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APPM 1350 Midterm 2 Solutions Spring 2012
- (a) Use the chain rule and quotient rule.
y =
5 x 1 − x^3 =
5 x 1 − x^3
y′^ =
5 x 1 − x^3
d dx
5 x 1 − x^3
5 x 1 − x^3
· 5(1^ −^ x
(^3) ) − 5 x(− 3 x (^2) ) (1 − x^3 )^2
=^1
5 x 1 − x^3
· 5(1 + 2x
(1 − x^3 )^2
(b) Use the product rule and chain rule.
y = tan θ sec^2 θ = tan θ (sec θ)^2
y′^ = tan θ(2)(sec θ) · d dx
(sec θ) + (sec θ)^2 · d dx
(tan θ)
= 2 tan θ sec θ(sec θ tan θ) + (sec θ)^2 (sec θ)^2 = 2 tan^2 θ sec^2 θ + sec^4 θ
- cos(xy) = y + 1
− sin(xy)
d dx (xy) =^
dy dx − sin(xy)(x
dy dx
dy dx −x sin(xy) dy dx
− y sin(xy) = dy dx dy dx
(−x sin(xy) − 1) = y sin(xy)
dy dx
= − y^ sin(xy) x sin(xy) + 1
- (^) (a) (^) f (0) = 1
f ′(x) = k(1 + x)k−^1 f ′(0) = k L(x) = f (0) + f ′(0)(x − 0) = 1 + kx
(b) We wish to find L(x) for x = 0. 012 and k = 1/ 3.
L(0.012) = 1 +^1
- f (x) =
x^2 4 − x^2 ,^ f^
′(x) = 8 x (4 − x^2 )^2 ,^ f^
′′(x) = 8(3x
(4 − x^2 )^3.
(a) f has vertical asymptotes at x = 2 and x = − 2.
lim x→ 2 +
x^2 4 − x^2
0 −^
lim x→− 2 +
x^2 4 − x^2
0 +^
f has a horizontal asymptote at y = − 1. Use dominance of pow- ers to evaluate the limits as x → ±∞. Since the degree of the numerator equals the degree of the denominator, the limit equals the ratio of the leading coefficients.
x^ lim→∞
x^2 4 − x^2
x→−∞lim^ x
2 4 − x^2
Because f has a horizontal asymptote, f has no slant asymptotes.
(b) Find the critical numbers of f. f ′^ is undefined at x = ± 2. f ′^ = 0 at x = 0.
Intervals f ′^ f x < − 2 – decreasing on (−∞, −2) − 2 < x < 0 – decreasing on (− 2 , 0) 0 < x < 2 + increasing on (0, 2) x > 2 + increasing on (2, ∞)
(c) f has a local minimum value at x = 0, y = 0. f has no local maximum value.
(d) f ′′^ is undefined at x = ± 2. f ′′^ = 0 has no solutions.
Intervals f ′′^ f x < − 2 – concave down on (−∞, −2) − 2 < x < 2 + concave up on (− 2 , 2) x > 2 – concave down on (2, ∞)
(e) f has no inflection points. Although f changes concavity at x = ± 2 , there are no inflection points at those values because f is not continuous there.
(f)
y á - 1
x á - 2 x á 2
y á
x^2 4 - x^2
2
4
y
r h
Let V represent the volume of the water, r represent the radius of the water, and h represent the height of the water. We are given that dV /dt = − 48. The ratio r/h = 45/5 = 9 ⇒ r = 9h. We wish to find dh/dt
when h = 4. V =
π 3
r^2 h
=
π 3
(9h)^2 h = 27πh^3 dV dt
= 81πh^2 dh dt −48 = 81π(4^2 )
dh dt dh dt
27 π
The water level is falling at
27 π
m/min
- Let V represent the volume of the hemisphere, r the radius, and dr the change in radius. We are given that r = 10 m and dr = 0. 2 cm = 0. 002 m. We wish to find dV.
V =
2 π 3
r^3
dV = 2πr^2 dr = 2π(10^2 )(0.002) = 0. 4 π m^3
We are given that xy = 20000 ⇒ y = 20000/x. The area R of the road equals the area of the large rectangle minus the small rectangle in
