# finalsol - Final Exam Solutions 1 Find the derivative of...

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Final Exam Solutions 1 . Find the derivative of f ( x ) = x tan x . Using a b = e b ln a , this function is f ( x ) = e (tan x ) ln x . So f ( x ) = e (tan x ) ln x (sec 2 x ) ln x + (tan x ) 1 x = x tan x (sec 2 x ) ln x + tan x x . (Here, we used the chain rule: d dx e u = e u du dx , where u = (tan x ) ln x has derivative du dx = (sec 2 x ) ln x + (tan x ) 1 x by the product rule.) 2 . Suppose we initially had 10 milligrams of Zirconium-95. How much will we have left in 100 days, assuming that it decays at a rate proportional to the amount present, with a half-life of 64 days? If y is the amount of the isotope, we have dy dx = ky where k is the unknown decay constant. This solves to give y = y 0 e kt . The initial amount is 10 milligrams, so y (0) = 10 = y 0 e k · 0 = y 0 . So y = 10 e kt . The half-life is 64 days, so y (64) = 5 or 10 e 64 k = 5. We solve e 64 k = 0 . 5 to obtain 64 k = ln 0 . 5 or k = (ln 0 . 5) / 64 = - 0 . 0108. Therefore y = 10 e - 0 . 0108 t and y (100) = 10 e - 1 . 08 = 3 . 38 milligrams. 3 . Find the limit: lim t 0 1 - cos t t 2 . Use L’Hospital’s rule twice: lim t 0 1 - cos t t 2 = lim t 0 sin t 2 t = lim t 0 cos t 2 = cos 0 2 = 1 2 . 4 . Find (2 x + 3) sin x dx . Use integration by parts: u dv = uv - v du , with u = 2 x + 3, du = 2 dx , dv = sin x dx and v = - cos x . We have (2 x +3) sin x dx = (2 x +3)( - cos x ) - 2( - cos x ) dx = - (2 x +3) cos x +2 sin x + C. 5 . Find sec x tan 3 x dx . This is tan 2 x sec x tan x dx = (sec 2 x - 1) sec x tan x dx (using the identity 1 + tan 2 x = sec 2 x ). Let u = sec x so du = sec x tan x dx . The integral becomes (sec 2 x - 1) sec x tan x dx = ( u 2 - 1) du = 1 3 u 3 - u + C = 1 3 sec 3 x - sec x + C .
6 . Find dx x 4 - x 2 . Use the trig substitution x = 2 sin θ , so dx = 2 cos θ dθ . The integral becomes dx x 4 - x 2 = 2 cos θ dθ 2 sin θ 4 - 4 sin 2 θ = 2 cos θ dθ 2 sin θ (2 cos θ ) = 1 2 csc θ dθ = 1 2 ln | csc θ - cot θ | + C. We can use a right triangle to find csc θ and cot θ . Now sin