quiz2-sol - M (15) = ( . 93) M (0), which implies that...

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AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I – Solution Notes Quiz # 2 1. Solve for x : 400 · 5 2 x = 30 · (1 . 22) x Taking the natural log of both sides, we get ln 400 + 2 x ln 5 = ln 30 + x ln 1 . 22 x (2 ln 5 - ln 1 . 22) = ln 30 - ln 400 x = ln 30 - ln 400 2 ln 5 - ln 1 . 22 = ln(3 / 40) ln(25 / 1 . 22) 2. Let f ( x ) = x 3 - 27 and g ( x ) = e x - 3 . (a). Determine g ( f ( x )) g ( f ( x )) = g ( x 3 - 27) = e ( x 3 - 27) - 3 = e x 3 - 30 (b). Which function has a larger value as x → ∞ : g ( f ( x )) or f ( g ( x ))? Note that f ( g ( x )) = ( e x - 3 ) 3 - 27 = e 3 x - 9 - 27, so f ( g ( x )) grows like e 3 x , which does not grow as quickly as e x 3 , which is how g ( f ( x )) grows. 3. The quantity of moisture in a frozen waFe decreases with time as it sits on the counter. Suppose that the moisture, M ( t ) , at time t minutes after being placed on the counter decreases according to the function M ( t ) = Qe - kt . If 7% of the moisture is gone at the end of 15 minutes, (a). What percentage of the original moisture is present after 30 minutes? We are told that
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Unformatted text preview: M (15) = ( . 93) M (0), which implies that Qe-15 k = ( . 93) Q-15 k = ln( . 93) k =-ln( . 93) 15 Now, we want to nd M (30) M (0) 100 = Qe-30 k Q 100 = 100 e-30-ln( . 93) 15 = 100 e 2 ln( . 93) = 100( . 93) 2 (b). How long will it take until the moisture is reduced to 50% of its original quantity? We want to nd t so that M ( t ) = 1 2 M (0) = 1 2 Q . Thus, we want t so that Qe-kt = 1 2 Q , i.e., so that e-kt = 1 2 . Thus, we want t so that-kt = ln 1 2 , i.e., t = ln 1 2-k = 15 ln 1 2 ln( . 93) 4. Plot f ( x ) = 2 sin( x )-1. Be sure to mark important points on the axes! The plot is a simple sine wave, oscillating between -3 and +1, with period 2 , since it has amplitude 2 and has been shifted downwards by 1....
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.

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