Unformatted text preview: The derivative of the ﬁrst equation gives (0) No. The solutions to the equations above are l = 6 and h = 12.5. The length 0:
a ladder that will reach needs to be at least 20%ft long. 5. (a) The mass is fa; $2015: 2 L3/3.
(b) The moment of the mass is If x3dm = 114/4. The center of mass is 3L/4.
(c) Closer to the right end 6. (a)
V(a) = 271' [434% “ﬁg?
(70) (1+a+a2) ((3) Yes. Consider the function f(a,) = V(a) — a. This function satisﬁes f(0) = 0,
f’(0) > O and lim f’(a) = ——1. Consequently, f is positive for small a > 0 and (Ii—’00
then decreases. It has no lower bound and becomes negative, so the Intermediate Value Theorem implies there is an a, with f (a) = 0.
7 (a) y=1~§($—1) (b) R 2: 515 and R is increasing at rate % V’(a) m 27r(1 + 2a) 3 3
a: :13: W 5%— + 0 (int. by parts) 1
b
( ) 4cos46 l :r:
(C) §(1 + $2
I
d
( ) 1 + x
(e) s/ 12 (subs, then trig subs) ”/4 “/4 dm 1—1—5li 4
9.Th 111'] (/1t2d=] =1——_——m’”=11 2
earc engt 18 0 + 311(3) 2: 0 n( “0 n( +\/_) COS m COS $ 8. (a) + C (substitution) + tan‘1 :12) + C (trig subs) + tau”1 :1: + 0 (partial fractions) 10. (a) This is not an indeterminate form. The limit is O.
(b) The limit is co. Expanding, all of the terms are positive and ew/rr —> do. ((3) The limit is 1/2. Diﬁ’erentiate numerator and denominator twice. ...
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 Fall '07
 BERMAN
 Math

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