Exam 2 Nov. 10, 2005 SHOW ALL WORK Math 25 Calculus I Either circle your answers or place on answer line.

[16] 1.) If f′(x) = 3x^{4} + 2 + 4x^{−1} and f(1) = 8, find f f(x) = \frac{3}{5}x^{5} + 2x + 4ln(x) + C for some C. f(1) = 8, f(1) = 8 = \frac{3}{5}(1)^{5} + 2(1) + 4ln(1) + C 8 = \frac{3}{5} + 2 + 0 + C 6 − \frac{3}{5} = \frac{30−3}{5} = \frac{27}{5} = C

Answer 1.) \underline{f(x) = \frac{3}{5}x^{5} + 2x + 4ln(x) + \frac{27}{5}}

[16] 2.)lim_{x→0^{+}}[x^{x^{3}}] = \underline{1} lim_{x→0^{+}}[x^{x^{3}}] = lim_{x→0^{+}}[e^{ln(x^{x^{3}} )}] = lim_{x→0^{+}}[e^{x^{3}ln(x)}] lim_{x→0^{+}}[x^{3}ln(x)] = lim_{x→0^{+}}[\frac{ln(x)}{x^{−3}}] = lim_{x→0^{+}}[\frac{\frac{1}{x}}{−3x^{−4}} ] = lim_{x→0^{+}}[\frac{x^{3}}{−3}] = 0 Hence lim_{x→0^{+}}[e^{x^{3}ln(x)}] = e^{lim_{x→0^{+}} [x^{3}ln(x)]} = e^{0} = 1

[10] 3a.) Given x^{2} + 2xy + y^{3} − x − 3 = 0, then y′ = \underline{\frac{1−2x−2y}{2x+3y^{2}}} \frac{d}{dx}(x^{2} + 2xy + y^{3} − x − 3) = \frac{d}{dx}(0), 2x + 2(y + xy′) + 3y^{2}y′ − 1 = 0, 2x + 2y + 2xy′ + 3y^{2}y′ − 1 = 0, (2x + 3y^{2})y′ = 1 − 2x − 2y, y′ = \frac{1−2x−2y}{2x+3y^{2}} ,

[6] 3b.) Find the equation of the tangent line to the curve x^{2} + 2xy + y^{3} − x − 3 = 0, at the point (-2, 1). y′ = \frac{1−2x−2y}{2x+3y^{2}} , Hence at (-2, 1), y′ = \frac{1−2(−2)−2(1)}{2(−2)+3(1)^{2}} = \frac{1+4−2}{−4+3} = −3. \frac{y−1}{x−(−2)} = −3, y = −3(x + 2) + 1 = −3x − 6 + 1 = −3x − 5

Answer 3b.) \underline{y = −3x − 5} [16] 4.) If g(3) = 4 and g′(x) ≤ 2, how large can g(8) be? \underline{14} Since g′ exists, g is differentiable and hence continuous. By the MVT since g continuous on [3, 8] and g differentiable on (3, 8), then there exists c ∈ [3, 8] such that

\frac{g(8) − g(3)}{8 − 3} = g′(c)

Hence \frac{g(8)−4}{5} = g′(c) ≤ 2. Thus g(8) ≤ 2(5) + 4 = 14.

[16] 5.) A tank is in the form of an inverted cone having a height of 16m and a diameter at the top of 8 m. Water is flowing into the tank at the rate of 2 m^{3}/min. How fast is the water level rising when the water is 5m deep? (Volume of cone = \frac{1}{3}πr^{2}h) \frac{dV}{dt} = 2, \frac{dh}{dt} =? when h = 5. NOTE: V, h, r change with respect to time. None of them are constant. V = \frac{1}{3}πr^{2}h, \frac{r}{h} = \frac{4}{16}. Hence r = \frac{1}{4}h. V = \frac{1}{3}π(\frac{1}{4}h)^{2}h = \frac{1}{3}π(\frac{h^{3}}{16} ) \frac{d}{dt}[V] = \frac{d}{dt}[\frac{1}{3}π(\frac{h^{3}}{16} )] \frac{dV}{dt} = \frac{1}{3}π(\frac{3h^{2}}{16} \frac{dh}{dt}) \frac{dV}{dt} = π(\frac{h^{2}}{16} \frac{dh}{dt}) 2 = π(\frac{5^{2}}{16} \frac{dh}{dt}). Hence \frac{32}{25π} = \frac{dh}{dt} Alternate answer: V = \frac{1}{3}πr^{2}h. NOTE: V, h, r change with respect to time. \frac{d}{dt}[V] = \frac{d}{dt}[\frac{1}{3}πr^{2}h] \frac{dV}{dt} = \frac{1}{3}π[2r \frac{dr}{dt} h + r^{2} \frac{dh}{dt}] When h = 5: 2 = \frac{1}{3}π[2r \frac{dr}{dt} (5) + r^{2} \frac{dh}{dt}] \frac{r}{h} = \frac{4}{16}. Hence r = \frac{1}{4}h. Thus \frac{dr}{dt} = \frac{1}{4} \frac{dh}{dt} and when h = 5, r = \frac{5}{4}. Thus, when h = 5: 2 = \frac{1}{3}π[2(\frac{5}{4})(\frac{1}{4})\frac{dh}{dt}(5) + (\frac{5}{4})^{2} \frac{dh}{dt}] = \frac{25π}{16} \frac{dh}{dt}. Hence \frac{32}{25π} = \frac{dh}{dt}

Answer 5.) \underline{\frac{dh}{dt} = \frac{32}{25π}}

6.) Find the following for f(x) = \frac{x}{(x−1)^{2}} (if they exist; if they don’t exist, state so). Use this information to graph f. Note f′(x) = \frac{−x−1}{(x−1)^{3}} and f′′(x) = \frac{2(x+2)}{(x−1)^{4}}

[1] 6a.) critical numbers: \underline{−1} [1.5] 6b.) local maximum(s) occur at x = \underline{none} [1.5] 6c.) local minimum(s) occur at x = \underline{−1} [1.5] 6d.) The global maximum of f on the interval [0, 5] is \underline{none} and occurs at x = \underline{none} [1.5] 6e.) The global minimum of f on the interval [0, 5] is \underline{0} and occurs at x = \underline{0} [1.5] 6f.) Inflection point(s) occur at x = \underline{−2}

[1] 6g.) f increasing on the intervals \underline{(−1, 1)}

[1] 6h.) f decreasing on the intervals \underline{(−∞, −1) ∪ (1, ∞)} [1.5] 6i.) f is concave up on the intervals \underline{(−2, 1) ∪ (1, ∞)} [1.5] 6j.) f is concave down on the intervals \underline{(−∞, −2)} [1.5] 6k.) Equation(s) of vertical asymptote(s) \underline{x = 1}

[4] 6l.) Equation(s) of horizontal and/or slant asymptote(s) \underline{y = 0}

[4] 6m.) Graph f