sol09term2

# sol09term2 - MAT 137Y 2008-2009 Winter Session Term Test 2...

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Unformatted text preview: MAT 137Y, 2008-2009 Winter Session, Term Test 2 Solutions 1. Evaluate the following limits, or show the limit does not exist. (8%) (i) lim x → cot x- 1 x . Let L be the limit. Then L = lim h → cos x sin x- 1 x = lim h → x cos x- sin x x sin x H = lim h → cos x- x sin x- cos x sin x + x cos x = lim h →- x sin x sin x + x cos x H = lim h →- sin x- x cos x cos x + cos x- x sin x = 2 = . (8%) (ii) lim x → ∞ x 2 + cos x x 2 + 2009 . Since- 1 ≤ cos x ≤ 1, we have x 2- 1 x 2 + 2009 ≤ x 2 + cos x x 2 + 2009 ≤ x 2 + 1 x 2 + 2009 , so by the squeeze theorem, lim x → ∞ x 2 + cos x x 2 + 2009 = 1. (Note, applying L’Hˆopital’s Rule twice yields lim x → ∞ 2- cos x 2 , which does not exist , but L’Hˆopital’s Rule does not apply in this situation.) (15%) 2. Recall that the volume of a right circular cone is V = 1 3 π r 2 h , where r is the radius of the (circular) base, and h is the height of the cone. Find the largest possible volume of a right circular cone that is inscribed in a sphere of radius R . Make sure that your answer yields an absolute maximum. This question is posed in SHE 4.5 #43, which provides a diagram. We maximize V = 1 3 π r 2 h . Since r 2 +( h- R ) 2 = R 2 by the Pythagorean Theorem, we solve for r 2 to get V ( h ) = 1 3 π R 2- ( h- R ) 2 ) h = 1 3 π R 2 h- 1 3 π ( h- R ) 2 h , where h ∈ ( , 2 R ) . We solve for the absolute maximum; differentiating, V ( h ) = 1 3 π R 2- 1 3 h ( h- R ) 2 + 2 h ( h- R ) i = = ⇒ R 2- ( h- R )...
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## This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto.

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sol09term2 - MAT 137Y 2008-2009 Winter Session Term Test 2...

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