Question

Find slope of the tangent line to the parametric curve \[x=t^2+1\space\space,\space\space y=\frac{t}{2}\space\text{ at} \space \space t=-1 \text{ and at} \space \space t=1 \text{ by eliminating parameters}\]

Answer 1

I know how to do it by without eliminating parameters. I want to know the method of elimination of parameters.

Answer 2

just solve both equations for t

*sorry my connection is bad right now

Answer 3

Why both?

Solve one and substitute that value of t in another equation. huh?

Answer 4

\[t=\sqrt{x-1}=2y\implies y=\frac12\sqrt{x-1}\]at\[t=-1\implies x=2\]so we now want\[\frac{dy}{dx}|_{x=2}\]

Answer 5

Sorry it is not square.

Answer 6

\[\large y^2=\pm\frac{\sqrt{x-1}}{2}\]
This is what I got.

Answer 7

we can only do one half of the parabola at a time, so you have two equations for the tangent lines

Answer 8

OK at t=1 and at t=-1 I have x=2 for both.

And I have equation look like this\[\large y=\pm\frac{\sqrt{x-1}}{2}\]Which one can I use equation with + sign or equation with -sign?

And also in this question we have t=1 and t=-1 both are giving same value x=2. What to do when I have two different values of x's on different t's, which one I have to choose then?

Answer 9

r(t) = <x(t),y(t)>

r'(t) = <x'(t),y'(t)>

right?

Answer 10

yes.

Answer 11

then r'(t) = <2t, 1/2>; at t=1 we have a tangent vector of <2,1/2> then

Answer 12

so what is it we are trying to accomplish here?

Answer 13

slope of tangent line.

Answer 14

Wait wait.. I think I got now.

Answer 15

slope = y/x

1/2
--- = 1/4
2

Answer 16

Yes right.

Answer 17

At t=-1?

Answer 18

-1/4 right?

Answer 19

1/2
---- is our general set up for this; so when t=-1, -1/4
2t

Answer 20

Yeah.. Thanks..

I want to ask you one thing. You used vector here?

Answer 21

In book nothing given as <> (with these brackets) . But you use these brackets. So that's why I am asking to you.

Answer 22

i did use vector yes; since a parametric defines a vector in terms of x and y components

Answer 23

OK..

Answer 24

if i were to define it another way:

x = t^2 + 1 ; y = t/2

t = 2y

x = (2y)^2 + 1

x = 4y^2 + 1 ; derive it implicitly

1 = 8y y'

y' = 1/8y ; y = t/2

y' = 1/(8t/2) = 1/4t

Answer 25

Yeah Thanks.. Now it is cleared.

Answer 26

@TuringTest I thank you too friend. :)

Answer 27

welcome, but amistre's answer was better :)