CALCULUS BC
WORKSHEET ON PARAMETRICS AND CALCULUS
Work these on notebook paper. Do not use your calculator.
dy
d2 y
and
On problems 1 – 5, find
.
dx
dx 2
1. x  t 2 , y  t 2  6t  5
2. x  t 2  1, y  2t 3  t 2
3. x  t , y  3t 2  2t
4. x  ln t, y  t 2  t
5. x  3sin t  2, y  4cost  1
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6. A curve C is defined by the parametric equations x  t 2  t  1, y  t 3  t 2 .
dy
in terms of t.
dx
(b) Find an equation of the tangent line to C at the point where t = 2.
(a) Find
7. A curve C is defined by the parametric equations x  2cost, y  3sin t .
(a) Find
dy
in terms of t.
dx
(b) Find an equation of the tangent line to C at the point where t =

.
4
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On problems 8 – 10, find:
dy
(a)
in terms of t.
dx
(b) all points of horizontal and vertical tangency
8. x  t  5, y  t 2  4t
9. x  t 2  t  1, y  t 3  3t
10. x  3  2cost, y  1 4sint
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On problems 11 - 12, a curve C is defined by the parametric equations given. For each problem,
write an integral expression that represents the length of the arc of the curve over the given interval.
11. x  t 2 , y  t 3 , 0  t  2
12. x  e2t  1, y  3t  1,  2  t  2
Answers to Worksheet on Parametrics and Calculus
3
 2
dy 2t  6
3 d2y
3
1.

 1 ;
 t  3
2
dx
2t
t
dx
2t
2t
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dy
d2y 3
2.
 3t  1;

dt
dx 2 2t
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1
18t 2

1
2
dy 6t  2
d y
 2t


 36t  4
1
1
2
dx 1  2
dx
1 2
t
t
2
2
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3
 12t 2
3.
1
 4t 2 ;
2
dy 2t  1
d 2 y 4t  1

 2t 2  t;

 4t 2  t
1
1
dx
dx 2
t
t
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4
 sec2 t
2
dy  4sin t
4
d y
4
5.

  tan t;
 3
  sec3 t
2
dx 3cos t
3
dx
3cos t
9
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dy 3t 2  2t

6. (a)
dx
2t  1
dy 312  2 1 8

 , x  5, y  4 so the tangent line equation is
(b) When t = 1,
dx
2 1  1
5
8
y4 x5
5
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dy 3cost
3
7. (a)

  cot t
dx  2sint
2
4.

(b) When t 


4
,
dy
3

3
3 2
so the tangent line equation is
  cot   , x  2, y 
dx
2
4
2
2


3 2
3
 x 2
2
2
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dy 2t  4
8. (a)

dx
1
dy
dx
(b) A horizontal tangent occurs when
 0 and
 0 so a horizontal tangent
dt
dt
occurs when 2t  4  0 which is at t= 2. When t = 2, x = 7 and y =  4 so a horizontal
dx
dy
tangent occurs at the point 7,  4 . A vertical tangent occurs when
 0 and
0
dt
dt
Since 1  0 , there is no point of vertical tangency on this curve.
y


9. (a)
dy 3t 2  3

dx 2t  1
dy
dx
 0 and
 0 so a horizontal tangent occurs when
dt
dt
3t 2  3  0 which is at t  1 . When t = 1 , x = 1 and y =  2, and when
t  1, x = 3 and y = 2 so a horizontal tangent occurs at the points 1,  2 and 3, 2
(b) A horizontal tangent occurs when


 
dx
dy
 0 and
 0 so a vertical tangent occurs when
dt
dt
1
3
11
1
2t  1  0 so t  . When t  , x  and y   so a vertical tangent occurs at the
2
4
8
2
 3 11
point  ,   .
4 8
A vertical tangent occurs when
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dy 4cost
10. (a)

dx 2sint
dy
dx
(b) A horizontal tangent occurs when
 0 and
 0 so a horizontal tangent
dt
dt

3

occurs when 4cost  0 which is at t  and
. When t =
, x = 3 and y = 3, and when
2
2
2
3
, x = 3 and y =  5 so a horizontal tangent occurs at the points 3, 3 and 3, 5 .
t
2
dx
dy
A vertical tangent occurs when
 0 and
 0 so a vertical tangent occurs when
dt
dt
2sin t  0 so t  0 and  . When t  0, x  5 and y  1 and when t   , x  1 and y  1
 






so a vertical tangent occurs at the points 5,  1 and 1, 1 .
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2
11. s 
b
a
2
 dx   dy 
 dt    dt  dt 
2
 0 2t   3t  dt or  0
2
2
2
2
4t 2  9t 4 dt
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2
12. s 
b
a
2
 dx   dy 
 dt    dt  dt 
2
2
2e   3 dt or 
2t 2
2
2
2
4e4t  9 dt