# Help!

• April 9th 2006, 08:23 PM
Kiwigirl
Help!
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.
• April 9th 2006, 09:13 PM
CaptainBlack
Quote:

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
• April 9th 2006, 09:23 PM
earboth
Quote:

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
• April 12th 2006, 01:29 PM
Kiwigirl
Wow, thank you both for your incredible help, especially for the graph CaptainBlack. I am very impressed :)