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APPM 1345 Exam 3 Solutions Spring 2006
- (a) Since y′^ = 6x^2 , the function is strictly increasing and therefore one-to-one so it has an inverse. (b) Solve for x, then exchange x and y. y = 2x^3 − 1 y + 1 = 2x^3 y + 1 2 = x^3
3
y + 1 2 = x
y−^1 = 3
x + 1 2
- (a) Use the product rule. y = 2t^ ln t y′^ = 2t^ ·
t
2 t t
(b) Use the change of base formula and product rule. y = esin^ t^ log 2 t = esin^ t^ · ln t ln 2 y′^ =
ln 2
esin^ t^ ·
t
y′^ =
ln 2
esin^ t t
(c)
y =
2 z − 3 (z^2 + 4)^7 sin^2 z
(2z − 3)^1 /^2 (z^2 + 4)^7 (sin z)^2 ln y = ln(2z − 3)^1 /^2 − ln(z^2 + 4)^7 − ln(sin z)^2 =
ln(2z − 3) − 7 ln(z^2 + 4) − 2 ln(sin z) 1 y
dy dz
=^1
2 z − 3
z^2 + 4 (2z) − 2 · 1 sin z · cos z
dy dz
2 z − 3 (z^2 + 4)^7 sin^2 z
2 z − 3
14 z z^2 + 4 − 2 cot z
- (a) Let u = ln t. Then du = dt t
ln^3 t t dt =
u^3 du
u^4 4 +^ C^ =^
ln^4 t 4 +^ C (b) Let u = sin t. Then du = cos t dt. ∫ esin^ t^ cos t dt =
eu^ du
= eu^ + C = esin^ t^ + C
- (a) 16 x^ = 2x+ 24 x^ = 2x+ 4 x = x + 1 x = 1 / 3 (b) ln e^6 x+4^ = 16 6 x + 4 = 16 x = 2
- (a)
x^ lim→ 07 −^ 7 cos^ x ex^ − x − 1 = lim x→ 0 7 sin^ x ex^ − 1 (by L’Hop)
= lim x→ 0 7 cos x ex^ (by L’Hop) = 7 (b)
L = lim x→ 0
x
)x
ln L = lim x→ 0 ln
x
)x
= lim x→ 0 x ln
1 +^5
x
= lim x→ 0
ln
1 + (^5) x
1 /x
= lim x→ 0
1 1+5/x
(− (^) x^52 ) − (^) x^12 (by L’Hop)
= lim x→ 0
1 + (^) x^5 = lim x→ 0 5 x x + 5
L = e^0 = 1
- (a) We examine limx→∞ f (x)/g(x).
x^ lim→∞
log 5 x log 10 x = (^) xlim→∞ (ln x)/(ln 5) (ln x)/(ln 10) = ln 10 ln 5 Therefore f and g grow at the same rate. (b) We examine (^) xlim→∞ f (x)/g(x).
x^ lim→∞
f (x) g(x) =^ xlim→∞
x/ 5 xe−x
= (^) xlim→∞^ e
x 5
Therefore f grows faster than g.
- Let y(t) represent the population of the bacteria after t years and y 0 be the original amount. We are given that y(1) = 0. 7 y 0. First we find the rate constant k.
y(t) = y 0 ekt
- 7 y 0 = y 0 ek 0 .7 = ek k = ln 0. 7
Then we find t when y(t) = 0. 4 y 0.
- 4 y 0 = y 0 ekt 0 .4 = ekt ln 0.4 = kt
t = ln 0. 4 k
ln 0. 4 ln 0. 7 ≈ 2. 57 years
