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**Kiwigirl** · April 9th 2006, 08:23 PM

A curve is defined by the parametric equations x = t² and y = 2t + 1
Find the equation of the tangent at the point where t = 2.

**CaptainBlack** · April 9th 2006, 09:13 PM

Originally Posted by Kiwigirl

A curve is defined by the parametric equations x = t² and y = 2t + 1
Find the equation of the tangent at the point where t = 2.

Let the equation of the tangent at the required point be:

$y=mx+c$ .

We need to find the slope of the tangent, which is $dx/dy$ , and
to find this we need to use the result that:

$ \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=2\times \frac{1}{2t}=1/t $

So at $t=2$ the slope of the tangent is $1/2$ .

So now we know that the equation of the tangent is of the form:

$y=x/2+c$ ,

and that at $t=2$ , $x=4$ and $y=5$ , so the tangent passes through the point
$x=4$ , $y=5$ .

Hence:

$ 5=4/2 +c $ ,

so:

$ c=3 $ ,

and the equation of the tangent is:

$y=2x-3$ .

RonL

**earboth** · April 9th 2006, 09:23 PM

Originally Posted by Kiwigirl

A curve is defined by the parametric equations x = t² and y = 2t + 1
Find the equation of the tangent at the point where t = 2.

Hello,

as you may know the derivation of parametric(?) functions is done with
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

So you need the coordinates of the tangent point, which is (4, 5) and the slope, which you calculate with the formula given above:
$\frac{dy}{dx}=\frac{2}{2t}$ . With t = 2 you get a slope of 1/2.
Now use the point slope formula of a line:
$\frac{y-5}{x-4}=\frac{1}{2} \Longleftrightarrow y=\frac{1}{2} x +3$

Greetings

EB

**Kiwigirl** · April 12th 2006, 01:29 PM

Wow, thank you both for your incredible help, especially for the graph CaptainBlack. I am very impressed

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