Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
x = t − t$^{−1}$, y = 6 + t$^2$, t = 1

The correct answer needs to be in the form of y=something... I have this (incorrect):
$$y=(\large\frac{2t^3}{t^2+1})x+7$$

@ t=1, x=0 and y=7

$\large\frac{dy}{dt}$=2t
$\large\frac{dx}{dt}$= 1+$\large\frac{1}{t^2}$ = $\large\frac{t^2+1}{t^2}$

Dividing by something, same as the product of its reciprocal, when doing $\frac{dy}{dx}$

Where's the error?

Question

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
x = t − t$^{−1}$,    y = 6 + t$^2$,    t = 1

The correct answer needs to be in the form of y=something...  I have this (incorrect):
$$y=(\large\frac{2t^3}{t^2+1})x+7$$

@ t=1, x=0 and y=7

$\large\frac{dy}{dt}$=2t
$\large\frac{dx}{dt}$= 1+$\large\frac{1}{t^2}$ = $\large\frac{t^2+1}{t^2}$

Dividing by something, same as the product of its reciprocal, when doing $\frac{dy}{dx}$

Where's the error?

anonymous · Answer

Fixed the typo with the fraction command, press F5 to refresh and make it display properly.  Am I understanding this question right @TuringTest ?

turingtest · Answer

It's looking correct to me...

turingtest · Answer

except that you can plug in the value for t...
which just makes y=x+7

anonymous · Answer

Wait what?  You mean just plug in 1?

anonymous · Answer

Ooooh!

turingtest · Answer

why not, you need to know the value of the slope at that point, no?

anonymous · Answer

2/2 = 1 and it goes away

turingtest · Answer

yeah, is that right? do you know?

anonymous · Answer

lol that was it

/facepalm at myself >_<

turingtest · Answer

haha sweet, it happens :)

anonymous · Answer

Well everybody got to see my work then lol

anonymous · Answer

i can't really read well your solution, but it is done like this:
find point value for t=1, which is (0,7)
find tanget vector at t=1 which is (2,2)
write the equation: (x,y) = (0,7)+v(2,2)
from here you can write it in the form y=something by eliminating v

anonymous · Answer

Ty myko :-)  Yeah that's basically my steps too

I'm doing...
1.  find dy & dx with respect to dt  (derivative of both)
2.  find dy/dx
3.  find the point for when t is as given, the x & y coords
4.  put in the form of y=mx+b, solve for b using the known x & y
5.  rewrite y=mx+b

anonymous · Answer

my way is shorter, :). You can avoid step 2. But ya, good job

anonymous · Answer

no need of that, becouse actually sometimes it will not work, if the dy/dx is infinity for example

anonymous · Answer

just keep y and x same level variables without considering one of them like the independent one till the end