## agentx5 3 years ago 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?

1. agentx5

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

2. TuringTest

It's looking correct to me...

3. TuringTest

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

4. agentx5

Wait what? You mean just plug in 1?

5. agentx5

Ooooh!

6. TuringTest

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

7. agentx5

2/2 = 1 and it goes away

8. TuringTest

yeah, is that right? do you know?

9. agentx5

lol that was it /facepalm at myself >_<

10. TuringTest

haha sweet, it happens :)

11. agentx5

Well everybody got to see my work then lol

12. myko

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

13. agentx5

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

14. myko

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

15. myko

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

16. myko

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