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.

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- Apr 9th 2006, 08:23 PM #1

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- Apr 9th 2006, 09:13 PM #2

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Originally Posted by**Kiwigirl**

$\displaystyle y=mx+c$.

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

- Apr 9th 2006, 09:23 PM #3Originally Posted by
**Kiwigirl**

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

- Apr 12th 2006, 01:29 PM #4

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