I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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 \(\left( {2,0} \right)\).
Show All Steps Hide All Steps
Start SolutionWe 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 2Now, 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 3Next, 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 4Finally, 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {2,0,2{{\bf{e}}^4}} \right\rangle + t\left\langle {1,0,5{{\bf{e}}^4}} \right\rangle = \left\langle {2 + t,0,2{{\bf{e}}^4} + 5{{\bf{e}}^4}t} \right\rangle }}\]