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I have not seen a problem like this so I have no idea what to do.

Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.

$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$

I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$

But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.

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4 Answers 4

One way to do this is by considering the parametric form of the curve: $(x,y)(t) = (1 + \log t, t^2 + 2)$, so $(x,y)'(t) = (\frac{1}{t}, 2t)$ We need to find the value of $t$ when $(x,y)(t) = (1 + \log t, t^2 + 2) = (1,3)$, from where we deduce $t=1$. The tangent line at $(1,3)$ has direction vector $(x,y)'(1) = (1,2)$, and since it passes by the point $(1,3)$ its parametric equation is given by: $s \mapsto (1,2)t + (1,3)$.

Another way (I suppose this is eliminating the parameter) would be to express $y$ in terms of $x$ (this can't be done for any curve, but in this case it is possible). We solve for $t$: $x = 1 + \log x \Rightarrow x = e^{x-1}$, so $y = t^2 + 2 = (e^{x-1})^2 + 2 = e^{2x-2} + 2$. The tangent line has slope $\frac{dy}{dx}=y_x$ evaluated at $1$: we have $y_x=2e^{2x-2}$ and $y_x(1)=2e^0 = 2$, so it the line has equation $y=2x +b$. Also, it passes by the point $(1,3)$, so we can solve for $b$: $3 = 2 \cdot 1 + b \Rightarrow b = 1$. Then, the equation of the tangent line is $y = 2x + 1$.

Note that $s \mapsto (1,2)t + (1,3)$ and $y = 2x + 1$ are the same line, expressed in different forms.

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I have no idea what a direction vector is. –  user138246 Jun 20 '12 at 23:42
    
Sorry : I am posting my answer when your was not edited yet ! –  Mohamed Jun 20 '12 at 23:46
    
If you have a line given in parametric form: $t \mapsto tA + B$, $A$ is called a direction vector of the line. Have you seen parametric equations of lines and curves? –  talmid Jun 20 '12 at 23:47
    
This is my first time working with parametric equations. –  user138246 Jun 20 '12 at 23:48
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@Jordon, in addition to what Talmid said, try this video youtube.com/watch?v=m6c6dlmUT1c –  sidht Jun 20 '12 at 23:56

2nd method: eliminating parameter:

$x=1+ \ln t , y=t^2+2 \Leftrightarrow t=\exp(x-1), y=2+\exp(2x-2)$

Consider the function: $f(x)=2+ \exp(2x-2)$, then: $f'(x)=2 \exp(2x-2)$

The tangent at $x=1$ has equation: $Y=f'(1)(X-1)+ f(1)$, thus: $Y=2(X-1)+3$, thus:$$Y=2X+1$$

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Method 1 Eliminating. I think they want to write everything in terms of x first.

$x = 1 + ln(t) \iff e^{x - 1} = t$

$y = t^2 + 2 = e^{2x -2} + 2 \implies y' = 2e^{2x -2}$

At (1,3) $y' = 2$. So the tangent line is $y = 2(x- 1) + 3$ or parametrically let $x - 1 = t \iff x = 1 + t$ and $y = 2t + 3$

Method 2 No eliminating.

Let $r = (1 + ln(t),2 + t^2) \implies r' = (1/t, 2t)$. At (1,3), $t = 1$, therefore $r'(1) = (1,2)$

$r = (1,3) + s(1,2)$ which gives you $x = 1 + s$ and $y = 3 + 2s$

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I do not understand the r subsitution, is this similar to finding $\frac{dy}{dx}$? –  user138246 Jun 20 '12 at 23:53
    
