] and exlain why or why not; if so, nd the values of all c guaranteed by the theorem. This f is a rational function without a zero of the denominator in the given interval [ ; ]. This imlies that the function is continuous there and dierentiable in ( ; ). Hence the conclusion of the theorem is true and there is at least one c in (; ) satisfying equation (). Note that 4 (; ) : f(x) x f 0 (x) x 4 f 0 (c) c 4 f() f() c 4 : 4 c4 c 7 4 : c 4

9 MATH LEC SECOND EXAM THU NOV 0 PROBLEM 9 Part (a) ( 5 oints ) Comute x 4 Using y g(x) x 4 and g 0 (x) the Substitution Rule for Indenite Integrals yields x 4 x 4 (x 4) 4 + C 4 8 (x 4)4 + C R x 4 8 (x 4)4 + C Part (b) ( 5 oints ) Comute x + x sin(x) Since (( x) + ( x) sin( x)) (x + x sin(x)), this integral vanishes as a denite integral of an odd function over an interval symmetric with resect to the origin. R (x + x sin(x)) 0