1. ## Help!

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

2. Originally Posted by Kiwigirl
A curve is defined by the parametric equations x = tē 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:

$\displaystyle y=mx+c$.

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

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

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

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

$\displaystyle y=x/2+c$,

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

Hence:

$\displaystyle 5=4/2 +c$,

so:

$\displaystyle c=3$,

and the equation of the tangent is:

$\displaystyle y=2x-3$.

RonL

3. Originally Posted by Kiwigirl
A curve is defined by the parametric equations x = tē 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
$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

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

Greetings

EB

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