Question

HELP PLEASE!

Answer 1

I am having trouble with #2.
@zepdrix @sweetburger @518nad

Answer 2

slope = dy/dx
point = (1-1^-1,7+1^2)

Answer 3

(dy/dt)/(dx/dt) = dy/dx

Answer 4

Eh. It's telling me that answer is wrong :/

Answer 5

How did you get those points?
and why didn't you take a derivative?

Answer 6

Please, would you share what you yourself have already done?

Here you're working with parametric equations. In this case the slope of the tangent line is given by \[\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt}}=\frac{ dy }{ dx }\]

Answer 7

Have you been able to calculate this derivative?

If so, let t=1 and figure out the particular slope for that parameter value.

Go back to the original equations for x and y and calculate their values at t=1.
This will give you a point on the curve.

From this point onward, just use the coordinates of the point and the slope to find the equation of the tangent line. Use the point-slope form of this equation to start with.

Answer 8

i did typo it shud be 1+1^-2 not 2

Answer 9

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Answer 10

This is what I did, but got the answer wrong.

Answer 11

@mathmale

Answer 12

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Answer 13

and oops.. i dunno why i wasnt getting rid of the constand in taking derivative ;)

Answer 14

Oh right.... Haha, I made a boo boo mistake. :/

Answer 15

(2t)/(1+t^-2)
t=1 (2)/(1+1^-2)= 1
y=x+8

Answer 16

still open?

Answer 17

Wait, so then the slop would be 2/0. Isnt that undefined?

Answer 18

@triciaal Sort of, just making silly mistakes.

Answer 19

Wouldn't the slope be...

\[\frac{ dy }{ dx} = \frac{ 2t }{ 1-\frac{ 1 }{ t^2 } }= \frac{ 2(1) }{ 1-\frac{ 1 }{ (1)^2 } }= \frac{ 2 }{ 0 }\]

Answer 20

Wait.... nevermind. I got it.... Thanks!

Answer 21

ok great