422 STEP 5. Build Your Test-Taking Confidence- The correct answer is (D).
The slope of a tangent line to a polar curve
r=f(θ)is
dy
dx=
dy
dθ
dx
dθ. Sincex=rcosθand
y=rsinθ, you have
dx
dθ
=f′(θ) cosθ−f(θ) sinθand
dy
dθ
= f′(θ) sinθ+ f(θ) cosθ, and thus
dy
dx=
f′(θ) sinθ+f(θ) cosθ
f′(θ) cosθ−f(θ) sinθ. Note that
r=f(θ)=2 cosθand f′(θ)=−2 sinθ.Atθ=
π
3,
dy
dx=
−2 sin(
π
3)
sin(
π
3)
+2 cos(
π
3)
cos(
π
3)−2 sin(
π
3)
cos(
π
3)
−2 cos(
π
3)
sin(
π
3).- The correct answer is (C).
Sincefis decreasing, f′<0 and sincefis
concave up,f′′>0. The graph also shows
thatf(0)<0. Thusf′′(0) has the largest
value. - The correct answer is (C).
Letu=x+2 and thus
du
dx=1ordu=dx.
Sinceu=x+2,u− 2 =xand therefore
fx√
x− 2 dxbecomes∫
(u−2)√
uduor
∫
(u−2)(u(^12)
)duor
∫
(u^3 /^2 − 2 u
(^12)
)du.
Integrating, you have
u^5 /^2
(^5) / 2 −
2 u
(^32)
3
2
+cor
2
5
(u)
(^5) / 2
−
4
3
(u)^3 /^2 +cor
2
5
(x+2)(^5) / 2
−
4
3
(x+2)(^3) / 2
+c.
- The correct answer is (B).
Apply partial fraction decomposition
5
(x−3)(x+2)
=
A
x− 3+
B
x+ 2=
A(x+2)+B(x−3)
(x−3)(x+2). Thus
A(x+2)+B(x−3)=5 and by lettingx=3,
you haveA=1 and lettingx=−2, you have
B=−1. Therefore,
∫
5
(x−3)(x+2)
dx=
∫ (
1
x− 3−
1
x+ 2)
dx=∫
1
x− 3dx−∫
1
x+ 2
dx=ln|x− 3 |−ln|x+ 2 |+c.- The correct answer is (B).
Note that for each column, all the tangents
have the same slope. For example, whenx= 0
all tangents are horizontal, which is to say their
slopes are all zero. This implies that the slope
of the tangents depends solely on the
x-coordinate of the point and is independent
of they-coordinate. Also note that whenx> 0
slopes are negative, and whenx<0 slopes are
positive. Thus, the only differential equation
that satisfies these conditions is
dy
dx
=− 2 x. - The correct answer is (C).
The Taylor series forexaboutx=0is
ex= 1 +x+
x^2
2!
+
x^3
3!
+···, and thus, fore^2 xise^2 x= 1 + 2 x+
(2x)^2
2!+
(2x)^3
3!
+···= 1 + 2 x+
(2^2 )x^2
2!+
(2^3 )x^3
3!
+···.Thereforexe^2 x=x+ 2 x^2 +
(2^2 )x^3
2!+
(2^3 )x^4
3!+···=
x+ 2 x^2 +
4 x^3
2!+
8 x^4
3!