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Parametric Equations - Have solution, need clarification.

  • #1

Homework Statement


Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 2cos(t)
y = sin(t)
z = t

At the point (0,1,pi/2)

Homework Equations





The Attempt at a Solution



r(t) = <2cos(t),sin(t),t>
r'(t) = <-2sin(t),cos(t),1>

The parameter value corresponding to the point (0,1,pi/2) is t=pi/2, so the tangent vector there is r'(pi/2) = <-2,0,1>. The tangent line through (0,1,pi/2) parallel to the vector <-2,0,1>, so its parametric equations are:

x=-2t
y=1
z = (pi/2)+t

I know this is correct, but the part in bold is what I do not understand. Why does the value of t=pi/2 correspond to the point (0,1,pi/2)?
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
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733

Homework Statement


Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 2cos(t)
y = sin(t)
z = t

At the point (0,1,pi/2)

Homework Equations





The Attempt at a Solution



r(t) = <2cos(t),sin(t),t>
r'(t) = <-2sin(t),cos(t),1>

The parameter value corresponding to the point (0,1,pi/2) is t=pi/2, so the tangent vector there is r'(pi/2) = <-2,0,1>. The tangent line through (0,1,pi/2) parallel to the vector <-2,0,1>, so its parametric equations are:

x=-2t
y=1
z = (pi/2)+t

I know this is correct, but the part in bold is what I do not understand. Why does the value of t=pi/2 correspond to the point (0,1,pi/2)?
What happens if you calculate r(pi/2) in r(t) = <2cos(t),sin(t),t>? Would any other value of t work in the 3rd component?
 
  • #3
Ooh okay, makes sense. Need to just find the component that makes r(t) equal the given point. Thanks!
 
  • #4
Actually no, that doesn't work for my problem.

(Now i'm doing the actual question I have to solve instead of the book example)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 1+2sqrt(t)
y = t^4 + t
z = t^4 - t

(3,2,0)

I have r(t) = <1+2sqrt(t),t^4 + t,t^4 - t>, but no point (3, 2 or 0) will make this vector (3, 2, 0). What am I doing wrong?
 
  • #5
Char. Limit
Gold Member
1,204
13
Have you tried t=1?
 
  • #6
Oh, so I can pick ANY number that makes r(t) equal to the given point?
 
  • #7
Char. Limit
Gold Member
1,204
13
well, yeah. Although if you have multiple t that make r(t) fit the point, you've got a pretty cool function.
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,795
925
In parametric equations, x= f(t), y= g(t), z= h(t), the parameter, t, is not necessarily equal to any of x, y, or z. It is a completely different parameter.


Sometimes, back in two dimensions, if we have, say y= f(x), we use x itself as parameter. That may have confused you.

Here, you want [itex](1+2\sqrt(t),t^4 + t,t^4 - t)= (3, 2, 0)[/itex] so we must have [itex]1+ 2\sqrt{t}= 3[/itex], [itex]t^4+ t= 2[/itex], and [itex]t^4- t= 0[/itex] all for the same t. We could, for example, solve [itex]t^4- t= t(t^3- 1)= t(t- 1)(t^2+ t+ 1)= 0[/itex] to see that t must be 0 or 1. t= 0 does not satisfy either [itex]1+ 2\sqrt{t}= 3[/itex] nor [itex]t^4+ t= 2[/itex] but t= 1 does.
 

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