# Parametric Equations - Have solution, need clarification.

## 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)

## 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)?

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LCKurtz
Homework Helper
Gold Member

## 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)

## 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?

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

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
Gold Member
Have you tried t=1?

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

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

HallsofIvy
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.