Parametric Equations - Have solution, need clarification.

[tangibleLime]

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)?

 

[LCKurtz]

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?

 

[tangibleLime]

Ooh okay, makes sense. Need to just find the component that makes r(t) equal the given point. Thanks!

 

[tangibleLime]

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?

 

[Char. Limit]

Have you tried t=1?

 

[tangibleLime]

Oh, so I can pick ANY number that makes r(t) equal to the given point?

 

[Char. Limit]

well, yeah. Although if you have multiple t that make r(t) fit the point, you've got a pretty cool function.

 

[HallsofIvy]

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