192 STEP 4. Review the Knowledge You Need to Score High
Example
Find the equation of the tangent line to the curver= 2 +2 sinθwhenθ=
π
4.
Step 1:
dr
dθ
=2 cosθ
Step 2:
dx
dθ
=−(2+2 sinθ) sinθ+cosθ(2 cosθ)=2(cos^2 θ−sin^2 θ−sinθ)
By the Pythagorean identity,
2(cos^2 θ−sin^2 θ−sinθ)=2(1−sin^2 θ−sin^2 θ−sinθ)
=2(1−sinθ−2 sin^2 θ)=2(1−2 sinθ)(1+sinθ).
Also,
dy
dθ=(2+2 sinθ) cosθ+sinθ(2 cosθ)=2 cosθ(1+2 sinθ).Step 3:
dy
dx=
2 cosθ(1+2 sinθ)
2(1−2 sinθ)(1+sinθ)=
cosθ(1+2 sinθ)
(1−2 sinθ)(1+sinθ)Step 4: Whenθ=
π
4,
dy
dx=
cos
π
4(
1 +2 sin
π
4)(
1 −2 sin
π
4)(
1 +sin
π
4).Evaluating,
dy
dx=
√
2
2(1+
√
2)(1−
√
2)(
1 +√
2
2)=− 1 −√
2.Step 5: When θ=
π
4
,r = 2 +√
2, sox=rcosθ =(
2 +√
2)√
2
2=
√
2 +1 andy=rsinθ=(
2 +√
2)√
2
2=
√
2 +1.Step 6: The equation of the tangent line isy−(√
2 + 1)
=(
− 1 −√
2)(
x−(√
2 + 1))
( ory=
− 1 −√
2)
x+ 4 + 3√
2.
Velocity and Acceleration of Vector Functions
A vector-valued function assigns a vector to each element in a domain of real numbers.
Ifr=〈x,y〉is a vector-valued function, limt→cr exists only if limt→cx(t) and limt→cy(t) exist.
limt→cr=〈
limt→cx(t), limt→cy(t)〉
=limt→cx(t)i+limt→cy(t)j.A vector-valued function is continuous
atc if its component functions are continuous atc. The derivative of a vector-valued
function is
dr
dt=
〈
dx
dt,
dy
dt〉
=i
dx
dt
+j
dy
dt.
Ifr =〈x,y〉is a vector-valued function that represents the path of an object in the
plane, andxandy are both functions of a variablet,x= f(t) andy =g(t), then the
velocity of the object isv=
dr
dt
=i
dx
dt
+ j
dy
dt=
〈
dx
dt,
dy
dt〉. Speed is the magnitude
of velocity, so|v|=√(
dx
dt) 2
+(
dy
dt) 2. The direction ofv is along the tangent to