Mathematics 136 – Advanced Placement Calculus Solutions to Ungraded Problems on Problem Set 3 September 29, 2009

2.8/10

(a) The rate of increase of the population is small at first, then much larger around t = 8, then smaller again (tending to zero, apparently) as t increases farther.

(b) The rate of increase is largest at t = 8.

(c) The graph of the population function is concave up for 0 < t < 8 and concave down for 8 < t < 18.

(d) At t = 8, the population is 350. The inflection point on the graph is at (8, 350).

2.8/16.

(a) f is increasing where f′ is positive, so on (1, 6) and for x > 8. f is decreasing where f′ is negative, so on (0, 1) and (6, 8).

(b) f has local minima at x = 1 and x = 8 and a local maximum at x = 6 (using what we stated as the First Derivative Test in class on 9/29).

(c) f is concave up where f′ is increasing, so on (0, 2), (3, 5), (7, ∞). f is concave down where f′ is decreasing, so on (2, 3) and (5, 7).

(d) and (e) The graph should follow parts (a), (b), (c). Note that f is increasing all the way from x = 1 to x = 6; what changes on that interval is the concavity (three changes!). There are inflection points at x = 2, 3, 5 and x = 7.

3.1/7. f′(x) = 3x^{2} − 4.

3.1/25. x = Ay^{−10} + Be^{y} (where A, B are constants). So \frac{dx}{dy} = −10Ay^{−11} + Be^{y}.

3.2/11. By the quotient rule:

y′ = \frac{(1 − x^{2})(3x^{2}) − x^{3} · (−2x)}{(1 − x^{2})^{2}} = \frac{3x^{2} − x^{4}}{(1 − x^{2})^{2}} .

3.2/22. By the quotient rule:

f′(x) = \frac{(x + e^{x})(−xe^{x} − e^{x}) − (1 − xe^{x})(1 + e^{x})}{(x + e^{x})^{2}} = \frac{−x^{2}e^{x} − e^{2x} − e^{x} − 1}{(x + e^{x})^{2}} after simplifying.

3.2/45. The idea of this problem is to use the product or quotient rule as appropriate, and read the values of f, g, and the derivatives of f, g from the graph, recalling that f′(a), g′(a) are slopes of tangent lines.

(a)

u′(1) = f(1)g′(1) + f′(1)g(1) = 2 · (−1) + 2 · 1 = 0.

(b)

v′(5) = \frac{g(5)f′(5) − f(5)g′(5)}{(g(5))^{2}} = \frac{2 · (−1/3) − 3 · (2/3)}{2^{2}} = −\frac{2}{3}.

3.3/5. By the product rule, y′ = sec θ · sec^{2} θ + tan θ · sec θ tan θ = sec θ(sec^{2} θ + tan^{2} θ). This can also be rewritten in various ways by using trig identities, recalling that 1 + tan^{2} θ = sec^{2} θ.

3.3/35.

(a) Since x(t) = 8 sin(t), v(t) = x′(t) = 8 cos(t) and a(t) = x′′(t) = −8 sin(t).

(b) At t = \frac{2π}{3} , v = 8 cos (\frac{2π}{3} ) = −4 and a = −8 sin (\frac{2π}{3} ) = −4\sqrt{3}. This means that the mass is moving to the left (v < 0).

3.4/15. The general derivative rule for the exponential functions is

\frac{d}{dx}a^{x} = a^{x} ln(a).

(This can be seen by rewriting a^{x} = e^{x ln(a)} and using the chain rule.) So h′(t) = 3t^{2} − 3^{t} ln(3).

3.4/30. Think of the function f(t) in the form f(t) = (\frac{t}{t^{2}+4})^{1/2} and use the chain rule and quotient rule:

f′(t) = \frac{1}{2} (\frac{t}{t^{2} + 4})^{−1/2} (\frac{(t^{2} + 4) − t · (2t)}{(t^{2} + 4)^{2}}) .

This can be simplied to:

f′(t) = \frac{4 − t^{2}}{2t^{1/2}(t^{2} + 4)^{3/2}} .

3.5/15. By implicit differentiation:

−e^{y} sin(x) + cos(x) · e^{y}y′ = cos(xy) · (y + xy′).

Solving for y′:

(cos(x)e^{y} − x cos(xy))y′ = y cos(xy) + e^{y} sin(x).

So

y′ = \frac{y cos(xy) + e^{y} sin(x)}{cos(x)e^{y} − x cos(xy)}.