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Parametric Equations and Calculus Worksheet

Parametric Equations and Calculus Worksheet
  • Page 1
 1.  
Find the arc length of the curve x = t2, y = t3 for 1 ≤ t ≤ 2.
a.
127(8010+1313) units
b.
127(8010-8) units
c.
127(22-1) units
d.
377 15 units
e.
127(8010-1313) units


Solution:

x = t2, y = t3, 1 ≤ t ≤ 2
[Write the equation of the curve.]

dxdt = 2t, dydt = 3t2
[Find dxdt and dydt.]

L = 12(2t)2+(3t2)2 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 124t2+9t4 dt

= 12t(4+9t2)12 dt

Let u = 4 + 9t2, du = 18t dt
At t = 1, u = 13 and t = 2, u = 40

L = 12t(4+9t2)12 dt

= 1 / 18 1340u du

= 1 / 18 [u3232]1340
[Integrate.]

= 1 / 27 [4032-1332]
[Substitute the limits.]

= 1 / 27 [8010-1313]
[Simplify.]

So, arc length of the given parametric curve is 1 / 27 [8010-1313] units.


Correct answer : (5)
 2.  
Determine the length of the parametric curve whose equations are x = 3sin t, y = 3cos t, 0 ≤ t ≤ 2π.
a.
6π2 units
b.
62π units
c.
6π units
d.
3π units


Solution:

x = 3sin t, y = 3cos t, 0 ≤ t ≤ 2π
[Write the equation of the curve.]

dxdt = 3cos t, dydt = - 3sin t
[Find dxdt and dydt.]

Arc length of the curve = L = 0(3cos t)2+(- 3sin t)2 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 03cos2t+sin2tdt

= 03dt
[Use cos2 t + sin2 t = 1]

= [3t]0
[Integrate.]

= 6π
[Apply the limits.]

So, the length of the parametric curve is 6π units.


Correct answer : (4)
 3.  
Find the length of the parametric curve x = 5sin 3t, y = 5cos 3t, 0 ≤ t ≤ 2π.
a.
10π units
b.
15π units
c.
450π units
d.
30π units


Solution:

x = 5sin 3t, y = 5cos 3t, 0 ≤ t ≤ 2π
[Write the equation of the curve.]

dxdt = 15cos 3t, dydt = - 15sin 3t
[Find dxdt and dydt.]

Arc length of the curve = L = 0(15cos 3t)2+(-15sin 3t)2dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 015cos23t+sin23t dt

= 015dt
[Use cos2 3t + sin2 3t = 1]

= [15t]0
[Integrate.]

= 30π - 0 = 30π
[Apply the limits.]

So, the length of the parametric curve is 30π units.


Correct answer : (4)
 4.  
Find the arc length of the curve defined by x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π.
a.
3 units
b.
6 units
c.
2 units
d.
1.5 units


Solution:

x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π
[Write the equation of the curve.]

dxdt=- 3cos2 t sin t, dydt = 3sin2 t cos t
[Find dxdt and dydt.]

Arc length of the curve = L = 0(- 3cos2 t sin t)2+(3sin2 t cos t)2dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 30cos t sin tcos2 t+sin2 t dt

= 30sin t cos t dt
[Use cos2 t + sin2 t = 1]

= 3 / 2 0sin 2t dt
[Use sin 2t = 2sin t cos t]

= 3 / 2 (2) 0πsin 2t dt
[sin 2t is a periodic function with period π.]

= 3(2) 0π/2sin 2t dt
[sin (π - 2t) = sin 2t]

= 6[- cos 2t2]0π/2

= 3[- (cos 2(π2) - cos 2(0))]
[Apply the limits.]

= 3[1 + 1] = 6
[Simplify.]

So, the arc length of the parametric curve is 6 units.


Correct answer : (2)
 5.  
Find the arc length of the parametric curve for x = t - sin t, y = 1 - cos t, 0 ≤ t ≤ 2π.
a.
4 units
b.
2 units
c.
22π units
d.
8 units
e.
42 units


Solution:

x = t - sin t, y = 1 - cos t, 0 ≤ t ≤ 2π
[Write the equation of the curve.]

dxdt = 1 - cos t, dydt = sin t
[Find dxdt and dydt.]

Arc length of the curve = L = 0(1-cos t)2+sin2 t  dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 01-2cos t+cos2 t+sin2 t dt
[Use (a - b)2 = a2 - 2ab + b2]

= 02 - 2cos t dt
[Use cos2 t + sin2 t = 1]

= 2 0sin t2 dt
[Use cos t = 1 - 2sin2t2]

= 2(- 2cos t2)|0
[Integrate.]

= - 4[cos 2π2 - cos 0]
[Substitute the limits.]

= - 4(- 2) = 8
[Simplify.]

So, the arc length of the parametric curve is 8 units.


