Parametric Equations (Calc.2)

[dolpho]

Homework Statement

Parametrize the curve by a pair of differentiable functions x = x(t), y = y(t) with [x '(t)]2 + [y '(t)]2≠0, then determine the tangent line at the origin.

y=2x^3

The Attempt at a Solution

Honestly I don't really understand what it's asking for. I assume it wants us to make 2 equations, x= something and y = something but I'm not quite sure how to get there. Then we can find the tangent line by taking the derivative.

Unfortunately I can't even show work on this problem since I don't even know where to start. Would appreciate any help on this question <3

 

[HallsofIvy]

"Parametric equations" for a curve in the xy-plane are two equation x= f(t), y= g(t) such that, for any t, the corresponding point (x(t), y(t)) is a point on the curve. In particular, if the curve is given a function, y= F(x), we can just write x= t, y= F(t).

 

[dolpho]

Hmmmm right... So I'm a bit confused on how we take our starting equation and turn it into two.

Would we just do

y=2x^3 and x = (y/2)^1/3 ?

 

[dolpho]

Hmmmm right... So I'm a bit confused on how we take our starting equation and turn it into two.

Would we just do

y=2t^3 and x = (t/2)^1/3 ?

 

[Mark44]

Hmmmm right... So I'm a bit confused on how we take our starting equation and turn it into two.

Would we just do

y=2x^3 and x = (y/2)^1/3 ?

No. Here you are apparently finding the inverse of the function. The first equation has y as a function of x, and the second has x as a function of y.

Your first equation can be symbolized as y = f(x), and the second as x = f^-1(y).
That's not what you need to do.

Would we just do

y=2t^3 and x = (t/2)^1/3 ?

No. How about x = t? What would y be then, as a function of t?