Download Parametric Equations and Calculus

Parametric Formulas
dy
dy dt

dx dx
dt
First derivative of a function in parametric form:
d  dy 
d y dt  dx 

dx
dx 2
dt
2
Second derivative of a function in parametric form :
2
2
 dx   dy 
Arc length of a function in parametric form: s 
   dt
a  dt 
 dt 

b
Parametric Equations and Calculus
Ex. 1 (Noncalculator)
Given the parametric equations x  2 t and y  3t 2  2t , find
dy
d2 y
and
.
dx
dx 2
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Ex. 2 (Noncalculator)
Given the parametric equations x  4cost and y  3sin t , write an equation of the tangent line to the
curve at the point where t 
3
.
4
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Ex 3 (Noncalculator)
Find all points of horizontal and vertical tangency given the parametric equations
x  t 2  t, y  t 2  3t  5.
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Ex. 4 (Noncalculator)
Set up an integral expression for the arc length of the curve given by the parametric
equations x  t 2  1, y  4t 3  1, 0  t  1. Do not evaluate.
Ex. 1 (Noncalculator)
Given the parametric equations x  2 t and y  3t 2  2t , find
dy
d2 y
and
.
dx
dx 2
Solution:
dy
To find
, we must differentiate both of the parametric equations with respect to t.
dx
1

dy d
dx d 
1  12
2




3t  2t   6t  2 and

2 t  2  t  t 2 so

dt dt 
dt dt 
2
dy
3
1
dy dt 6t  2
2  2t 2



6t
1
dx dx
t 2
dt
d2 y
dy
dy
To find
, we must differentiate
with respect to x so [thinking of
2
dx
dx
dx
as a function of t and t as a function of x] that
1
1
1
1
1 dt
 1 dt
d 2 y d  dy 
 9t 12  t  12  dt  9t 2  t 2  9t 2  t 2  9t  1
2
2


9t

t

 
1
 dx
dx
dx
dx 
dx 2 dx  dx 
t 2
dt
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Ex. 2 (Noncalculator)
Given the parametric equations x  4cost and y  3sin t , write an equation of the tangent line to the
curve at the point where t 
3
.
4
Solution:
dy
dy dt


dx dx
dt
d
3sin t 
3cost
3
dt 

  cot t
d
 4sin t
4
 4cost 
dt

 2 3 2
3 dy
3
3
2
When t 
,
  1  , x  4  
   2 2, and y  3

4 dx
4
4
2
 2 
 2 
 


3 2 3
 x2 2 .
2
4
(Remind students that they may leave their tangent line equations in point-slope form.)
Therefore the tangent line equation is: y 
Ex 3 (Noncalculator)
Find all points of horizontal and vertical tangency given the parametric equations
x  t 2  t, y  t 2  3t  5.
dy
d 2
t  3t  5

 2t  3
dy dt
dt
Solution:



d 2
dx dx
2t  1
t  t 


dt
dt
dy
A horizontal tangent will occur when
 0 , which happens when 2t  3  0 (and 2t  1  0)
dx
3
3
so a horizontal tangent occurs at t  . Substituting t  into the given equations, we find that a
2
2
 15 11
dy
horizontal tangent will occur at  ,  . A vertical tangent will occur when
is undefined,
dx
 4 4
1
. Substituting
2
 1 27 
1
t   into the given equations, we find that a vertical tangent will occur at   ,  .
2
 4 4
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Ex. 4 (Noncalculator)
Set up an integral expression for the arc length of the curve given by the parametric
equations x  t 2  1, y  4t 3  1, 0  t  1. Do not evaluate.
which happens when 2t  1  0 (and 2t  3  0) so a vertical tangent occurs at t  
Solution:
For parametric equations, the formula for arc length is:
2
s
b
a
2
 dx   dy 
 dt    dt  dt
For our problem,
dx d 2
dy d
 t  1  2t and
  4t 3  1  12t 2 so the arc length is given by the
dt dt
dt dt
integral expression s  
1
0
2t   12t 2  dt
2
2
or s  
1
0
4t 2  144t 4 dt .
CALCULUS BC
WORKSHEET ON PARAMETRICS AND CALCULUS
Work the following 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 , 0  t  2
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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
dy d  2
dx d  2 

t  6t  5  2t  6 and

t  2t

dt dt 
dt dt  
dy
dy dt 2t  6
3


 1
dx dx
2t
t
dt
d2y
dy
To find
, we must differentiate
with respect to x so that
2
dx
dx
3
3
 2  2
d 2 y d  dy  d  3 
3 dt
t  t  3