What substitution? You mean me writing it as r(t)? –  sidht Jun 20 '12 at 23:54
    
You said "Let r =" –  user138246 Jun 20 '12 at 23:55
    
Yup, that's the vector valued equation of a line. –  sidht Jun 20 '12 at 23:55
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I have no idea what a vector value equation is. –  user138246 Jun 21 '12 at 0:01
  1. Eliminating the parameter $t$. The given system of two parametric equations $$\begin{eqnarray*}\left\{ \begin{array}{c} x=1+\ln t \\ y=t^{2}+2 \end{array} \right. \end{eqnarray*} \tag{A}$$ is equivalent successively to $$\begin{eqnarray*} \left\{ \begin{array}{c} x-1=\ln t \\ y=t^{2}+2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} t=e^{x-1} \\ y=t^{2}+2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} t=e^{x-1} \\ y=\left( e^{x-1}\right) ^{2}+2 \end{array} \right. \end{eqnarray*}$$ and finally to $$\left\{ \begin{array}{c} t=e^{x-1} \\ y=e^{2\left( x-1\right) }+2. \end{array} \right. \tag{B} $$ We have thus eliminated the parameter $t$. Here is a plot of $y=e^{2\left( x-1\right) }+2$ enter image description here Differentiating the equation of the curve $$y=e^{2\left( x-1\right) }+2\tag{C}$$ gives by the chain rule applied to $e^u$, with $u=2(x-1)$ $$\begin{eqnarray*} \frac{dy}{dx} &=&\frac{d}{dx}\left( e^{2\left( x-1\right) }+2\right) =\frac{d }{dx}e^{2\left( x-1\right) }=\left( \frac{ d}{du}e^{u}\right) \frac{du}{dx} =e^{u}\times 2=2e^{2\left( x-1\right) }. \end{eqnarray*}$$ At $x=1$ this derivative has the value $$\begin{equation*} \left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2e^{2\left( x-1\right) }\right\vert _{x=1}=2e^{2(1-1)}=2.\tag{1} \end{equation*}$$ Since at $x=1$, $$y=e^{2(1-1)}+2=3,$$ we have confirmed that the point $(x,y)=(1,3)$ belongs to the graph of the curve given by equation $(C)$. Hence the equation of the tangent line to the graph of the curve at $(1,3)$ is $$\begin{equation*} y-3=2(x-1)\Leftrightarrow y=2x+1\tag{2} \end{equation*}$$

  2. Without eliminating the parameter $t$. (Reformulated in view of OP's comment.) To compute the derivative we use now the parametric equations $(A)$ and the formula $$\frac{dy}{dx} =\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}/\frac{dx}{dt}.\tag{D}$$ We have $$\begin{eqnarray*} \frac{dy}{dx} &=&\frac{dy}{dt}/\frac{dx}{dt}= \left( \frac{d}{dt}\left( t^{2}+2\right) \right) /\frac{d}{dt}\left( 1+\ln t\right) \\ &=&2t/\frac{1}{t}=2t^2,\tag{3} \end{eqnarray*}$$ which confirms your result. Since for $x=1$ the equation $$x=1+\ln t $$ gives $$1=1+\ln t\Leftrightarrow 0=\ln t \Leftrightarrow t=1,$$ we get the same value as in $(1)$ for the derivative $$\begin{equation*} \left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2t^2\right\vert _{t=1}=2\cdot 1^2=2.\tag{3a} \end{equation*}$$ The equation of the tangent line is as above $$\begin{equation*} y=2x+1.\tag{4} \end{equation*}$$ In terms of the parameter $t$ the tangent line at $t=1$, i.e. at $(x,y)=(1,3)$ is given by the parametric equations $$\begin{equation*} \left\{ \begin{array}{c} x=t \\ y=2t+1, \end{array} \right.\tag{4a} \end{equation*}$$ because $$\begin{equation*} \left. \frac{dx}{dt}\right\vert _{t=1}=\left. \frac{1}{t}\right\vert _{t=1}=1 \end{equation*}$$ and $$\begin{equation*} \left. \frac{dy}{dt}\right\vert _{t=1}=\left. 2t\right\vert _{t=1}=2. \end{equation*}$$

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I don't quit follow part 2, in my book the equation is t$\frac{dy}{dt}/ \frac{dx}{dt}$ is this the same thing as that and why did you use this other identity? –  user138246 Jun 21 '12 at 0:38
    
@Jordan Yes. We can write $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$ –  Américo Tavares Jun 21 '12 at 0:41
    
@Jordan I added the computation with $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$ –  Américo Tavares Jun 21 '12 at 1:00

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