Correct answer : (4)
 6.  
Find the arc length of the curve between 0 and 1 for x = 3 - 3t, y = 4t.
a.
5 units
b.
7 units
c.
1 unit
d.
25 units


Solution:

x = 3 - 3t, y = 4t where 0 ≤ t ≤ 1
[Write the equation of the curve.]

dxdt = - 3, dydt = 4
[Find dxdt and dydt.]

Arc length of the curve = L = 01(- 3)2+42 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 015 dt
[Simplify.]

= 5[t]01
[Integrate.]

= 5(1 - 0) = 5
[Substitute the limits and simplify.]

So, the arc length of the parametric curve is 5 units.


Correct answer : (1)
 7.  
Find the arc length of the curve x = t2, y = 4t3 - 1 for - 1 ≤ t ≤ 1.
a.
127(3737-1) units
b.
904 15
c.
127(3737+1) units
d.
136(3737-1) units
e.
29(3737-1) units


Solution:

x = t2, y = 4t3 - 1, - 1 ≤ t ≤ 1
[Write the equation of the curve.]

dxdt = 2t, dydt = 12t2
[Find dxdt and dydt.]

L = - 11(2t)2+(12t2)2 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= - 114t2+144t4 dt

= - 112 | t | 1+36t2 dt

= 2012t 1+36t2 dt
[As - 112 | t | 1+36t2 dt is an even funcion.]

Let u = 1 + 36t2, du = 36(2)tdt
At t = 0, u = 1 and t = 1, u = 37

L = 2012t 1+36t2 dt

= 2 (1 / 36)137u du

= 1 / 18[u3/23/2]137
[Integrate.]

= 2 / 54 [3732-132]
[Apply the limits.]

= 1 / 27 [3737-1]
[Simplify.]

So, arc length of the parametric curve is 1 / 27 [3737-1] units.


Correct answer : (1)
 8.  
Find the arc length of the parametric curve for x = 1 - sin t, y = 2 + cos t, - π2  t π2.
a.
π units
b.
π24 units
c.
2π units
d.
π2 units


Solution:

x = 1 - sin t, y = 2 + cos t, - π2  t π2
[Write the equation of the curve.]

dxdt = - cos t, dydt = - sin t
[Find dxdt and dydt.]

Arc length of the curve = L = - π/2π/2(- cos t)2+(- sin t)2 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= - π/2π/2 dt
[cos2 t + sin2 t = 1]

= [t]- π/2π/2
[Integrate.]

= π2 - (-π2)
[Apply the limits.]

= π
[Simplify.]

So, the arc length of the parametric curve is π units.


Correct answer : (1)
 9.  
Calculate the arc length of the curve on [- 1 2, 1 3] for x = sin- 1 t, y = ln 1-t2.
a.
1 2(3ln 2 - 3ln 3 + ln 4) units
b.
1 2(ln 6) units
c.
1 2(ln 5) units
d.
ln 6 units
e.
1 2(ln 2 - ln 3 - ln 4) units


Solution:

x = sin- 1 t, y = ln 1-t2, - 1 / 2t1 / 3
[Write the equation of the curve.]

dxdt =11-t2, dydt =11-t2 (121-t2) (- 2t) = -t1-t2
[Find dxdt and dydt.]

Arc length of the curve, L = - 1/21/3(11-t2)2+(- t1-t2)2 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= - 1/21/311-t2+t2(1-t2)2 dt

= -1/21/31(1-t2)2 dt

= -1/21/3 11-t2 dt
[Simplify.]

= 1 / 2 - 1/21/3 [11-t+11+t] dt
[Write 11-t2 as 1 / 2(11-t+11+t)]

= 1 / 2 [- ln |1 - t| + ln |1 + t|]- 1/21/3
[Integrate.]

= ln 2 + ln 32
[Apply the limits and simplify.]

= 1 / 2 (ln 6)
[Use ln a + ln b = ln (ab)]

So, the arc length of the parametric curve is 1 / 2(ln 6) units.


Correct answer : (2)
 10.  
Find the arc length of the curve defined by the parametric equations x = t, y = 4-t2, 0 ≤ t ≤ 2.
a.
π2 units
b.
π6 units
c.
π4 units
d.
π units
e.
π3 units


Solution:

x = t, y = 4-t2, 0 ≤ t ≤ 2
[Write the equation of the curve.]

dxdt = 1, dydt =- 2t24-t2 =- t4-t2
[Find dxdt and dydt.]

Arc length of the curve, L = 0212+(- t4-t2)2 dt
[Use L = ab(dxdt)2+(dydt)2 dt]

= 021+t24-t2 dt

= 0224-t2 dt

= 02221-(t2)2 dt

= 2[sin- 1 (t2)]02
[ 11-x2 dx =sin- 1 x]

= 2[sin- 1 (1) - sin- 1 (0)]
[Apply the limits.]

= 2(π2 - 0)

= 2(π2) = π

So, the arc length of the parametric curve is π units.


Correct answer : (4)

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