1








2t
dx 2 dx  dx  dx  t 
t 2 dx dx
2t 3
dt
1.
______________________________________________________________________________
dy d  3 2 
dx d  2 
  2t  t   6t 2  2t and

t  1  2t
dt dt
dt dt 
dy
dy dt 6t 2  2t


 3t 1
dx dx
2t
dt
d2y
dy
To find
, we must differentiate
with respect to x so that
2
dx
dx
d 2 y d  dy  d
dt 3
3
    3t 1  3 

2
dx  dx  dx
dx dx 2t
dx
dt
2.
________________________________________________________________________________
dy d  2
dx d   1  12

3.

3t  2t  6t  2 and

t  t

dt dt 
dt dt   2
dy
3
1
dy dt 6t  2
2


 12t  4t 2
dx dx 1  12
dt 2 t
d2y
dy
To find
, we must differentiate
with respect to x so that
2
dx
dx
1
1 
1
2
1 
d 2 y d  dy  d  32
2   18t 2  2t 2  dt  18t  2t



12
t

4
t
 dx
dx
dx 2 dx  dx  dx 
 

dt
1
2
1
18t 2  2t

1  12
t
2
1
2
 36t  4
dy d  2 
dx d   1

t  t  2t  1 and

ln t 


dt dt
dt dt   t
dy
dy dt 2t  1


 2t 2  t
1
dx dx
t
dt
2
d y
dy
To find
, we must differentiate
with respect to x so that
2
dx
dx
d 2 y d  dy  d  2 
dt 4t  1 4t  1
  
2t  t   4t  1 

 4t 2  t
2


1
dx
dx  dx  dx
dx
dx
t
dt
4.
________________________________________________________________________________
dy d
dx d 
  4cos t 1   4sin t and

3sin t  2  3cos t
dt dt
dt dt 
dy
dy dt  4sin t
4


  tan t
dx dx 3cos t
3
dt
d2y
dy
To find
, we must differentiate
with respect to x so that
2
dx
dx
4
4
 sec2 t  sec2 t
  4

d 2 y d  dy  d  4
dt
4
      tan t     sec2 t   3
 3
  sec3 t
2
dx
dx  dx  dx  3
3cos t
9
dx
  3
 dx
dt
5.
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6. (a)
dy 3t 2  2t

dx
2t  1
(b) When t = 1,
dy 3 12  2 1 8

 , x  5, y  4 so the tangent line equation is
dx
2 1  1
5
8
 x  5
5
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dy 3cos t
3
7. (a)

  cot t
dx  2sin t
2
 dy
3

3
3 2
  cot   , x  2, y 
(b) When t  ,
so the tangent line equation is
2
4
2
2
4 dx
3 2
3
y
  x 2
2
2
y4 


8. (a)
dy 2t  4

dx
1
dy
dx
 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 tangent occurs at the point 7,  4 . A vertical tangent occurs
dy
dx
when
 0 and
 0.
dt
dt
Since 1  0 , there is no point of vertical tangency on this curve.
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dy 3t 2  3

9. (a)
dx 2t 1
dy
dx
(b) A horizontal tangent occurs when
 0 and
 0 so a horizontal tangent
dt
dt
occurs when 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 
dy
dx
A vertical tangent occurs when
 0 and
 0 so a vertical tangent occurs
dt
dt
1
1
3
11
when 2t 1  0 so t  . When t  , x  and y   so a vertical tangent
2
2
4
8
 3 11 
occurs at the point  ,   .
8
4
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dy 4cos t
10. (a)

dx 2sin t
dy
dx
(b) A horizontal tangent occurs when
 0 and
 0 so a horizontal tangent
dt
dt


3
occurs when 4cos t  0 which is at t  and
. When t = , x = 3 and y = 3,
2
2
2
3
and when t 
, x = 3 and y =  5 so a horizontal tangent occurs at the
2
points  3, 3 and  3,  5 .
dy
dx
A vertical tangent occurs when
 0 and
 0 so a vertical tangent occurs
dt
dt
when 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 .
(b) A horizontal tangent occurs when
2
2
2
 dx   dy 
11. s 
   dt 


a  dt 
0
 dt 

b
  2t 
2
 
 3t 2
2
dt or
2
0
4t 2  9t 4 dt
______________________________________________________________________
2
2
2
 dx   dy 
12. s 
   dt 


a  dt 
2
 dt 

b

 
2e2t
2
  3 dt or
2
2
2
4e4t  9 dt