Question

Tangent lines to parametric curves

Answer 1

I guessed

Answer 2

I know you first take the dy/dt and divide by dx/dt

Answer 3

Chain rule:\[
\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}
\]

Answer 4

well, a parametric gives us a vector:

r = (x,y)

yes, x',y' gives us the direction of the tangent.

Answer 5

Results in: \[
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]

Answer 6

x' = 1 - 1/t^2
y' = 2t - 3/t^3

given that x = t+1/t = 3, what is our value of t?

Answer 7

and for that matter, what is x,y

Answer 8

y'=2t -2/t^3

Answer 9

yeah, had a 3 on the brain :)

Answer 10

I'm having trouble with finding what t= when x=3

Answer 11

cause you have t+(1/t)=3

what??

Answer 12

you could multiply both sides by t

Answer 13

it turns into an ugly quadratic

Answer 14

quads are never ugly :)

Answer 15

maybe changing it into cartesian makes it simple

Answer 16

notice t^2 + 1/t^2 = (t+1/t)^2 - 2

Answer 17

Oh, did you complete the square there?

Answer 18

oh very nice @ganeshie8
you can square x...
then subtract 1 one on both sides
and bingo have y

Answer 19

`t^2 + 1/t^2` = ( `t+1/t`)^2 - 2
y = x^2 - 2

Answer 20

\[x=t+\frac{1}{t} \\ x^2=(t+\frac{1}{t})^2=t^2+2+\frac{1}{t^2}\]
yeah oops 2 not 1

Answer 21

NOw that I have y=x^2
y'=2x
y'(3)=6

y-y_0=6(x-x_0)

Answer 22

oh x=3 and y=(3)^2-2=7

y-7=6(x-3)
y=6x-18+7
y=6x-11?

Answer 23

Oh yeah I got it, thanks everyone! :)