Section 10.2 Calculus with Parametric Curves EX.2

Section 10.2 Calculus with Parametric Curves

EX.2

x = ¹_t ⇒ ^dx_dt = ⁻¹_t2 and y =√

te^−t ⇒ ^dy_dt = e^−tt⁻¹² (¹₂ − t). So ^dy_dx = e^−tt³²(t − ¹₂)

EX.5

x = t cos t ⇒ ^dx_dt = cos t − t sin t and y = t sin t ⇒ ^dy_dt = sin t + t cos t. So t = π ,^dy_dx = ^−π₋₁ = π Moreover, the tangent line is y − y(π) = π(x − x(π) ⇒ y = πx + π²

EX.6

x = cos θ + sin 2θ ⇒ x(0) = 1 and y = sin θ + cos 2θ ⇒ y(0) = 1.

dx

dθ = − sin θ + 2 cos θ and ^dy_dt = cos θ − 2 sin 2θ. So t = 0 ,^dy_dx = ¹₂. So the tangent line is y = ¹₂x +¹₂.

EX.11

x = t² + 1 and y = t²+ t ⇒ ^dx_dt = 2t and ^dy_dt = 2t + 1. So ^dy_dx = ^2t+1_2t . Moreover, ^d

dy dx

dt = ⁻¹_2t2, ^d_dx^2y2 =

ddy dx dt dx dt

=

−1 2t2

2t = _4t⁻¹3. So the curve is concave upward as _dx^d2y2 > 0 ⇒ t < 0

EX.16

x = cos 2t and y = cos t ⇒ ^dx_dt = −2 sin 2t and ^dy_dt = − sin t. So ^dy_dx = _{4 cos t}¹ = ^{sec t}₄ . Moreover, ^d

dy dx

dt = sec t tan t

4 , _dx^d2y2 =

ddy dx dt dx dt

=

sec t tan t 4

−2 sin 2t = ^{− sec}₁₆^3t. So the curve is concave upward as ^d_dx^2y2 > 0 ⇒

π

2 < t < π

EX.30

x = 3t²+ 1 and y = 2t³+ 1 ⇒ ^dx_dt = 6t and ^dy_dt = 6t² ⇒ _dx^dy = t

The tangent line is y − (2t³+ 1) = t(x − (3t²+ 1)) and passes through x = 4, y = 3.

So 2 − 2t³ = 3t − 3t³ ⇒ t³− 3t + 2 = 0 ⇒ (t + 2)(t − 1)² = 0 ⇒ t = −2 or t = 1.

The tangent line is y − 3 = x − 4 or y − 3 = −2(x − 4).

1

EX.31

The area is 4Ra

0 y dx x=a cos θ,y=b sin θ

===========⇒ 4R0

π 2

b sin θa(− sin θ) dθ = 4abR ^π₂

0 sin²θdθ = abπ

EX.41

dx

dt = 6t , ^dy_dt = 6t², Arc length=R1

0 p(6t)²+ (6t²)² dt ⇒ 6R1 0 t√

t²+ 1 dt = 2(t²+ 1)(³²) 

1 0 = 4√

2 − 2

EX.44

dx

dt = −3 sin t + 3 sin 3t and ^dy_dt = 3 cos t − 3 cos 3t, Arc length =Rπ

0 3p(sin 3t − sin t)²+ (cos 3t − cos t)²dt =Rπ 0 3√

2 − 2 sin 3t sin t − 2 cos 3t cos t dt

=Rπ

0 3p2 + cos 4t − cos 2t − (cos 4t + cos 2t) dt = 6 R₀^πq

1−cos 2t 2

= 6Rπ

o sin t dt = −6 cos t 

π 0 = 12

EX.62

By the formula:

The area of surface =R1

0 2π 3t²p(3 − 3t²)²+ (6t)²dt = 18πR1

0 t²p(t²− 1)²+ (2t)²dt

= 18πR1

0 t²(t²+ 1) dt = 18π(¹₅ +¹₃) = ⁴⁸₅π

EX.63

By the formula:

The area of surface =R ^π₂

0 2π a sin³θp

(−3a cos²θ sin θ)²+ (3a cos θ sin²θ)²dθ

= 2πaR^π₂

0 sin³θp

(3a cos θ sin θ)(cos θ²+ sin²θ) dθ

= 6πa²R^π₂

0 sin⁴θ cos θ dθ = ⁶₅πa²

2