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I need to find an equation of the tangent line to the curve $x=5+t^2-t$, $y=t^2+5$ at the point $(5,6)$.

Setting $x=5$ and $y = 6$ and solving for $t$ gives me $t=0,1,-1$. I know I have to do y/x, and then take the derivative. But how do I know what $t$ value to use?

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Please put the actual definition of your function into the question. You can typeset formulas by enclosing them in $-symbols, and using latex notation. – fgp Oct 11 '12 at 17:51
up vote 0 down vote accepted

You have

i) $x=5+t^2-t$ and ii) $y=t^2+5$ and $P=(5,6)$.

From $P=(5,6)$ you get $x=5$ and $y=6$.

From ii) you get now $t^2=1\Rightarrow t=1$ or $t=-1$ and from i) you get (knowing that $t\in${1,-1} ):$\ $ $t=1$

Your curve has the parametric representation

$\gamma: I\subseteq\mathbb{R}\rightarrow {\mathbb{R}}^2: t\mapsto (5+t^2-t,t^2+5)$.

Therefore, $\frac{d}{dt}\gamma=(2t-1,2t)$.

Put $t=1$ into the derivative of $\gamma$ and from this you can get the slope of the tangent in P.

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@ZankFrappe but this is wrong, see: i) solved – Charlie Yabben Oct 11 '12 at 17:55
If you put $t=1$ into the derivative of $\gamma$, you get $(1,2)$, therefore, your slope is $\frac{2}{1}=2$. Alternatively, as Ganesh suggested: $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t}{2t-1}$. If you put in $t=1$, you'll get the same result. – Bernhard Boehmler Oct 11 '12 at 18:01
not at all what I was asking, but I understand now – Charlie Yabben Oct 11 '12 at 18:07
But from this you can get your tangent equation: your slope m is 2 and you know that the point P lies on the tangent, so: $y=mx+b$ and $m=2 \Rightarrow y=2x+b$ and now $6=2\cdot 5 + b\Rightarrow b=-4\Rightarrow y=2x-4$ is the equation of your tangent line. – Bernhard Boehmler Oct 11 '12 at 18:13

Please type up or paste the image text in your problem statement.

Solving for $t: x(t)=5$ gives $t \in \{0,1\}$ and solving for $t:y(t) = 6$ gives $t \in \{ 1,-1\}$. The common $t$ is $t=1$. Compute $dy/dx$ and not $y/x$ at $t=1$ , and substitute this into the slope of the tangent.

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I don't understand how 5=5+t^2-t => t=1, it should be t=0,1. Oh wait i think i just got it, i) t= 0, 1; ii) t=-1,1; but we pick the common t right? – Charlie Yabben Oct 11 '12 at 17:58
Correct. And your x(t) and y(t) are a parametrisation of a curve in ${\mathbb{R}}^2$ (and t is the parameter), look at parametric plot in^2%E2%88%92t%2Ct^2%2B5%29 – Bernhard Boehmler Oct 11 '12 at 18:07

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