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Posted

I know to find the slope of parametric equations you find dy/dx.

 

If you are given the parametric equations and you find an answer in terms of t for the gradient and you are given points in terms of x and y.

 

E.G.

 

Find the tangent of

 

x= 1 + lnt, y= t^2 +2 ; (1,3)

 

I get dy/dx = 2t^2

 

Now the text book is not very clear...

 

What should I do with the 2t^2 and what should I do with this to find the equation of the tangent?

 

Must I find what t is equal to at that given point or should I eliminate the parameter?

 

at y = 3

3 = t^2 +2

therefore t = 1 or t = -1 (but t cannot = -1 because of the terms of x for real numbers)

 

therefore you substitute t = 1 into dy/dx and find the gradient to be 2

 

I then get y - 3 = 2(x -1)

and get an answer of y=2x +1

 

I am not sure if its right and if the reasoning is correct or not

Posted (edited)

I know to find the slope of parametric equations you find dy/dx.

 

 

That is one way.

 

A more geometric approach involves finding the tangent vector to the parameterized curve.

 

If [math]\gamma (t) = (x(t), y(t)) [/math] is your curve then the tangent vector (velocity if [math]\gamma[/math] represents position) is just [math]\gamma '(t) = (x'(t),y'(t))[/math]. The slope of the tangent line at some point [math]\gamma (t_0)[/math] is then just [math]\frac{y'(t_0)}{x'(t_0)}[/math]

 

This generalizes immediately to curves in 3 or more dimensions, except that "slope" of a tangent is not a defined notion.

Edited by DrRocket
Posted

Is that not essentially the same thing though?

You still need to define t from the given points and then substitute.

 

 

 

Posted

Is that not essentially the same thing though?

You still need to define t from the given points and then substitute.

 

 

 

 

No, you simply need to know the point as a function of t, then differentiate. You do not need to find y as a function of x. In fact the parameterized curve need not be the graph of a function -- for instance it can have closed loops.

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