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# Parametric equation problem watch

1. the parametric eqns of a curve are:

prove that the equation of the tangent at the point with parameter is

_________________________

so, eliminating between the two equations gives

differentiating to obtain the gradient of the tangent

so the gradient of the tangent is

but how do I get the x and y values to get to the equation of the line?
2. (Original post by Plato's Trousers)
...
You know the gradient is so you put that into the standard equation of a line (y=mx+c)

So we have and sub in t from the original equations to find the value of c.

3. So dy/dx = 1/t

Use y-y1=m(x-x1) with (x1,y1) = (t^2, 2t) to find tangent.
4. (Original post by Plato's Trousers)
the parametric eqns of a curve are:

prove that the equation of the tangent at the point with parameter is

_________________________

so, eliminating between the two equations gives

differentiating to obtain the gradient of the tangent

so the gradient of the tangent is

but how do I get the x and y values to get to the equation of the line?
This is how I'd do it:

They've asked you to find the equation of the tangent to the curve with the parameter t, do I'd work out dy/dx in terms of t (using the chain rule; no need to eliminate t). "the equation of the tangent at the point with parameter " Essentially tells us at what is the equation for the tangent? You have dy/dx in terms of t and a point for y and a point for x in terms of t, so you can find the equation of the tangent, in terms of t, using the standard method.
5. (Original post by Plato's Trousers)
the parametric eqns of a curve are:

prove that the equation of the tangent at the point with parameter is

_________________________

so, eliminating between the two equations gives

differentiating to obtain the gradient of the tangent

so the gradient of the tangent is

but how do I get the x and y values to get to the equation of the line?
You're given the equations which paramatrically describe the curve. The question then asks for the tangent at the point where the parameter is t.

Also you'll want to express your gradient in terms of t, not y.

EDIT: What the others said while I was typing
6. ok, so I got it now. Thanks everyone

(dknt and EEngWillow - I have repped you too recently and can't rep you again)
7. (Original post by Plato's Trousers)
ok, so I got it now. Thanks everyone

(dknt and EEngWillow - I have repped you too recently and can't rep you again)
Aha, I know that feeling. I'm not allowed to rep any of the people I want to rep when they help me! Silly forum. We don't help for the rep anyway, we help for the sake for helping
8. (Original post by Plato's Trousers)
ok, so I got it now. Thanks everyone

(dknt and EEngWillow - I have repped you too recently and can't rep you again)

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