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 and
y=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