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### Section 13.3 : Interpretations of Partial Derivatives

3. Write down the vector equations of the tangent lines to the traces for $$f\left( {x,y} \right) = x\,{{\bf{e}}^{2x - {y^{\,2}}}}$$ at $$\left( {2,0} \right)$$.

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Start Solution

We know that there are two traces. One for $$x = 2$$ (i.e. $$x$$ is fixed and $$y$$ is allowed to vary) and one for $$y = 0$$ (i.e. $$y$$ is fixed and $$x$$ is allowed to vary). We also know that $${f_y}\left( {2,0} \right)$$ will be the slope for the first trace ($$y$$ varies and $$x$$ is fixed!) and $${f_x}\left( {2,0} \right)$$ will be the slope for the second trace ($$x$$ varies and $$y$$ is fixed!).

So, we’ll need the value of both of these partial derivatives. Here is that work,

\begin{align*}{f_x}\left( {x,y} \right) & = \,{{\bf{e}}^{2x - {y^{\,2}}}} + 2x\,{{\bf{e}}^{2x - {y^{\,2}}}} & \hspace{0.05in} & \to \hspace{0.25in}{f_x}\left( {2,0} \right) = 5{{\bf{e}}^4} = 272.9908\\ {f_y}\left( {x,y} \right) & = - 2yx\,{{\bf{e}}^{2x - {y^{\,2}}}} & \hspace{0.05in} & \to \hspace{0.25in}{f_y}\left( {2,0} \right) = 0\end{align*} Show Step 2

Now, we need to write down the vector equation of the line and so we don’t (at some level) need the “slopes” as listed in the previous step. What we need are tangent vectors that give these slopes.

Recall from the notes that the tangent vector for the first trace is,

$\left\langle {0,1,{f_y}\left( {2,0} \right)} \right\rangle = \left\langle {0,1,0} \right\rangle$

Likewise, the tangent vector for the second trace is,

$\left\langle {1,0,{f_x}\left( {2,0} \right)} \right\rangle = \left\langle {1,0,5{{\bf{e}}^4}} \right\rangle$ Show Step 3

Next, we’ll also need the position vector for the point on the surface that we are looking at. This is,

$\left\langle {2,0,f\left( {2,0} \right)} \right\rangle = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle$ Show Step 4

Finally, the vector equation of the tangent line for the first trace is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle + t\left\langle {0,1,0} \right\rangle = \left\langle {2,t,2{{\bf{e}}^4}} \right\rangle }}$

and the trace for the second trace is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle + t\left\langle {1,0,5{{\bf{e}}^4}} \right\rangle = \left\langle {2 + t,0,2{{\bf{e}}^4} + 5{{\bf{e}}^4}t} \right\rangle }}$